# Abstract Nonsense

## 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.