weakly locally connected

Question

A space $X$ is said to be weakly locally connected at $x$ if for every neighborhood of $x$, there is a connected subspace of $X$ contained in $U$ that contains a neighborhood of $x$. Show that if $X$ is weakly locally connected at each of its points, then $X$ is locally connected.(hint: Show that components of open sets are open.)

Proving the hint:

Let, $C$ be a component of a open set of $X$. Let $x\in C$. Now, let $U$ be any neighborhood of $x$, then being weakly locally connected, you can find another nbh. of $x$ which contains inside a connected subspace, hence that nbh is also connected open in $X$. This is true for each $x$. Hence $C$ must be open set.

But, how this proves the main result? Thanks.

Answer 1

The fact that $X$ is locally connected iff for all open subsets $O$ of $X$, all components of $O$ (as a subspace) are open in $O$ (and $X$ too), should be known to you. It's Munkres theorem 25.3

So we will show $X$ to be locally connected using the right to left direction of this theorem.: let $O$ be open in $X$ and let $C$ be a component of $O$ (in the subspace topology).

Let $x \in C$. Then $x \in O$ and $O$ is a neighbourhood of $x$ so by assumption of weak locally connectedness there is a connected set $C_x \subseteq O$ such that $x \in \operatorname{int}(C_x)$ (as $C_x$ is a a neighbourhood of $x$).

Then $C$ and $C_x$ are both connected subsets of $O$ that intersect (in $x$) so $C \cup C_x$ is also a connected subset of $O$. As $C$ is a component of $O$ and as such is a maximally connected subset of $O$,

$$C \cup C_x = C \text{ so } x \in \operatorname{int}(C_x) \subseteq C$$

showing finally that $x$ is an interior point of $C$. As this holds for all $x \in C$, $C$ is open, as required.

Answer 2

From the fact that the connected components of opens are open then given a point $x \in X$ and an open neighbourhood $ X \supset U \ni x $ then given the connected components of $x$ inside $U$, it is an open connected subset containing $x$, so a connected open neighbourhood, and then $X$ is locally connected in $x$.

Note that the key property used to showing the hint lies in the fact that weakly locally connected in $x$ means that you can find, for every open neighbourhood of $x$, a subset of that neighbourhood such that it is connected and such that $x$ is an interior point of that subset.

As said in wikipedia's article, there are spaces which are weakly locally connected in a point but not locally connected in that point, e.g. the infinite broom (here is why ). But I want to make a clarification about this space: why it is not weakly locally connected?

One can argue as this: "It cannot be, because it is not locally connected and we have the just-prooved theorem which would return a contradiction".

This is ok, but then we can try to find which are the points of this infinite broom, in which the space is not weakly locally connected.

So, which is a point in this space such that it is not weakly locally connected there?

Let's take a point $a_n , n>1$, then there exists $r \in \mathbb{R}^+$ such that $a_{n-1}-a_n > r$ and $a_n - a_{n+1}>r$. Now we can take a ball centered in $a_n$ and of radius $0<\epsilon < r$, $B_\epsilon$ , then we want to show that there cannot be a connected subset of this open ball such that $a_n$ is an interior point of this subset. This can be done by noting that if $a_n$ is an interior point of this hypothetic connected subset, say $N$, then it means that there is an open set contained in $N$ which contains $a_n$. But which are the opens of the infinite broom? They are obviously intersection of the infinite broom with open balls of $\mathbb{R}^2$, and so if we do this intersections with balls contained in $B_\epsilon$ then we have that $N$, if it exists, contains at least one of this intersections, which, by construction (note that $0<\epsilon<r$) are made of parts of the broom from $a_n$ but also some stalks of the broom of $a_{n-1}$, but not the complete $n-1$th-broom. In particular there cannot be such $N$, because if it exists, it needs to contain the connected component of the $n-1$th-broom, which is an absurd by the choice of $\epsilon$.