Abstract Nonsense

Homotopy and the Homotopy Category (Pt. III)

Point of Post: This is a continuation of this post.

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August 30, 2012

Homotopy and the Homotopy Category (Pt. II)

Point of Post: This is a continuation of this post.

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August 30, 2012

Homtopy and the Homotopy Category (Pt. I)

Point of Post: In this post we motivate, define, and discuss the notion of homotopy. We then introduce the homotopy category as a quotient category of $\mathbf{Top}$.

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Motivation

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In topology we care about the geometry of a space–how we can describe all of the geometric properties that a space has. For example, we care if the space is compact, we care if it locally connected, we care if it $T_4$. All of the properties factor into whether or not two given topological spaces are to be considered “the same”.  That said, for large portions of mathematics some of the topological invariants of a space are less important than others. This can occur either because they literally matter less to us (for example, compactness is something that, while nice, isn’t an absolute necessity for a topological space to be nice). This can also be true because the subject area we are are working in contains spaces which necessarily already satisfy some of the properties (for example, metric spaces already satisfy all of the nice separation axioms).

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August 29, 2012

Path Connectedness

During our discussion of connectedness of topological groups I alluded to the fact that I would not be talking about path connectedness, but why? It seems natural that the two concepts, ostensibly so connected ($\overset{. .}{\smile}$), should be discussed at the same time. While I cannot refute that logic, my reason for separating (I’m on fire) them was purely a personal matter of connotation. To me connectedness while very geometric is much more aligned in the depths of point-set topology. It is a fundamental property of spaces. Path connectedness (while admittedly possessing all of the aforementioned properties) on the other hand has a very geometric flavor to it. Also, path connectedness will be an integral part of our discussion of manifolds. Thus, without further justification to a non-existent audience let us begin:

Path: Let $X$ be a topological space, then we call a continuous map $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$ a path from $x_0$ to $x_1$.

Path Connected: We call a topological space $X$ path connected if for any $x_0,x_1\in X$ there exists a path between them.

Intuitively we can think about a path between two points as just connecting them with a “squiggly line”.  So, it is natural to assume that every path connected space is connected but as some wise old guy once said “What is asserted without proof may be dismissed without proof”:

Theorem: Let $X$ be path connected, then $X$ is path connected.

Proof: Suppose that $\varphi:X\to D$ is a continuous surjection (where $D$ as always is the two-point discrete space). Then, choose $x_0\in\varphi^{-1}(\{0\}),x_1\in\varphi^{-1}(\{1\})$. By assumption there exists a path $\alpha:[0,1]\to X$ between them. Clearly then, $\alpha:[0,1]\to\alpha(X)$ is a continuous surjection and so it follows that $\varphi\circ\alpha:[0,1]\to D$ is also a continuous surjection, but this clearly contradicts the connectedness of $[0,1]$. $\blacksquare$

So, what are some examples of path connected sets you may ask? We can find a plethora, but first we need a little itsy bitsy lemma (barely worth saying).

Theorem: Every convex subset of $\mathcal{V}$ (a normed vector space over $\mathbb{R}$)  is path connected.

Proof: Clearly given any $\bold{x}_0,\bold{x}_1\in C$ where $C$ is the convex set the map $\alpha:[0,1]\to L(\bold{x}_0,\bold{x}_1):t\mapsto \bold{x}_0t+(1-t)\bold{x}_1$ is continuous obviously and by assumption $L(\bold{x}_0,\bold{x}_1)\subseteq C$, from where the conclusion follows. $\blacksquare$

So, from this we can derive zillions and zillions (technical term of course) of path connected sets, for if you’ll remember we proved earlier that every open ball in a normed vector space $\mathcal{V}$ is convex. So, in particular every open ball in $\mathbb{R}^n$ is path connected. Also, realizing that $\mathbb{R}^n$ themselves are convex (obviously) and in fact so is the punctured plane $\mathbb{R}^2-\{\bold{0}\}$ (just take a polygonal line if needed). Slightly less obvious is that $\mathbb{R}^n-\{p_1,\cdots,p_n\}$ for some arbitrary $p_1,\cdots,p_n\in\mathbb{R}^n$ (once again take a polygonal line)

So, we begin to prove some pretty obvious theorems.

Theorem: Let $X$ be path connected and $\varphi:X\to Y$ a continuous map. Then, $\varphi(X)$ is a path connected subspace.

Proof: Given $\varphi(x_0),\varphi(x_1)\in \varphi(X)$ it is evident that $\varphi\circ\alpha:[0,1]\to\varphi(X)$ where $\alpha$ is the path from $x_0$ to $x_1$ in $X$ is the required path.  $\blacksquare$

Theorem: Let $\left\{X_\beta\right\}_{\beta\in\mathcal{B}}$ be a collection of path connected topological spaces, then $\displaystyle X=\prod_{\beta\in\mathcal{B}}X_\alpha$ with the product topology  is path connected. The converse is also true.

Proof: Recall that if $\left\{\varphi_\beta\right\}_{\beta\in\mathcal{B}}$ is a collection of continuous maps $\varphi_\beta:E\to X_\beta$ then the map

$\displaystyle \bigoplus_{\beta\in\mathcal{B}}\varphi_\beta:E\to\prod_{\beta\in\mathcal{B}}:x\mapsto\left\{\varphi_\beta(x)\right\}$

is connected (since each of the projections is connected). So, for any $\left\{x_0^\beta\right\},\left\{x_1^\beta\right\}\in X$ we merely take the guaranteed guaranteed path $\alpha_\beta:[0,1]\to X_\beta$ from $x_0^\beta$ to $x_1^\beta$ and note that

$\displaystyle \bigoplus_{\beta\in\mathcal{B}}\alpha_\beta:[0,1]\to\prod_{\beta\in\mathcal{B}}X_\beta$

is the desired path.

Conversely, if $X$ is path connected then so are each of it’s coordinate spaces since the projection onto them is continuous. $\blacksquare$

Theorem: Let $X$ be path connected and $Y$ a quotient space of $X$, then $Y$ is path connected.

Proof: This is blindingly obvious since the quotient map $\xi:X\leadsto Y$ which defines the topology on $Y$ is automatically a continuous surjection. $\blacksquare$

Theorem: If $X,Y$ are two non-empty topological spaces, then $X\amalg Y$ is not path connected.

Proof: This obviously follows since we proved earlier that they aren’t even connected.

That said, there are some fairly neat theorems which are rooted very closely in geometric intuition. If for some fixed point $x_0$ we can connect it to any other point by a path we should be able to connect any two points $x_1,x_2$ by connecting the first point $x_1$ to $x_0$, the second point $x_2$ to $x_0$, and then “connecting” the paths. We formalize in the following paragraphs, but first we need a very important lemma.

Theorem (Gluing Lemma): Let $X$ be a topological space and $\left\{X_1,\cdots,X_n\right\}$ a closed cover of it. Furthermore, suppose that for each $k=1,\cdots,n$ there exists a continuous map $\varphi_k:X_k\to Y$ such that $\varphi_k\mid_{X_k\cap X_j}=\varphi_j\mid_{X_k\cap X_j}$ for each $k,j=1,\cdots,n$. Then, there exists a  unique continuous map $\varphi:X\to Y$ such that $\varphi\mid_{X_k}=\varphi_k$.

Proof: First note that if $x\in X_{j_1},\cdots,X_{j_m}$ then

$\varphi_{j_1}(x)=\varphi\mid_{X_{j_1}\cap X_{j_2}}(x)=\varphi_{j_2}(x)=\cdots=\varphi\mid_{X_{j_{m-1}}\cap X_{j_m}}(x)=\varphi_{j_m}(x)$

and so the map

$\displaystyle \varphi_1\sqcup\cdots\sqcup\varphi_n= \varphi:X\to Y:x\mapsto\begin{cases}\varphi_1(x)\quad\text{if}\quad x\in X_1\\ \vdots \\ \varphi_n(x)\quad\text{if}\quad x\in X_n\end{cases}$

is at least well-defined, and in fact this map clearly satisfies $\varphi\mid_{X_k}=\varphi_k$. To see that it’s continuous we let $C\subseteq Y$ be closed. Clearly

$\varphi^{-1}(C)=\bigcup_{j=1}^{n}\varphi_j^{-1}(C)$

but each $\varphi_j^{-1}(C)$ is a closed subset of a closed subspace of $X$ and thus closed in $X$. Thus, $\varphi^{-1}(C)$ is the finite union of closed sets and thus closed. Lastly, to see that $\varphi$ is unique we merely note that if $\varphi'$ where another such map then given $x\in X$ we have that $x\in X_k$ for some $k=\{1,\cdots,n\}$ and so

$\varphi(x)=\varphi\mid_{X_k}(x)=\varphi_k(x)=\varphi'\mid_{X_k}(x)=\varphi'(x)$

from where the conclusion follows. $\blacksquare$

Corollary: By an infinitesimal modification the above proves the same if $\{X_1,\cdots,X_n\}$ is replaced by an arbitrary open cover $\left\{X_\beta\right\}_{\beta\in\mathcal{B}}$

Theorem (*): Let $X$ be a topological space and suppose that for some fixed $x_0\in X$ there is a path from $x_0$ to any other point of $X$, then $X$ is path connected.

Proof: Let $x_1,x_2\in X$ be arbitrary. By assumption there exists a path $\alpha$ from $x_0$ to $x_1$. Define a “reverse path” from $x_1$ to $x_0$ as follows

$\overset{\leftarrow}{\alpha}:[0,1]\to X:t\mapsto \alpha(1-t)$

This is readily verified to be path from $x_1$ to $x_0$. So, let $\gamma$ be the path from $x_0$ to $x_2$. Clearly the maps $\overset{\leftarrow}{\alpha}^*:[0,\tfrac{1}{2}]\to X:t\mapsto \overset{\leftarrow}{\alpha}(2t)$ and $\gamma^*:[\tfrac{1}{2},1]\to X:t\mapsto \gamma(2t-1)$ are continuous and since

$\overset{\leftarrow}{\alpha}^*(\tfrac{1}{2})=\overset{\leftarrow}{\alpha}(\frac{1}{2})=\alpha(0)=x_0=\gamma(0)=\gamma^*(\tfrac{1}{2})$

It follows from the gluing lemma that $\overset{\leftarrow}{\alpha}^*\sqcup\gamma^*:[0,1]\to X$ is continuous and so noting that

$\left(\overset{\leftarrow}{\alpha}^*\sqcup\gamma^*\right)(0)=\overset{\leftarrow}{\alpha}^*(0)=\overset{\leftarrow}{\alpha}( 0)=\alpha(0)=x_1$

and

$\left(\overset{\leftarrow}{\alpha}^*\sqcup\gamma^*\right)(1)=\gamma^*(1)=\gamma(1)=x_1$

and so the theorem follows. $\blacksquare$

With this we have an easy way to prove another nice theorem

Theorem: Let $X$ be a topological space and $\left\{U_\beta\right\}_{\beta\in\mathcal{B}}$ a collection of path connected subspaces such that $\displaystyle \bigcap_{\beta\in\mathcal{B}}U_\beta\ne\varnothing$, then $\displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta=\Omega$ is path connectected.

Proof: Let $\displaystyle x_0\in\bigcap_{\beta\in\mathcal{B}}U_\beta$, then given any $x_1\in \Omega$ we have that $x_1\in U_\beta$ for some $\beta\in\mathcal{B}$ and so by assumption there exists some path $\alpha:[0,1]\to U_\beta$ from $x_0$ to $x_1$. Clearly, if we extend the codomain we see that $\alpha:[0,1]\to \Omega$ is also a path from $x_0$ to $x_1$ in $\Omega$. Considering our previous theorem finishes the argument. $\blacksquare$

With this we can prove a satisfying theorem, which in a very real sense is just mimicry of a previous theorem.

Theorem: Let $\mathcal{V}$ be any normed vector space over $\mathbb{R}$. Then, $\mathcal{V}$ is path connected.

Proof: We merely need to note that the map $\varphi_{\bold{v}}:\mathbb{R}\to\mathcal{V}:\gamma\mapsto \gamma\bold{v}$ is continuous. Thus, by previous theorem $\varphi_{\bold{v}}\left(\mathbb{R}\right)$ is path connected. But,

$\displaystyle \mathcal{V}=\bigcup_{\bold{v}\in\mathcal{V}}\varphi_{\bold{v}}\left(\mathbb{R}\right)$

and thus noting that

$\displaystyle \bold{0}=\varphi_{\bold{v}}(0)\in\bigcap_{\bold{v}\in\mathcal{V}}\varphi_{\bold{v}}\left(\mathbb{R}\right)$

and appealing to the previous theorem finishes the argument. $\blacksquare$

We finish our discussion with a very, very nice theorem which in essence will show that every manifold is path connected. But, first we need to define the local analogue of path connectedness.

Locally Path Connected: Let $X$ be a topological space, then $X$ is said to be locally path connected if for each $x\in X$ there exists a path connected neighborhood $U$ of it.

The role of local path connectedness plays a much more important role that it’s plain old connectedness analogue. This can be seen from the following theorem.

Theorem: Let $X$ be a locally path connected connected space, then $X$ is path connected.

Proof: Let $x_0\in X$ and let $\Omega$ be the set of all points in $X$ for which there is a path from $x_0$ to that point. Since $\Omega\ne\varnothing$ ($x_0\in \Omega$) it suffices to prove that it is both open and closed (and the result will then follow from $X$‘s connectedness). So, let $x\in\Omega$. By assumption there exists some neighborhood $U$ of it such that $U$ is path connected, we claim that $U\subseteq\Omega$. To see this, let $x'\in U$. Using the exact same argument as in theorem (*) we can then construct a path from $x'$ to $x_0$ and so $x'\in \Omega$. It follows that $\Omega$ is open. Now, to show that $\Omega$ is closed we show it is invariant under closure, and sine $\Omega\subseteq\overline{\Omega}$ it suffices to show the reverse inclusion. So, let $x\in\overline{\Omega}$. Then, by $X$‘s local connectedness there exists a neighborhood $U$ of $x$ which is path connected. But, since $x\in\overline{\Omega}$ there must be some point $x'\in\Omega\cap U$. So, using the exact same technique described in theorem (*) again we may connect the path from $x$ to $x'$ and  the path from $x'$ to $x_0$ to get a path from $x$ to $x_0$. It follows that $x\in\Omega$.

Thus, $\Omega$ is open and closed and since it’s non-empty it must be that $\Omega=X$. But, that means that every point of $X$ may be connected to $\Omega$ by a path. Thus, by theorem (*) $X$ is path connected. $\blacksquare$

As stated this will prove that every manifold is path connected, but maybe a little closer to home is that this implies that every open connected set in $\mathbb{R}$ is path connected, we state this in the following theorem.

Theorem: Let $\mathcal{V}$ be a normed vector space and $O\subseteq\mathcal{V}$ open and connected. Then, $O$ is path connected.

Proof: By assumption for each point $x\in O$ there exists some open ball $B_{\delta}(o)\subseteq O$, but this open ball is convex and thus path connected. So, $O$ is locally path connected and connected thus by the previous theorem path connected. $\blacksquare$.

This confirms everyone’s intuition from back in complex analysis or calculus when it “seemed” that every connected “blob” in $\mathbb{R}^2$ could have a  “line drawn in it connecting two points”.

This ends our brief discussion of path connectedness.

April 29, 2010

Thoughts about connectedness (Locally connected spaces)

Just as with compactness there is a certain local connectedness property that spaces may posses without being themselves connected.

Locally Connected: Let $X$ be a topological space, then we call $X$ locally connected if for every $x\in X$ and every neighborhood $N$ of $x$ there exists some neighborhood connected $C$ of $x$ such that $C\subseteq N$.

Remark: It is apparent that this is equivalent that $X$ has a connected open base.

We have already said that every normed vector space with the usual topology is locally connected (since every open ball is convex and thus connected). Also, it is clear that our example $(-\pi,-e)\cup (e,\pi)$ is locally connected but not connected and that the rationals are neither connected nor locally connected. In fact, you can have a locally connected space which is totally disconnected, take any discrete space.

But, unlike local compactness which is implied by compactness local connectedness is not implies by connectedness as the following example shows.

Example (Topologist’s Sine Curve): Define $f:(0,1]\to\mathbb{R}$ by $x\mapsto \sin(\frac{1}{x})$. So, define $\mathcal{T}=\Gamma_f\cup \{0\}\times [-1,1]$. It is clear that $\mathcal{T}$ is closed since it is equal to $\overline{\Gamma_f}$ which is the graph of a continuous function on a connected space. But it is fairly easy to see that $\mathcal{T}$ is not locally connected.

We now prove some facts about locally connected spaces.

Theorem: Let $X$ be a locally connected space. If $Y$ is an open subspace of $X$, then each component of $Y$ is open in $X$. In particular each component of $X$ is open.

Proof: Let $C$ be a component in $Y$ and let $x\in C$. Since $Y$ is a neighborhood of $x$ and $X$ is locally connected there exists some connected neighborhood $N$ of $x$ contained in $Y$. But, since $C$ is a component we must have that $N\subseteq C$. The conclusion follows. $\blacksquare$.

Theorem: Let $X$ be a topological space where all the components of every open subspace of $X$ are open, then $X$ is locally connected.

Proof: Let $x\in X$ and let $N$ be any neighborhood of it. Since $N$ is an open subspace of $X$ it’s components are also open in $X$. In particular, the component $C$ which contains $x$ is open in $X$. Thus, we have found a connected neighborhood of $x$ contained in $N$. The question is whether or not it is connected with respect to $X$. But, since the topology of $x$ as a subspace of $N$ is the same as that with respect to $X$ there is nothing to prove.

Theorem: Let $X$ be a compact locally connected space. Then, $X$ has finitely many components.

Proof: Suppose not. Then we have that the class $\Omega$ of components of $X$ is infinite. Since the formation of components is a partition we see in particular that $\Omega$ covers $X$ and it admits no finite subcovering. But, $\Omega$ is in fact an open cover of $X$ by previous theorem. It follows that $\Omega$ is an open cover of $X$ which admits no finite subcover, which of course contradicts its compactness. $\blacksquare$

Theorem: Let $X$ be locally connected and $\varphi:X\to Y$ open, surjective, and continuous. Then, $Y$ is locally connected.

Proof: Let $\varphi(x)\in Y$ be arbitrary and $N$ any neighborhood of $\varphi(x)$. Since $\varphi$ is continuous we have that $\varphi^{-1}(N)$ is a neighborhood of $x$. Now, by $X$‘s local connectedness there then exists some connected neighborhood $U$ of $x$ such that $U\subseteq N$. It follows that $\varphi(U)$ is connected and open and $\varphi(x)\in \varphi(U)\subseteq N$. The conclusion follows. $\blacksquare$.

Corollary: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of topological spaces and $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ locally connected. Then, $X_j$ is locally connected for every $j\in\mathcal{J}$.

We know answer the question of whether or not local connectedness is invariant under products.

Theorem: Let $X_1,\cdots,X_n$ be a finite collection of locally connected spaces. Then, $X=X_1\times\cdots\times X_n$ is locally connected.

Proof: Let $\bold{x}\in X$ be arbitrary and $N$ any neighborhood of $\bold{x}$. Clearly then $\pi_j(N)$ is a neighborhood of $\pi_j(\bold{x})$ for each $j=1,\cdots,n$. Using $X_j$‘s local connectedness we are furnished with some open connected $U_j$ such that $\pi_j(\bold{x})\in U_j\subseteq \pi_j(N)$. Clearly then, $U_1\times \cdots\times U_n$ is a connected open subspace of $X$ such that $\bold{x}\in U_1\times\cdots\times U_n\subseteq N$. The conclusion follows. $\blacksquare$.

The general case for the product of arbitrarily many locally connected spaces need not be locally connected. With a little bit of unjustified knowledge we can see that $\{0,1\}$ under the discrete topology is locally connected but $\{0,1\}^{\aleph_0}\approx\mathfrak{C}$ where $\mathfrak{C}$ is the Cantor set. Clearly the Cantor set is not locally connected though.

If we strengthen our conditions though we’re right as rain.

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of connected and locally connected spaces. Then, $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ is locally connected under the product topology.

Proof: Let $\bold{x}\in X$ be arbitrary and $N$ any neighborhood of $\bold{x}$. We of course may assume WLOG that $N$ is basic open. So, let $j_1,\cdots,j_n$ be the finitely many indices such that $\pi_{j_k}(N)\ne X_{j_k}$. For each $j_k$ we have that $\pi_{j_k}(N)$ is a neighborhood of $\pi_{j_k}(\bold{x})$. And thus, by $X_{j_k}$‘s local connectedness there exists a connected neighborhood $U_{j_k}$ such that $\pi_{j_k}\in U_{j_k}\subseteq\pi_{j_k}(N)$. Clearly then if

$G_j=\begin{cases} U_{j_k} & \mbox{if} \quad j=j_1,\cdots,j_n \\ X_j & \mbox{if} \quad j\ne j_1,\cdots, j_n\end{cases}$

Then $\displaystyle \prod_{j\in\mathcal{J}}G_j$ is open and connected and $\displaystyle \bold{x}\in\prod_{j\in\mathcal{J}}G_j\subseteq N$. The conclusion follows. $\blacksquare$.

Remark: The necessity for each $X_j$ to be connected was used when we were able to conclude that the product of the $G_j$‘s was connected.

March 28, 2010

In the last post we gave quite a few equivalent definitions of a connected space, we now prove some basic theorems relating to connectedness.

Perhaps the easiest to relate to  is the following.

Theorem: Let $X$ be a subspace of $\mathbb{R}$ then it is connected if and only if it is an interval.

Suppose that $X\subseteq \mathbb{R}$ is connected but it is not an interval. Then, there exists some $y\in \mathbb{R}$ such that for some $x,z\in X$ we have that $x but $y\notin X$. Clearly then $\left(X\cap\left(-\infty,y\right)\right)\cup\left(X\cap\left(y,\infty\right)\right)$ is a disconnection of $X$. This is of course a contradiction.

Conversely, suppose that $I\subseteq\mathbb{R}$ is an interval and $A\cup B$ is a disconnection of $I$. Since $A,B\ne\varnothing$ we may choose $a\in A$ and $b\in B$ and we may clearly assume that $a. Since $[a,b]$ is an interval and each point in $[a,b]$ is either in $A$ or $B$ we define $\gamma=\sup\left\{x:x\in A\cap [a,b]\right\}$. Evidently we have that $a\leqslant \gamma\leqslant b$ and so $\gamma\in X$ (since it is an interval). Since $A$ is closed in $X$ it is clear from basic real line topology that $\gamma\in A$. Thus, we conclude that $\gamma. Again, by basic real line topology we know $y+\varepsilon\in B$ for every $\varepsilon>0$ such that $y+\varepsilon\leqslant z$, and since $B$ is closed we must have that $\gamma\in B$. This contradicts the assumption that $A\cap B=\varnothing$. $\blacksquare$

Corollary: Every subset $E\subseteq\mathbb{R}$ such that $E'$ is dense is not connected.

Remark: Note that the only properties that were used above is the idea of what an interval is and the L.U.B. property of $\mathbb{R}$. It is not surprising then that this extends to the idea of a linear continuum (an order space with the L.U.B. property), but considering I haven’t defined and am not going to define this…there is no point in merely introducing this.

We next prove a mean feature of connectedness.

Theorem: Let $X$ be connected and $\varphi:X\to Y$ be continuous, then $\varphi(X)$ is a connected subspace of $Y$.

Proof: Suppose not, and there exists some $G,H\subseteq Y$ which are non-empty and open and such that $\left(G\cap\varphi(X)\right)\cap\left(H\cap\varphi(X)\right)=\varnothing$ and $\left(G\cap\varphi(X)\right)\cup\left(H\cap\varphi(X)\right)=\varphi(X)$. Then we have that $\varphi^{-1}\left(G\cap\varphi(X)\right)=\varphi^{-1}(G),\varphi^{-1}(G\cap\varphi(X))=\varphi^{-1}(H)$ are non-empty disjoint open sets in $X$ such that $\varphi^{-1}(G)\cup\varphi^{-1}(H)=\varphi^{-1}(G\cup H)=X$. This obviously contradicts $X$‘s connectedness. The conclusion follows $\blacksquare$

This is the standard proof, but there is a slightly more eloquent one. (it is essentially the same, but worded differently).

Proof: For the sake of notational convenience let us assume that $Y=\varphi(X)$. Now, let us assume that $\varphi(X)$ is not connected. Then, there exists some $\psi:\varphi(X)\to D$ which is continuous and surjective. Clearly then $\psi\circ\varphi:X\to D$ is continuous and surjective. This is a contradiction $\blacksquare$

Corollary: If $X\approx Y$ then $X$ is connected if and only if $Y$ is .

From this we can derive one of the most widely used theorems in calculus and analysis.

Theorem (Generalized Intermediate Value Theorem-IVT): Let $X$ be a connected space and $\varphi:X\to \mathbb{R}$ be continuous. Then $\varphi(X)$ is an interval. In particular, given any $c\in\mathbb{R}$ such that there exists $x,y\in X$ such that $\varphi(x) there exists some $c_0\in X$ such that $\varphi(c_0)=\varphi(c)$.

Proof: This follows from the fact that $\varphi(X)$ is a connected subspace of $\mathbb{R}$ and thus an interval. $\blacksquare$.

Corollary: If $\varphi:[a,b]\to\mathbb{R}$ is continuous then given any $c$ between $\varphi(a)$ and $\varphi(b)$ then there exists some $c_0\in[a,b]$ such that $\varphi(c_0)=c$.

The next is a neat other theorem consequent of the generalized mean value theorem.

Theorem: Let $X$ be connected and suppose there exists some non-constant continuous map $\varphi:X\to\mathbb{R}$ then $X$ is uncountably infinite.

Lemma: Any interval of the form $(a,b),\text{ }a and $\mathbb{R}$ are equipotent.

Proof: Define $f:(a,b)\to (-1,1)$ by $x\mapsto \displaystyle \frac{2}{b-a}(x-a)-1$ then $f$ is easily verified to be a bijection. Now, consider $w:(-1,1)\to\mathbb{R}$ by $\displaystyle x\mapsto \frac{x}{1+|x|}$. This is equally easily verified to be a bijection. Thus, $w\circ f:(a,b)\to\mathbb{R}$ is a bijection. The conclusion follows $\blacksquare$

Now by assumption we know that $X$ is connected and $\varphi$ continuous, and so by previous comment $\varphi(X)$ is an interval. But, by the assumption that $\varphi$ is non-constant we know that it is an interval of either the form $[a,b]$ or $(a,b)$. Since these two are equipotent we may assume WLOG that $\varphi(X)=(a,b),\text{ }a. Thus, by previous comment this implies that $\varphi(X)\simeq\mathbb{R}$. It follows that there exists some $\eta:\varphi(X)\to\mathbb{R}$ which is a bijection. Now, if we assume that $X$ is countable then there exists some $\psi:\mathbb{N}\mapsto X$ which is bijective. It follows that $\eta\circ\varphi\circ\psi:\mathbb{N}\to\mathbb{R}$ is a surjection, but this is clearly absurd. The conclusion follows $\blacksquare$

The next two theorems are just consequences of something I have forgot to say in general.

Theorem: Let $X$ be connected and $\varphi:X\to Y$ a continuous map. Then, $\Gamma_\varphi\subseteq X\times Y$ is connected.

Proof: As was proved earlier the map $\iota_X\oplus\varphi:X\to X\times Y$ given by $x\mapsto(x,\varphi(x))$ is continuous. Also, it is quite clear that $\Gamma_\varphi=\left(\iota_X\oplus\varphi\right)$ and so $\Gamma_\varphi$ is the continuous image of a connected set, and thus connected.

Theorem: Let $X$ be a connected space, then $\Delta_X\subseteq X\times X$ is connected.

Proof: The map  $\iota_X\oplus \iota_X:X\to X\times X$ given by $x\mapsto (x,x)$ is trivially continuous and it is clear that $\Delta_X=\left(\iota_X\oplus\iota_X\right)\left(X\right)$ and so $\Delta_X$ is the continuous image of a connected set, and thus connected. $\blacksquare$.

Theorem: Suppose that $\left\{X_j\right\}_{j\in\mathcal{J}}$ is an arbitrary collection of non-empty topological space. Then $\displaystyle \prod_{j\in\mathcal{J}}X_j$ is connected implies that $X_\ell$ is connected for each $\ell\in \mathcal{J}$.

Proof: Clearly we have that $\displaystyle \pi_\ell:\prod_{j\in\mathcal{J}}X_j\mapsto X_\ell$ (the canonical projection) is continuous and surjective. Following the same logic as the previous problems we may conclude. $\blacksquare$

We now prove the converse of the above.

Theorem: Suppose that $\left\{X_j\right\}_{j\in\mathcal{J}}$ is an arbitrary collection of non-empty connected space, then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ is connected under the product topology.

Proof: Assume not, then there exists some $f:X\to D$ (where $D$ is the two-point discrete space) which is continuous and non-constant. So, let $\displaystyle \bold{x}=\prod_{j\in\mathcal{J}}\{x_j\}$ be a fixed element of $X$ and consider some $\ell\in\mathcal{J}$. Define $w:X_\ell\to X$ by $\displaystyle x\mapsto\prod_{j\in\mathcal{J}}\{y_j\}$ where

$\displaystyle y_j=\begin{cases} x_j & \mbox{if} \quad j\ne \ell \\ x & \mbox{if} j=\ell \end{cases}$

Clearly $w$ is continuous , and so $f\circ w:X_\ell\to D$ is continuous.  But, by the assumption that $X_\ell$ is continuous this means that $f\circ w$ is constant, let’s assume everything maps to $0$. It follows that $f(\bold{y})=0$ for every $\bold{y}\in X$ such that $\pi_j(\bold{x})=\pi_j(\bold{x})$ for every $j\in\mathcal{J}-\{\ell\}$. Repeating this process with a different index $k\in\mathcal{J}-\{\ell\}$ we arrive at another set of $\bold{y}$ such that they differ from $\bold{x}$ in all but finitely many coordinates,  which also map under $f$ to zero. Doing this for each $j\in\mathcal{J}$ we arrive at a dense subset of $X$ which all map to zero under $f$. It follows that the zero function $0:X\mapsto D$ and $f$ are both continuous mappings into a Hausdorff space such that their agreement set $A(0,f)$ is dense. It follows from a previous problem that $f=0$. This contradicts the assumption that $f$ was constant. The conclusion follows $\blacksquare$

Before we make a corollary about this we prove something that everyone reading this (if there is anyone) knows but I haven’t actually taken the time to write down. Namely:

Theorem: $\mathbb{C}^n\approx\mathbb{R}^{2n}$ under the usual metric topologies.

Proof: We use the canonical mapping $\alpha:\mathbb{C}^n\to\mathbb{R}^{2n}$ given by $(a_1+ib_1,\cdots,a_{2n-1}+ib_{2n})\mapsto (a_1,b_1,\cdots,a_{2n-1},b_{2n-1})$. It is easily verified that $\alpha$ is bijective. And in fact we see that $\|\alpha(z)\|=\|z\|$ and so it’s an isometry, thus clearly bicontinuous. The conclusion follows.

We are now prepared to state that

Corollary: $\mathbb{R}^n$ and so $\mathbb{C}^n$ are connected.

March 28, 2010

In this post we begin a series of discussions about one of the easiest to define concepts in topology. That said, it is one of the most important. We begin with some motivation.

If you were a calc II student and asked to visualize these two sets $(0,1)$ and $(-\pi,-e)\cup (\pi,e)$ and say what is the most obvious difference between them you most likely respond  “The first one is in one piece and the second one isn’t”.  This idea of something being in one piece is a fundamental concept. It intuitively fits with our idea of homeomorphisms that two spaces cannot be homeomorphic of one space is in two pieces and the other one.

There is also the vague notion of “reachability”. Imagine you took two people and placed one on a random spot in $(0,1)$ and the other in a random spot in $(-\pi,-e)\cup (e,\pi)$. Then, the first person could walk to any other point of the set he’s on, while the other cannot. This idea of when two things can be traveled between without leaps or jumps is a huge concept in more advanced topology, that we will eventually hit on.

So we want to formalize this idea of being in one piece. Using our original example again we may be able to model this idea, which is commonly called connectedness.

Connected: Let $X$ be a topological space. Then, $X$ is called connected if it cannot be expressed as $X=E\cup G$ where $E,G$ are open, non-empty and disjoint.

We begin by proving a couple of equivalent definitions.

Theorem: Let $X$ be a topological space. Then, $X$ is connected if and only if the only sets which are simultaneously open and closed are $X$ and $\varnothing$.

Proof:

Suppose that $X$ is connected, but there existed some $\varnothing\subset E\subset X$ which is both open and closed. Clearly then $E'$ is both open and closed. Thus, $E,E'$ are disjoint non-empty open sets such that $E\cup E'=X$. This is a contradiction.

Conversely, suppose that the only sets which are both open and closed are $\varnothing$ and $X$. Next suppose that $A,B$ were disjoint non-empty open sets such that $A\cup B=X$. Then since $A=B'$ (this is easy to check) and $B$ is open it follows that $A$ is closed. But, since $A\cup B=X$ and $A,B\ne\varnothing$ it follows that $\varnothing\subset A\subset X$ and in particular $A\ne \varnothing, X$. Thus, $A$ is a set that is both open and closed which is neither $X$ or $\varnothing$. Contradiction. $\blacksquare$

Theorem: Let $X$ be a topological space. Then $E\subseteq X$ is both open and closed if and only if $\partial E=\varnothing$.

Proof:

Suppose that $E$ is both open and closed but $\partial E\ne\varnothing$. Then, let $x\in \partial E$. We have two choices, if $x\in E$ then every neighborhood of $x$ intersects $E'$ and so $x\notin E^{\circ}$ but since $E=E^{\circ}$ it follows that $x\notin E$, contradiction. If $x\notin E$ then every neighborhood of $x$ must contain a point of $E$ different from $x$ and so $x\in D(E)$, but since $E$ is closed we know that $D(E)\subseteq E$ and so $x\in E$. Contradiction. It follows that $\partial E=\varnothing$

Conversely, suppose that $\partial E=\varnothing$. Then, clearly $\partial E\subseteq E$ and so from basic topology $E$ is closed.  But, let $x\in E$ then since $x\notin \partial E$ there exists some neighborhood $N$ of $x$ such that $N$ does not intersect both $E,E'$. But since every neighborhood contains $x$ itself it follows that there exists some neighborhood $U$ of $x$ such that $U\cap E'=\varnothing\implies U\subseteq E$. It follows that $x\in E^{\circ}$ and appealing once again to basic topology we see that $E$ is open. $\blacksquare$.

Corollary: A topological space $X$ is connected if and only if the only sets with empty interior are $X$ and $\varnothing$.

Before we continue with need a definition.

Separated sets: If $X$ is a topological space and $A,B\subseteq X$ are such that $A\cap\overline{B}=\overline{A}\cap B=\varnothing$ we call $A,B$ separated set.

Theorem: A topological space $X$ is connected if and only if it cannot be written as the union of none-empty separated sets.

Proof:

Suppose that $X$ is connected but $A\cup B=X$ where $A,B$ are separated and non-empty. Clearly, $A\cup B\subseteq A\cup\overline{B}$ and since $A\cup B=X$ it follows that $\overline{A}\cup B=X$. But, $\overline{A}\cap B=\varnothing$ and so $B=X-\overline{A}$ and so $B$ is open. But, it is also true that $A\cup\overline{B}=X$ and that $A\cap\overline{B}=\varnothing$ and so $A=X-\overline{B}$ and so $A$ is open. Thus, $A\cup B=X$, and $A,B$ are both non-empty and open. Lastly noting that $A\cap B\subseteq A\cap\overline{B}=\varnothing$ we arrive at our contradiction.

Conversely, notice that if $A,B$ are open and disjoint then $A\cap\overline{B}=B\cap\overline{A}=\varnothing$ (this is easy to prove). Thus, if $A,B$ are disjoint non-empty open sets, then they are disjoint separated sets and thus $A\cup B\ne X$. The conclusion follows. $\blacksquare$

This last one gives, in my opinion, the most elegant way of thinking about connectedness.

Theorem: If $X$ is a topological space, then it is connected if and only if the only continuous $\varphi:X\mapsto D$ where $D=\{0,1\}$ with the discrete topology is a constant map.

Proof:

Suppose that $X$ is connected but $\varphi:X\mapsto D$ is non-constant. Clearly then $\varphi^{-1}(\{0\}),\varphi^{-1}(\{1\})$ are open and non-empty. They are also clearly disjoint, for if  $x\in \varphi^{-1}(\{0\})\cap\varphi^{-1}(\{1\})$ then

$\varphi(x)\in\varphi\left(\varphi^{-1}(\{0\})\cap\varphi^{-1}(\{1\})\right)\subseteq \varphi\varphi^{-1}(\{0\})\cap\varphi\varphi^{-1}(\{1\})\subseteq \{0\}\cap\{1\}=\varnothing$

Thus, $\varphi^{-1}(\{0\}),\varphi^{-1}(\{1\})$ are disjoint non-empty open sets and

$\varphi^{-1}(\{0\})\cup\varphi^{-1}(\{1\})=\varphi^{-1}(\{0\}\cup\{1\})=\varphi^{-1}(\{0,1\})=X$

This of course contradicts that $X$ is connected.

Conversely, suppose that the only $\varphi:X\mapsto D$ which is continuous is a constant map, but $A,B$ are non-empty disjoint open sets such that  $A\cup B=X$. Define $\varphi:X\mapsto D$ by

$\varphi(x)=\begin{cases} 0 & \mbox{if} \quad x\in A \\ 1 & \mbox{if} \quad x\in B\end{cases}$.

It is clear by the assumption that these two sets are non-empty, disjoint, and their union equals $X$ that this mapping is well define. Also,

$\varphi^{-1}(\{0\})=A,\varphi^{-1}(\{1\})=B,\varphi^{-1}(\varnothing)=\varnothing,\varphi^{-1}(\{0,1\})=X$.

Thus, $\varphi$ is continuous. It follows that $\varphi$ is a non-constant continuous map from $X$ to $D$. This contradicts our assumption. $\blacksquare$

Remark: Clearly every two point discrete space is homeomorphic, so the usage of $0,1$ opposed to $a,b$ or $\Delta,\Sigma$ is merely notational. Also, notice, by virtue that $\text{card }D=2$, that we could have rephrased the above as saying “A space is connected if and only if every continuous map from $X$ to $D$ is not surjective”

March 28, 2010

It is not surprising that if one cannot speak of a connected space that they would like to at least speak of the “biggest” connected subspaces. To motivate our point let us consider our example we used in the first post. $(-\pi,-e)\cup(e,\pi)$. It is intuitively clear once again what a component of this space would be, but defining it is a little trickier than may seem. We clearly don’t want to say a component is a connected subspace, for clearly $\left(e,e+\frac{1}{2}\right)$ is a connected subspace but not a “component”. It is clear once again that our minds are drawn to maximality. We want to describe our space by as few connected subspaces as possible.  Defining this maximality is what requires finesse though. Usually in math when we think of biggest thing that contains something we usually think of the intersection of all things containing it. That is a problem here though, since the intersection of two connected sets need not be connected (think about the intersection of a line and a circle)….

With this in our noodles as motivation we begin the actual work.

Component: If $X$ is a topological space we define $C\subseteq X$ to be a component if it is connected and it is not contained in any other connected subspace.

Before we get into the meat of the argument we prove some facts regarding unions of connected sets.

Theorem: Let $X$ be a topological space and $\left\{E_j\right\}_{j\in\mathcal{J}}$ and arbitrary collection of non-empty connected subspaces. Then if $\displaystyle \bigcap_{j\in\mathcal{J}}E_j$ is non-empty then $\displaystyle E-\bigcup_{j\in\mathcal{J}}E_j$ is connected.

Proof: Suppose not, then there exists some $A,B\subseteq X$ which are open and non-empty and such that $(A\cap E)\cup (B\cap E)=\varnothing$ and $(A\cap E)\cup (B\cap E)=E$ and $A\cap E,B\cap E\ne\varnothing$. Now, clearly no one particular $E_j$ may be in both $A\cap E,B\cap E$ otherwise $A\cap E_j,B\cap E_j$ so by assumption that $A\cap E, B\cap E\ne \varnothing$ we may conclude in particular that there exist some $E_k$ such that $E_k\not\subseteq A\cap E$, and there is some $E_\ell\subseteq B$. But, since $E_k\cap E_\ell\ne\varnothing$ we see that $A\cap E_k,B\cap E_k$ is a disconnection of $E_k$ and this is a contradiction. $\blacksquare$

We can actually prove some pretty cool theorems using this. For example:

Theorem: Let $\left(\mathcal{V},\|\cdot\|\right)$ be a normed vector space  over $\mathbb{R}$. Then $\mathcal{V}$ is connected.

Proof: Fix $\vec{v}\in\mathcal{V}$ and define $\lambda_{\vec{v}}:\mathbb{R}\to \mathcal{V}$ by $\alpha\mapsto\alpha\vec{v}$.

Lemma: $\lambda_{\vec{v}}:\mathbb{R}\to\mathcal{V}$ is continuous.

Proof: Since these are both metric spaces we use the more metric space oriented definition of continuity.If $\vec{v}=\bold{0}$ this is trivial, so assume not. Let $\varepsilon>0$ be given, then choosing $\alpha,\beta\in\mathbb{R}$ such that $\displaystyle |\beta-\alpha|<\frac{\varepsilon}{\|\vec{v}\|}$ we see that $\displaystyle \|\beta\vec{v}-\alpha\vec{v}\|=|\beta-\alpha|\|\vec{v}\|<\frac{\varepsilon}{\|\vec{v}\|}\|\vec{v}\|=\varepsilon$. The conclusion follows. $\blacksquare$

Now clearly $\lambda_{\vec{v}}\left(\mathbb{R}\right)$ is a connected subspace of $\mathcal{V}$ since it is the continuous image of a connected space. But, seeing that $\displaystyle \bigcap_{\vec{v}\in\mathcal{V}}\lambda_{\vec{v}}\left(\mathbb{R}\right)\supseteq\{\bold{0}\}$ and $\displaystyle \mathcal{V}=\bigcup_{\vec{v}\in V}\lambda_{\vec{v}}\left(\mathbb{R}\right)$ appealing to the previous theorems finishes the argument. $\blacksquare$

Corollary: Every Banach space is connected.

Corollary: $\mathbb{R}^n$ and thus $\mathbb{C}^n$ are connected.

Corollary: $\mathcal{C}\left[X,\mathbb{R}\right]$ with the norm $\|\cdot\|_{\infty}$ is connected for any topological space $X$.

We next wish to prove something that might have been intuitively obvious.

Theorem: Let $X$ be a topological space and $A$ be a connected subspace then given any $A\subseteq B\subseteq \overline{A}$ we have that $B$ is connected, in particular $\overline{A}$ is connected.

Proof: Suppose that $G\cup H$ is a disconnection of $B$ in $X$ (that is there restrictions to $B$ is a disconnection of $B$ in the normal sense). Since $A$ is connected we clearly must have that either $A\subseteq G$ or $A\subseteq H$.  Now, it is relatively easy to show that this implies that $\overline{A}\cap G=\varnothing$ and thus $B\cap G=\varnothing$ which contradicts our choice of $B$. $\blacksquare$

Remark: There is a particularly nice way of proving that $A$ is connected implies that $\overline{A}$ is connected. To see this consider $\varphi:\overline{A}\to D$ then from basic topology we have that $\varphi\left(\overline{A}\right)\subseteq\overline{\varphi(A)}$ and since $\varphi(A)$ must be either $\{0\}$ or $\{1\}$ which are both closed in $D$ it follows that $\varphi(\overline{A})=\{0\},\{1\}$. From where the conclusion is immediate.

Corollary: If $X$ is a topological space with a connected dense subset then $X$ is connected.

We can now prove the main properties of components which interest us.

Theorem: If $X$ is a topological space and $x\in X$ then it is contained in exactly one component.

Proof: Define $\mathfrak{C}_x$ to be the set of all connected subsets of $X$ which contain $x$. Clearly then $\displaystyle \bigcap_{C\in\mathfrak{C}_x}C\supseteq\{x\}$ and so by a previous theorem we see that $\displaystyle \bigcup_{C\in\mathfrak{C}_x}C$ is connected. This is clearly a component which contains $x$ Also, it is clearly the only one. For, suppose that $C'$ is a component which contains $X$ then $C\in\mathfrak{C}_x$ and so $\displaystyle C'\subseteq\bigcup_{C\in\mathfrak{C}_x}C$ and by $C'$‘s maximality it follows that $\displaystyle C'=\bigcup_{C\in\mathfrak{C}_x}C$. The conclusion follows. $\blacksquare$.

Theorem: Any connected subspace of $X$ is contained in exactly one component.

Proof: The fact that if they are contained in one it is unique is obvious. To see that it is contained in at least one merely note that it is contained in the component constructed in the above theorem. $\blacksquare$.

Theorem: The components of $X$ are closed.

Proof: Let $C$ be a component of $X$, then by previous theorem we have that $\overline{C}$ is a connected subspace of $X$ which contains $C$, and by $C$‘s maximality it follows that $C=\overline{C}$. $\blacksquare$.

Theorem: A connected subspace of $X$ which is both open and closed is a component of $X$.

Proof: Suppose that $C$ is such a subspace and $C\subset C'$ where $C'$ is another connected subspace of $X$. Then $C'-C\cup C$ is clearly a disconnection of $C'$. It follows that no connected subspace of $X$ may properly contain $C$. The conclusion follows. $\blacksquare$.

One of the important facts about components is the following.

Theorem: Let $X$ be a topological space and $\mathfrak{C}$ the set of components of $X$, then $\mathfrak{C}$ forms a partition of $X$.

Proof: Since every point of $x$ belongs to at least one component it’s clear that $\displaystyle \bigcup_{C\in\mathfrak{C}}C=X$. Also, suppose that $C,C'\in\mathfrak{C}$ were such that $C\cap C'\ne\varnothing$ then by a previous theorem we have that $C\cup C'$ is a connected subspace of $X$ which properly contains $C$ and $C'$, which of course contradicts their maximality. The conclusion follows. $\blacksquare$

Corollary: If $C'$ is a component of $X$ then $\displaystyle C'=X-\bigcup_{C\in\mathfrak{C}-\{C'\}}C$

From this we can prove the following theorem.

Theorem: Let $X$ have finitely many components, then they are open subsets of $X$.

Proof: Let $\mathfrak{C}$ be as above and let $C'\in\mathfrak{C}$ then $\displaystyle X-C'=\bigcup_{C\in\mathfrak{C}-\{C'\}}C$ which is the finite union of closed subspaces of $X$ and thus a closed subspace. The conclusion follows. $\blacksquare$

This finished up what I wanted to say about components. So, I am just going to prove some related theorems. Most of them relate to union of connected subspaces.

Theorem: Let $X$ be a topological space and $\left\{U_n\right\}_{n\in\mathbb{N}}$ a sequence of closed subspaces of $X$ such that $U_n\cap U_{n+1}\ne\varnothing$. Then $\displaystyle U= \bigcup_{n=1}^{\infty}U_n$ is a connected subspace of $X$.

Proof: Suppose there existed non-empty open subsets $G,H$ of $X$ such that their restrictions to $U$ form a disconnection. Let $A=G\cap U,B=H\cap U$. Clearly we must have that either $U_1\subseteq A$ or $U_1\subseteq B$ (it obviously can’t have some in one and some in the other or the restrictions of $A,B$ to $U_1$ would be a disconnection). WLOG assume that $U_1\subseteq A$ and let $\Omega=\left\{n\in\mathbb{N}:U_n\not\subseteq A\right\}$. If $\Omega=\varnothing$ we are done, so assume not. Then by the well-ordering principle $k=\min\text{ }\Omega$ exists. Clearly we must have that $k\geqslant 2$.

Now, assume that $U_{k-1}\subseteq A$ then since $U_{k-1}\cap U_k\ne\varnothing$ we have that $A\cap U_k,B\cap U_k$ are both non-empty open subspaces of $U_k$ and their union is clearly $U_k$. This of course contradicts $U_k$‘s connectedness. It follows that $U_{k-1}\not\subseteq A\implies k-1\in\Omega$ which contradicts the minimality of $k$. It follows that $\Omega=\varnothing$. The conclusion follows. $\blacksquare$

Theorem: Let $X$ be a topological space and $\left\{U_j\right\}_{j\in\mathcal{J}}$ a collection of connected subspaces of $X$ such that $U_k\cap U_\ell\ne\varnothing,\text{ }k\ne\ell$. Then, $\displaystyle U=\bigcup_{j\in\mathcal{J}}U_j$ is a connected subspace of $X$.

Proof: Suppose not, and there existed some $G,H$ were are non-empty and open in $X$ and $\left(G\cap U\right)\cup\left(H\cap U\right)$ is a disconnection of $U$. Then, since $A=G\cap U,B=H\cap U$ are non-empty and each $U_j$ connected we must have that $U_k\subseteq A$ and $U_\ell\subseteq B$ for some $k,\ell$ but since $U_k\cap U_\ell\ne\varnothing$ it is pretty plain that $G\cap U_\ell, H\cap U_\ell$ is a disconnection of $U_\ell$. The fact that $U$ is disconnected is therefore untenable. $\blacksquare$.

We now give an alternate theorem that the product of finitely many connected spaces is connected.

Theorem: Let $X$ and $Y$ be connected topological spaces, then $X\times Y$ is connected under the product topology.

Proof: Fix $x_0\in X$ and $y_0\in Y$. Clearly $Y\approx \{x\}\times Y$ for any $x\in x$ and $X\approx X\times \{y_0\}$. Thus, given any $x\in X$ we have that $\Omega_x=\left(X\times\{y_0\}\right)\cup\left(\{x\}\times Y\right)$ is the union of two interesting connected sets, thus connected. Also, we have that $\Omega_x\cap \Omega_{x'}\supseteq\{(x_0,y_0)\}$ for any $x,x'\in X$. It follows that $\displaystyle \bigcup_{x\in X}\Omega_x=X\times Y$ is connected. $\blacksquare$.

Corollary: It follows from induction that the product of finitely many connected spaces is connected.

We now need a definition. Note though that the following may be extended considerably.

Convex set: Let $\mathcal{V}$ be a normed vector space over $\mathbb{R}$. We then call $C\subseteq V$ convex if $u,v\in C\implies (1+t)v+ut\in C$ for every $t\in[0,1]$.

Theorem: Let $\mathcal{V}$ be a normed vector space and $C$ a convex subset, then $C$ is connected.

Proof:

Lemma: Let $u,v\in C$ and define $L(u,v)=\left\{(1-t)v+ut:t\in[0,1]\right\}$. Then if $E=\left\{(t,1-t):t\in[0,1]\right\}$ the mapping $\varphi:E\to L(u,v)$ given by $(t,1-t)\mapsto (1-t)v+ut$ is continuous. (it is assumed that $E$ is a subspace of $\mathbb{R}^2$ under the usual topology)

Proof: Notice that if , $(\alpha,1-\alpha),(\beta,1-\beta)\in E$ that

$d\left(\left(\alpha,1-\alpha\right),\left(\beta,1-\beta\right)\right)=\sqrt{\left(\alpha-\beta\right)^2+\left(\left(1-\alpha\right)-\left(1-\beta\right)\right)^2}=\sqrt{2}|\beta-\alpha|$

Now it is assume that $u\ne v$ we see that $\|u-v\|\ne 0$. So, letting $\delta=\frac{\sqrt{2}\varepsilon}{\|u-v\|}$ we see that

$d\left(\left(\alpha,1-\alpha\right),\left(\beta,1-\beta\right)\right)\implies\left \|\varphi\left(\left(\alpha,1-\alpha\right)\right)-\varphi\left(\left(\beta,1-\beta\right)\right)\right\|$

$\leqslant |\beta-\alpha|\|u-v\|$$=\frac{d(\left(\left(\alpha,1-\alpha\right),\left(\beta,1-\beta\right)\right)}{\sqrt{2}}\|u-v\|<\varepsilon$

The conclusion follows. $\blacksquare$

Now that would help us if we knew that $E$ is connected, so we must prove it. Don’t fret though, we must merely note that if $\ell:[0,1]\to\mathbb{R}$ is given by $x\mapsto 1-x$ then it is evident that $\ell$ is continuous and $E=\Gamma_\ell$. It follows from an earlier question that $E$ and thus $L(u,v)$ is connected.

So, assume that $\varphi: C\to D$ is continuous and surjective. Let $a\in\varphi^{-1}(0),b\in\varphi^{-1}(1)$ then clearly the mapping $\varphi\mid L(a,b):L(a,b)\to D$ is also continuous and surjective. This contradicts that $L(a,b)$ is connected. The conclusion follows.

From this we can derive a theorem which will become mainly interesting to us in the next post.

Theorem: Let $\mathcal{V}$ be a normed vector space. Then the open (or closed) ball $B_{\varepsilon}(x)$ is convex for every $\varepsilon>0$ and for every $x\in\mathcal{V}$.

Proof: Let $u,v\in B_{\varepsilon}(x)$ then

$\|x-(1-t)v-ut\|=\|(1-t)(x-v)+t(x-u)\|$

$\leqslant (1-t)\|x-v\|+t\|x-u\|$ $<(1-t)\varepsilon+t\varepsilon=\varepsilon$

it follows that $(1-t)+ut\in B_{\varepsilon}$. The conclusion follows. $\blacksquare$.

Corollary: Every open (or closed) ball in a normed vector space is connected.

March 16, 2010