Suppose that $X$ is Hausdorff. Show that $X$ is locally path connected.

Question

Let $f:[0,1]\rightarrow X$ be a continuous surjective function to a Hausdorff space $X$. Prove that $X$ has the following property:

For every $x\in X$ and every neighborhood $U$ of $x$, there exists a neighborhood $V$ of $x$ such that for all $a,b\in V$ there is a path from $a$ to $b$ contained in $U$.

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I have been going through this thinking of things that could potentially help. I realize that what we are trying to show is equivalent to showing that $X$ is locally path connected.

I was thinking about using that a space $X$ is locally path connected if and only if for every open set $U$ of $X$, each path component of $U$ is open in $X$.

I was also thinking of using that since $X$ is Hausdorff, we know that any closed subset is going to be compact, and then somehow intersecting the finite subcover with an open neighborhood of $x\in X$ to get a smaller neighborhood. (Edit: the comments have pointed out that this is false. I was thinking compact in Hausdorff is closed. That being said it made me think of how the image of a compact set under a continuous map is compact) Edit: using that the image of [0,1] will be compact, can we still use the idea of a finite subcover to find a smaller neighborhood of $x$?

I was also thinking of how the image of a connected space under a continuous map is connected.

With all of these thoughts together, I was unsure of how to proceed. Any suggestions for how to proceed would be appreciated.

Answer 1

Lemma: $f$ is a closed map.
Proof: Let $C \subset [0, 1]$ be closed. Because $[0, 1]$ is compact, $C$ is compact. Then $f(C)$ is compact. Since $X$ is Hausdorff, this means that $f(C)$ is closed.

Now, let $x$ be any element of $X$ and $A$ be any open neighborhood of $x$. Choose an open interval $U_t$ around each element $t \in f^{-1}(x)$ such that $f(U_t) \subset A$. Then $U = \bigcup U_t$ is open. What do we know about $$X \setminus f\left([0, 1] \setminus U\right)?$$

Answer 2

Let me not just write down a proof, but instead explain how by "following your nose" and doing "the only possible thing" at each stage we can wind up with a proof at the end of the day.

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So, the only data we are given is a surjective continuous map $f : [0, 1] \to X$ and an open neighborhood $U$ in $X$ of some fixed $x \in X$. We need to find another neighborhood $V$ of $x$, necessarily contained in $U$ (good luck finding a path in $U$ to a point not in $U$!). This $V$ is supposed to satisfy some nice property related to path-connectedness, and constructing paths in an arbitrary topological space seems very hard, so the only hope I have for finding paths is by restricting the domain of $f$ to intervals $[a, b]$.

In order to build these paths, let $T := f^{-1}(U) \subset [0, 1]$, which is just some open subset. We care about points in $[0, 1]$ which $f$ maps to $x$, so also set $T_0 := f^{-1}(\{x\})$ (of course $T_0 \subset T$). Since $T$ is open, for each $t \in T_0$ there is an interval $(a_t, b_t) \subset T$ containing $t$. Since $f^{-1}(\{x\})$ is a closed subset of $[0, 1]$ it is compact, so there is a finite subset $\{t_1, \ldots, t_n\} \subset T_0$ so that $T_0 \subset \bigcup_{j = 1}^n (a_{t_j}, b_{t_j})$ (i.e. these intervals form a finite cover of $T_0$).

As a shorthand let's define $S := \bigcup_{j = 1}^n (a_{t_j}, b_{t_j})$. This seems pretty great: if $x' \in f(S)$ then there exists some $t' \in S$ such that $f(t') = x'$, and thus there is some $j$ so that $t' \in (a_{t_j}, b_{t_j})$. If $t' < t_j$ then the restriction of $f$ to $[t', t_j]$ is a path from $f(t') = x'$ to $f(t) = x$, and if $t' > t_j$ then likewise the restriction of $f$ to $[t_j, t']$ is a path the other way. At any rate, $f(S)$ has the desired property.

The only concern is whether $V := f(S)$ is actually a neighborhood of $x \in X$, which is something which we can just check. (Suspiciously we haven't used the hypothesis of Hausdorffness yet, so at this stage we hope it comes into play.) Now we use a standard trick: $S \subset [0, 1]$ is open, so $S^\text{c} := [0, 1] \setminus S$ is a closed subset of $[0, 1]$, hence compact. Thus $f(S^\text{c})$ is compact, too. Of course $f(S) := f(S^\text{c})^\text{c}$, so all we need is that $f(S^\text{c})$ must now be closed in $X$.

Fortunately for us, this is true by Hausdorffness of $X$ (which is what we were lead to believe by the hypotheses which we see that we have used so far)! Indeed, let $C \subset X$ be any compact subset and fix any $y \in X \setminus C$ (obviously there is no problem if $C$ is all of $X$). Then by Hausdorffness of $X$ for each $x \in C$ there is an open neighborhood $A_x$ of $x$ in $X$ and likewise an open neighborhood $B_x$ of $y$ such that $A_x \cap B_x = \emptyset$. Of course the union $\bigcup_{x \in C} A_x$ covers the compact set $C$, so there is a finite collection $\{x_1, \ldots, x_k\}$ such that the open sets $\{A_{x_j}\}_{1 \leq j \leq k}$ still cover $C$. Now of course $y$ belongs to the finite intersection $B := \bigcap_{j = 1}^k B_{x_j}$, which is still open. Moreover, this intersection necessarily has empty intersection with $C$ (by the pairwise disjointness of each $A_{x_j}$ and $B_{x_j}$). Thus $B$ is an open neighborhood of $y$ wholly contained in the complement $X \setminus C$. Therefore this complement is open, so $C \subset X$ is closed, as desired.