Abstract Nonsense

Crushing one theorem at a time

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.