Abstract Nonsense

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.