I know this has a solution here:

Showing that every connected open set in a locally path connected space is path connected

But I would like to see if there is a fault in this simpler proof:

Let $C \subset X$ be open and connected. I know that every path component of $C$ is open in $X$ by the fact that $X$ is locally path connected. Since the path components of $C$ form a disjoint partition of $C$, and since they are open in $X$, and since $C$ is non empty, $C$ must equal one and not more of those path components. So $C$ is path connected.

  • $\begingroup$ this correct except the fact that you justify why the the path components are open. this i guess follows from the locally path connectedness of X. $\endgroup$ – Guy Fsone Jul 26 '17 at 12:26
  • $\begingroup$ Note that the answer in the linked question justifies exactly the part that you did not justify, namely why path components are open. $\endgroup$ – Lee Mosher Jul 26 '17 at 12:41

If you have Munkres (2nd ed.): Thm 25.4 states:

A space $X$ is locally path-connected iff for every open set $U$ of $X$, each path-component of $U$ is open in $X$.

So if indeed $U$ is open and connected it has (as all spaces) a decomposition into path-components, which are open in $X$ and thus open in $U$ too, and by being a partition, they are also closed (the complement is also a union of open sets) in $U$. So by connectedness there can be only one path-component.

25.5 even says (part 2 of it)

If $X$ is a topological space, and $X$ is locally path-connected, its components and path-components coincide.

Apply this to $X=U$ (which is locally path-connected as an open subspace of a locally path-connected space) and you're done right away.

Note, this is assuming you use the same definition of local path-connectedness as Munkres does (which is non-standard): every neighbourhood $U$ of $x$ contains a path-connected neighbourhood $V$ of $x$.

The definition in e.g. Engelking is:

for every open set $U$ and every $x \in U$ there is an open neighbourhood $V$ of $x$ such that for any $y \in V$ there is a path $p: [0,1] \to U$ connecting $x$ to $y$.

Note that $V$ is not supposed to be itself path-connected, as Munkres does. So the latter has a stronger notion, so maybe this fact only holds for the stronger notion; at least the proof does.

  • $\begingroup$ +1 nice answer very easy to understand@ henno sir $\endgroup$ – lomber Apr 11 '18 at 14:01
  • $\begingroup$ Dear Henno! Could you explain one moment of your proof where you wrote " So by connectedness there can be only one path-component"? $\endgroup$ – ZFR Oct 6 at 14:20
  • 1
    $\begingroup$ @ZFR The path component is open and closed (as I showed just before) and in a connected space $X$ the only open-and-closed (or clopen) sets are $\emptyset$ and $X$. (if $A$ were another, $\{A, X\setminus A\}$ would be a decomposition of $X$ in non-trivial parts) $\endgroup$ – Henno Brandsma Oct 6 at 14:33

Your proof is (almost) right. Besides the fact that you should state that locally path connectedness implies that path components of $C$ are open in $X,$ you used another fact that I think should be stated explicitly in your proof to be complete: the fact that $C$ is open in $X.$ Every path component of $C$ is open in $X.$ Since $C$ is open in $X,$ every path component of $C$ is open in $C$ as well. Then each nonempty path component of $C$ is both open and closed in $C,$ and connectedness of $C$ implies that $C$ has only one path component: itself.


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