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?

  • $\begingroup$ What is your definition of connected space? $\endgroup$
    – svelaz
    Commented Feb 13, 2023 at 3:13
  • $\begingroup$ @svelaz $X$ is open if there does not exist open subsets $A$ and $B$ of $X$ such that $X = A \cup B$ and $A \cap B = \emptyset$ $\endgroup$
    – CBBAM
    Commented Feb 13, 2023 at 3:15
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    $\begingroup$ 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 $U=X$. $\endgroup$ Commented Feb 13, 2023 at 3:34
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    $\begingroup$ @SamuelAdrianAntz That is a pretty clever argument, thanks for your help. $\endgroup$
    – CBBAM
    Commented Feb 13, 2023 at 3:46
  • $\begingroup$ You're welcome! Also a little correction for me: Of course I meant "every space would not be connected" when mentioning the partition. $\endgroup$ Commented Feb 13, 2023 at 12:46

2 Answers 2


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.)

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    $\begingroup$ 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. $\endgroup$
    – ronno
    Commented Feb 15, 2023 at 10:46
  • $\begingroup$ 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! $\endgroup$ Commented Feb 15, 2023 at 13:50
  • $\begingroup$ Oh I misread something, ignore the first part of my previous comment. $\endgroup$
    – ronno
    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 $y$.


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