# When is the closure of a path connected set also path connected?

What are the most general criteria we can impose on a locally path connected Hausdorff space $X$ and a path connected subset $A$ such that $\overline{A}$ is path connected? Do more restrictions need to be imposed on $X$ or $A$?

For instance, I know that if $\overline{A}$ is locally path connected then $\overline{A}$ is path connected; for all $x \in \overline{A}$ and some neighborhood $U$ of $x$ that is open in $\overline{A}$, there must be some path connected neighborhood $U' \subseteq U$ of $x$ that is open in $\overline{A}$. That is, there is some open subset $V'$ of $X$ such that $U' = V' \cap \overline{A}$. Since $x$ is a point of closure of $A$, $V'$ must contain some point $x' \in A \subseteq \overline{A}$, i.e. $x' \in U'$, so $x$ is path connected to $x'$ and hence also to $A$. This holds for all $x \in \overline{A}$ so $\overline{A}$ is path connected.

However, the tough part is proving that $\overline{A}$ is locally path connected, because $\overline{A}$ is probably (?) not open in $X$. I'm a complete novice so the only useful thing I know from browsing definitions is that all open subsets of a locally path connected space inherit the local path connectivity. Are there more ways to prove that a subspace inherits local path connectedness?

This is more specific, but would it help if I knew that $A$ was the set difference of two closed sets (i.e. the intersection of a closed set and an open set)?

I've been looking at stronger restrictions such as $X$ being locally simply connected, but the online documentation is scarce. Would local simple connectivity be "inherited more easily" by subspaces?

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I don't think you'll get far imposing conditions on $X$: $\mathbb{R}^2$ is about as nice a space as you could wish for, but $A=\{(x,\sin(1/x))| x \in (0,1) \}$ still gives a counterexample. Here $A$ is homeomorphic to $\mathbb{R}$, another very nice space, suggesting imposing conditions on the topology of $A$ won't help either. It looks like you really do need conditions specifically on how $A$ sits in $X$. – Chris Eagle Jun 25 '12 at 14:34
@ChrisEagle: IMO that example is decisive enough to be an answer... oh except that now the question has been edited. I wonder if this is not really just two questions, now. – Ben Millwood Jun 25 '12 at 14:49
Good point, I guess I really have to prove some restrictions on $\overline{A}$ to exclude pathological cases such as the Topologist's Sine Curve (in fact, I'd prefer working on $A \cup \{x\}$, where $x \in \overline{A}$), such as local path connectivity (which I can't prove easily) or local simple path connectivity (which I don't fully understand). – Herng Yi Jun 25 '12 at 14:50
Considering the edit, the problem in Chris' counterexample is not that the function doesn't exist, but that it fails to be continuous, which sounds like it would be hard to prove in generality. – Ben Millwood Jun 25 '12 at 14:50
Sorry for the change, but the initial question was to give a sufficient condition to construct the $p$ I mention in my edits. The second half of the question now contains another attempt at the same construction. – Herng Yi Jun 25 '12 at 14:52

Let $S \subseteq X$ and $x^1 \in \overline{S}$. Suppose $S$ is path-connected and there exists a countable decreasing (i.e. $U_{i+1} \subseteq U_i$) neighborhood base $(U_i)_{i=1}^{\infty}$ a $x^1$ s.t. for each $i$, whenever $s^i \in U_i \cap S$ then there exists a path in $S \cap U_i$ from $s^i$ to some element of $S \cap U_{i+1}$. Then $S^1 = S \cup \left\lbrace x^1 \right\rbrace$ is path-connected.
Note that we are not assuming that for all $i$, there exists a path between any two point of $S \cap U_i$. The sets $S \cap U_i$ need not even be connected so this is weaker than requiring local connectivity of $\overline{S}$ at $x^1$. If this condition is satisfied at each $x^1 \in \overline{S}$ (or more generally, if such a point exists in each path-component of the boundary of $S$) then $\overline{S}$ is path-connected.
Proof: Let $0 = t_0 < t_2 < \cdots < 1$ be s.t. $t_i \to 1$. Pick $s^0 \in S$ and $s^1 \in S \cap U_1$. Let $\gamma_0 : [t_0, t_1] \to S$ be a path from $s^0$ to $s^1$. Suppose for all $1 \leq l \leq i + 1$ we've picked $s^l \in S \cap U_l$ and for all $1 \leq l \leq i$ we've constructed paths $\gamma_l : [t_l, t_{l+1}] \to S \cap U_l$ from $s^l$ to $s^{l+1}$. By assumption, we can pick $s^{i+2} \in S \cap U_{i+2}$ and a path $\gamma_{i+1} : [t_{i+1}, t_{i+2}] \to S \cap U_{i+1}$ from $s^{i+1}$ to $s^{i+2}$.
After starting this inductive construction at $i = 0$ we can define $\gamma : [0, 1] \to S^1$ on $[0, 1)$ in the obvious way and then declare $\gamma(1) = x^1$. For any integer $N$, $l \geq N$ implies $\operatorname{Im} \gamma_l \subseteq U_l \subseteq U_N$ so that $\gamma([t_N, 1]) \subseteq U_N$. Thus $\gamma$ is continuous at $1$, which shows that $x^1$ belongs to the path-component of $\overline{S}$ containing $S$.