About locally path-connected spaces 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.
 A: 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 $C = C'$.
