# Proving/disproving a claim on the relationship between the kinds of connectedness (the usual, and path and local)

Context: I am given, as an exercise, the task of proving this claim:

For $$X$$ a topological space, show it is path-connected if and only if $$X$$ is connected and each $$x \in X$$ lies in some path-connected open neighborhood (presumably of $$x$$).

Proving the forward implication is straightforward enough, but I've been having difficulties with the converse.

First, some definitions for context:

$$\newcommand{\nc}{\newcommand} \nc{\T}{\mathcal{T}} \nc{\p}[1]{\left(#1_1,#1_2\right)}$$ Definitions: We assume throughout $$(X,\mathcal{T})$$ is a topological space.

• $$X$$ is path-connected if $$\forall x,y \in X$$ there exists $$\gamma : [0,1] \to X$$ continuous such that $$\gamma(0) = x$$ and $$\gamma(1) = y$$.

• $$X$$ is connected if it is not disconnected.

• That is, $$X$$ is connected if we cannot write it as a disjoint union of nonempty open sets.
• That is, symbolically, $$X$$ is connected if we cannot write $$X = A \cup B$$ for $$\varnothing \ne A,B \in \mathcal T$$ with $$A \cap B = \varnothing$$.
• $$X$$ is locally connected if $$\mathcal{T}$$ has a basis of connected sets.

Note that if $$X$$ is path-connected, it is also connected.

Elaboration: As I looked into the question, I came across a case (thanks to this post) that would contradict the possible alternative claim where $$\text{locally connected & connected} \implies \text{path-connected}$$ This counterexample would arise with the lexicographic ordering on $$[0,1]^2$$.

For $$x := \p \xi,y := \p \eta \in [0,1]^2$$ we say $$x \prec y \iff (\xi_1 < \eta_1) \lor (\xi_1 = \eta_1, \xi_2 < \eta_2)$$ This induces a topology $$\T$$ on $$[0,1]^2$$ with the basis formed by open "intervals"

$$(x,y) := \left\{ z := \p \zeta \in [0,1]^2 \, \middle| \, x \prec z \prec y \right\}$$

To my understanding, this topology $$\T$$ is locally-connected and connected, yet not path connected -- at least from what I can look up.

Why is this relevant? Well, my thinking is this: if each $$x \in X$$ is in some path-connected open neighborhood, I feel like that might imply that $$X$$ is locally connected. That's where my instincts lie. Then the above example would give a counterexample.

My Question: Does this prove to be a suitable counterexample for the converse of the proposed claim I'm told to prove?

More directly, then: is it true to say that, if for each $$x \in X$$ (for $$(X,\T)$$ any connected topological space), there exists $$C_x \in \T$$ path-connected containing $$x$$, then we have local connectedness?

If not (and there is no suitable counterexample), what might be a suitable way to indeed prove the converse of the claim? I've been having trouble doing so -- in particular, I have no clue how I should establish an $$x,y$$-path function $$\gamma$$ in $$X$$ if the claim is true (how would one deal with the case when $$x,y$$ are in different such neighborhoods in the hypothesis?).

Well, my thinking is this: if each $$x\in X$$ is in some path-connected open neighborhood, I feel like that might imply that $$X$$ is locally connected. That's where my instincts lie. Then the above example would give a counterexample.

Not really. First of all locally connected is stronger then your property. Locally connected means that every open neighbourhood has open connected subneighbourhood. In your property we only care about existance of path-connected neighbourhood, it doesn't imply that every open neighbourhood has a path-connected subneighbourhood (a counterexample will be given later).

Secondly, the counterexample has weaker assumptions. So you say that there is a counterexample for a space having property $$P1$$ and $$P2$$. Then you have stronger property $$P1'$$ that implies $$P1$$. It may happen that in such case there is no counterexample, because now you deal with $$P1'$$ and $$P2$$ which is stronger.

My Question: Does this prove to be a suitable counterexample for the converse of the proposed claim I'm told to prove?

No.

More directly, then: is it true to say that, if for each $$x \in X$$ (for $$(X,\mathcal{T})$$ any connected topological space), there exists $$C_x \in \mathcal{T}$$ path-connected containing $$x$$, then we have local connectedness?

No. The standard counterexample is any path-connected space that is not locally connected, e.g. closed infinite broom. Note that in that case $$C_x$$ is simply equal to entire $$X$$.

If not (and there is no suitable counterexample), what might be a suitable way to indeed prove the converse of the claim?

For any $$x\in X$$ consider $$C_x$$ to be the the union of all open path-connected neighbourhoods of $$x$$. I leave it as an exercise that it exists, it is open, path-connected and maximal (among open path-connected subsets). I also leave as an exercise that this implies that $$\{C_x\}_{x\in X}$$ is a partitioning of $$X$$ into open disjoint subsets. But since $$X$$ is connected then no such partitioning can exist, except for a singleton, i.e. $$\{C_x\}_{x\in X}=\{C_{x_0}\}$$ for some (any) $$x_0\in X$$. Thus $$X=C_{x_0}$$ is path connected.

• I see. Thanks for the insights and help, I'll try proving it with your tip. Sorry for the trouble, now it feels like a bit of a silly question. Commented Oct 6, 2021 at 7:46
• @EeveeTrainer its not silly, it is perfectly valid question. You have a problem and we try to help you. Noone is saying you've done something wrong or silly. Everything is easy once you understand it, but not before. :) Feel free to ask more questions if you have any issue. Commented Oct 6, 2021 at 7:47