# Abstract Nonsense

## Thoughts about connectedness (Pt. II)

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.