# Locally path connected implies connected if and only if path connected

In Hall's book on Lie groups he claims that if a topological space $$X$$ locally path connected then it is connected if and only if it is path connected. For the proof, he refers the reader to a text that I do not have so I am trying to prove it myself. The converse is true because any path connected space is also connected.

The answer to this question says that if $$X$$ is locally path connected and connected then one can prove it is also path connected by fixing $$a \in X$$ and showing that the set $$U = \{x \in X | \text{ there exists a path connected a and x}\}$$ is both open and closed. How does this lead to the conclusion that $$X$$ is path connected?

• What is your definition of connected space? Commented Feb 13, 2023 at 3:13
• @svelaz $X$ is open if there does not exist open subsets $A$ and $B$ of $X$ such that $X = A \cup B$ and $A \cap B = \emptyset$ Commented Feb 13, 2023 at 3:15
• A little correction: The definition has to include both $A$ and $B$ being non-empty, otherwise every space would be connected because of the disjoint partition $X=X\cup\emptyset$. Concerning the upper problem: You have a disjoint partition $X=U\cup U^\complement$ into open sets ($U^\complement$ is open because $U$ is closed), therefore one of them has to be empty. Since $a\in U$, we have $U^\complement=\emptyset$ and therefore $U=X$. Commented Feb 13, 2023 at 3:34
• @SamuelAdrianAntz That is a pretty clever argument, thanks for your help. Commented Feb 13, 2023 at 3:46
• You're welcome! Also a little correction for me: Of course I meant "every space would not be connected" when mentioning the partition. Commented Feb 13, 2023 at 12:46

As requested by ronno, here is my comment as a full answer with more details (like both directions) and references (as a more general view on the concepts might be helpful). For completeness, I have also included the backwards direction:

Lemma: $$X$$ connected $$\Leftarrow X$$ path-connected

(The backwards direction does indeed not need the condition of $$X$$ being locally path-connected.)

Proof: Assume $$X$$ is path-connected, but not connected, then there are non-empty, disjoint and open subsets $$U,V\subset X$$ with $$U\cup V=X$$. Since they are non-empty, there are points $$x\in U$$ and $$y\in V$$ and since $$X$$ is path-connected, there is a path $$\gamma\colon[0;1]\rightarrow X$$ with $$\gamma(0)=x$$ and $$\gamma(1)=y$$. Consider $$\gamma^{-1}(U),\gamma^{-1}(V)$$:

1. They are non-empty as $$0\in\gamma^{-1}(x)\subseteq\gamma^{-1}(U)$$ and $$1\in\gamma^{-1}(y)\subseteq\gamma^{-1}(V)$$.
2. They are disjoint as $$\gamma^{-1}(U)\cap\gamma^{-1}(V)=\gamma^{-1}(U\cap V)=\gamma^{-1}(\emptyset)=\emptyset$$.
3. They are open as preimages of open sets under a continuous map.
4. They fulfill $$\gamma^{-1}(U)\cup\gamma^{-1}(V)=\gamma^{-1}(U\cup V)=\gamma^{-1}(X)=[0;1]$$

Hence $$[0;1]$$ would not be connected, which is not the case and hence yields a contradiction. $$\square$$

The last step has to be proven the hard way: If it could be parted into two non-empty disjoint and open subsets $$U,V\subset[0;1]$$ with $$U\cup V=[0;1]$$ (with $$0\in U$$ without loss of generality), then we would get the contradiction $$\sup U\in U$$ and $$\sup U\in V$$, so $$U\cap V\neq\emptyset$$.

Be aware: It might be tempting at first, but $$[0;1]$$ being connected must not be proven using that it is obviously path-connected and therefore connected because of the upper lemma as this is a circular conclusion.

Lemma: $$X$$ connected and locally path-connected $$\Rightarrow X$$ path-connected

Proof: Let $$x\in X$$ be a point and denote by $$U=[x]_\sim\subseteq X$$ its path-connected component, where $$x\sim y$$ iff there is a path $$\gamma\colon[0;1]\rightarrow X$$ with $$\gamma(0)=x$$ and $$\gamma(1)=y$$. (The relation is reflexive because of the constant path $$\epsilon_x\colon[0;1]\rightarrow X,t\mapsto x$$, symmetric because of the inverse path $$\overline\gamma\colon[0;1]\rightarrow X,t\mapsto\gamma(1-t)$$ and transitive because of the path composition.)

Let $$y\in U$$, so there is a path $$\gamma$$ from $$x$$ to $$y$$. Since $$X$$ is locally path-connected, there is an open and path-connected neighborhood $$V$$ of $$y$$, so for every point $$z\in V$$, there is a path $$\delta$$ from $$y$$ to $$z$$. With path composition, $$\gamma*\delta$$ is a path from $$x$$ to $$z$$, hence $$V\subset U$$ and $$U$$ is open.

Let $$z\in\overline U$$. Since $$X$$ is locally path-connected, there is an open and path-connected neighborhood $$V$$ of $$z$$ with $$U\cap V\neq\emptyset$$. Choose $$y\in U\cap V$$. Because of $$y\in U$$, there is a path $$\gamma$$ from $$x$$ to $$y$$ and because of $$y\in V$$, there is a path $$\delta$$ from $$y$$ to $$z$$. With path composition, $$\gamma*\delta$$ is a path from $$x$$ to $$z$$, hence $$z\in U$$ and $$U$$ is closed, meaning $$U^\complement$$ is open.

We get the partition $$X=U\cup U^\complement$$ into disjoint open subsets and since $$X$$ is connected, they cannot both be non-empty. Since $$x\in U$$ (as $$\sim$$ is reflexive) and $$U$$ is therefore non-empty, $$U^\complement=\emptyset$$ must be empty, so $$X=U$$, which means, that $$X$$ is path-connected. $$\square$$

Be aware: Since I sometimes encounter people believing the following myth, I want to add an important remark for dealing with local properties in topology: (Path-)connectedness does not imply local (path-)connectedness! The Warsaw sine curve is connected, but not locally connected (See "Counterexamples in Topology" by Steen and Seebach found here, Remark 6 on page 138.) The Comb space is path-connected, but not locally path-connected. (It is also contractible, but not locally contractible.)

• One typo, I think $z$ should be in $U^\complement$ in paragraph 3 of the second proof. Alternatively you could use that complement of $U$ is union of other path components, which are each open by the previous paragraph, so must be open. Commented Feb 15, 2023 at 10:46
• You mean $z\in U$ in the last line there? That is not a typo, as I wanted to show $\overline U\subseteq U$ to show that $U$ is closed. You are right, that is also a possibility and indeed the easier one. I also thought about writing it down, but then thought, that giving another possibility (as the equivalence class was already used before) would be more helpful. I could have written both down though, so thanks for adding that! Commented Feb 15, 2023 at 13:50
• Oh I misread something, ignore the first part of my previous comment. Commented Feb 15, 2023 at 13:56

If we know that $$U$$ is both open and closed, we also know that $$V = X \setminus U$$ is open. You have $$a \in U$$, thus $$U \ne \emptyset$$. Since $$U \cup V = X$$ and $$U \cap V = \emptyset$$, we conclude that $$V = \emptyset$$ because otherwise $$X$$ would not be connected. Thus $$U = X$$ which means that $$X$$ is path-connected because any point $$x \in X$$ admits a path connecting it with $$a$$.

Note that the above argument shows that a space $$X$$ is connected iff the only clopen (= open and closed) subsets of $$X$$ are $$X, \emptyset$$.

Let us now prove that $$U$$ is clopen.

1. Let $$x \in U$$. There exists a path-connected open neigborhood $$U(x)$$ of $$x$$ in $$X$$. Clearly $$U(x) \subset U$$ since for each $$y \in U(x)$$ there exists a path in $$X$$ from $$a$$ to $$x$$ and a path in $$U(x)$$ from $$x$$ to $$y$$, thus a path in $$X$$ from $$a$$ to $$y$$.

2. Let $$x \in X \setminus U$$. Again there exists a path-connected open neigborhood $$U(x)$$ of $$x$$ in $$X$$. Assume $$U(x) \cap U \ne \emptyset$$. Pick $$y \in U(x) \cap U$$. There exists a path in $$X$$ from $$a$$ to $$y$$ and a path in $$U(x)$$ from $$y$$ to $$x$$, thus a path in $$X$$ from $$a$$ to $$x$$ which means $$x \in U$$, a contradiction. Thus $$U(x) \subset X \setminus U$$.

Note that the assumption of local path-connectedness can be weakened to prove that $$X$$ is path connected. It suffices to require that each $$x \in X$$ has an open neigborhood $$U(x)$$ such that for each $$y \in U(x)$$ there exists a path in $$X$$ from $$x$$ to $$y$$.