Let $$X$$ be a locally path-connected space such that every point $$x$$ has a neighbourhood $$U$$ with $$\overline{U}$$ compact. I need to show that the path-components of $$X$$ are open and are the same as the connected components of $$X$$.

I just know that in every locally connected topological space $$A$$ every component $$B$$ is open. But I don't know how to use it. Help me.

• Linear connected = path connected? "It is mind" = it is known/given? components of connect = connected components? – Henno Brandsma Dec 20 '18 at 22:49
• I just look Wikipedia, yes any is such – Andrey Komisarov Dec 20 '18 at 22:53
• I offer a better formulation of the question. Agree? – Henno Brandsma Dec 21 '18 at 5:44
• Your question was put on hold, the message above (and possibly comments) should give an explanation why. You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Dec 27 '18 at 14:47

Compactness of certain sets is not needed.

For the first part of your question you will find an answer in Definition of locally pathwise connected. But for the sake of completeness let us prove once more that the following are equivalent:

(1) $$X$$ is locally connected (locally path connected), i.e. has a base consisting of open connected (open path connected) sets.

(2) Components (path components) of open sets are open.

(1) $$\Rightarrow$$ (2): Let $$\mathcal{B}$$ be a base for $$X$$ consisting of open connected (open path connected) sets. Let $$U \subset X$$ be open and $$C$$ be a component (path component) of $$X$$. Consider $$x \in C$$. By assumption there exists $$V \in \mathcal{B}$$ such that $$x \in V \subset U$$. Since $$V \cap C \ne \emptyset$$, we see that $$V \cup C$$ is a connected (path connected) subset of $$U$$ which contains $$C$$. By definiton of $$C$$ we see that $$V \cup C = C$$, i.e. $$V \subset C$$. Hence $$C = \bigcup_{V \in \mathcal{B}, V \subset C} V$$. In particular, $$C$$ is open in $$X$$.

(2) $$\Rightarrow$$ (1): Let $$U \subset X$$ be open. For any $$x \in U$$ the component (path component) of $$U$$ containing $$x$$ is open, hence $$U$$ is the union of open connected (open path connected) sets.

Now, if $$X$$ is locally path connected, then it is also locally connected. Hence components and path components of open sets are open. Moreover, components and path components of open sets agree (this applies in particular to $$X$$ itself). To see this, consider an open $$U \subset X$$. Each path component $$C$$ of $$U$$ is contained in a component $$C'$$ of $$U$$. Assume $$C \subsetneqq C'$$. Let $$C_\alpha$$ be the path components of $$C'$$. They are again open, and we must have more than one. Then $$C'$$ can be decomposed as the disjoint union of two non-empty open sets (e.g. $$C_{\alpha_0}$$ and $$\bigcup_{\alpha \ne \alpha_0} C_\alpha$$). This means that $$C'$$ is not connected, a contradiction. We conclude $$C = C'$$.