# A locally connected topological space has open connected components

i’m studying from An Introduction To Manifold of Loring W. Tu. In one of its exercises, the text gives the following definition of locally connected topological space:

A topological space $$S$$ is said to be locally connected at $$p\in S$$ if for every neighborhood $$U$$ of $$p$$ there exists a connected neighborhood V of $$p$$ such that $$p \in V\subset U$$.The space $$S$$ is locally connected if it is locally connected at every point. Prove that if S is locally connected, then the connected components of S are open.

Here is thepart of my attempt with the doubt i wanted to ask.

Let us take $$x$$ in $$S$$ its connected component $$C_x$$. If $$p \in C_x$$ then there exists a connected subset $$W$$ such that $$p\in W$$. Since S is locally connected then i can take a neighborhood $$U$$ containing $$p$$ such that $$p \in U \subset W$$. Here is my questions, is my last claim true? I think so because i am specializing the definition of locally connectedness.

Thanks for any hints or answers.

• Hello, $W$ is just a connected set containing $p$ but there is no reason that $W$ is in fact a neighborhood of $p$, so you can't use the local connectedness assumption. Instead, take a neighborhood of your point, use the fact that $S$ is locally connected and then just say that $C_x$ is the union over every connected set containing your point. Commented Jul 24 at 10:33
• Thanks. I thought that i could say that it were possibile to find a neighborhood U contained in W containing p then by locally connectedness say there is a connected neighborhood V such that $p \in V \subset U \subset W$. But I see it is easier the way you say Commented Jul 24 at 10:52
• Something to be aware, there is a subtle difference between open and non-open neighbourhoods here. When for every neighbourhood $U$ of $x$ there is a connected neighbourhood $V$ of $x$, we say $X$ is connected im kleinen at $x$. If there is an open connected neighbourhood, we say that $X$ is connected at $x$. The two aren't equivalent, but $X$ is connected at every $x\in X$ iff its connected im kleinen at every $x\in X$. Commented Jul 24 at 14:24
• Commented Jul 25 at 15:44

Note: in the following, I use "neighborhood of $$x$$" to refer to an open set containing $$x$$. A locally connected space is one in which for every point $$x$$ and every neighborhood $$U$$ of $$x$$, there exists a connected neighborhood $$V$$ of $$x$$ such that $$V \subseteq U$$.

Let $$X$$ be a locally connected space, and let $$C$$ be a connected component of $$X$$. Let $$c$$ be some element of $$C$$. Since the whole space $$X$$ is a neighborhood of $$c$$, and $$X$$ is locally connected, there is some connected neighborhood $$W$$ of $$c$$. Since any connected set containing $$c$$ must be contained in $$C$$, then $$W \subseteq C$$. Since $$c \in C$$ was arbitrary, this shows $$C$$ is open.

EDIT: If "neighborhood of $$x$$" is instead taken to mean a set $$C$$ containing $$x$$ such that there is an open set $$U$$ with $$x \in U \subseteq C$$, then the definition above leads to a different conception of connectedness called weak local connectedness (sometimes referred to as "connectedness im kleinen"). Weak local connectedness is, as the name suggests, generally weaker than local connectedness (consider the infinite broom space, which is weakly locally connected but not locally connected at the origin).

However, if a space $$X$$ is weakly locally connected, then it is also locally connected, which can be proven as follows. Consider the following fact:

If for each open set $$U$$ of $$X$$, each component of $$U$$ is open, then $$X$$ is locally connected.

This is trivial. Let $$x \in X$$, and let $$U$$ be an open neighborhood of $$x$$. Let $$C$$ be the component of $$U$$ that contains $$x$$. The set $$C$$ is open and connected by hypothesis. Since $$x$$ and $$U$$ were arbitrary, $$X$$ is locally connected.

Now, let $$U$$ be an open set in the weakly locally connected space $$X$$. Let $$C$$ be a component of $$U$$, and let $$x \in C$$. Since $$X$$ is weakly locally connected, there is a connected subspace $$A \subseteq U$$ that contains an open neighborhood $$V$$ of $$x$$. Since $$A$$ is connected, it must be contained in $$C$$; therefore, $$V \subseteq C$$. Since $$x$$ was arbitrary, $$C$$ is open. Since $$C$$ and $$U$$ were arbitrary, by the fact above, $$X$$ is locally connected.

• Thanks. I use the following definition of neighbourhood: $U$ is a neighbourhood of $x$ iff there is an open set $A$ such that $x \in A \subset U$ but the equivalence between the two definitions is quite clear Commented Jul 24 at 16:30
• @loyiideruu The two definitions are not the same. And in this case this is pretty important, since this leads to two different definitions of local connectedness at a point Commented Jul 24 at 16:37
• Why are they different? I know that given the definition I gave of neighbourhood, it can be shown that a set is open iff it is a neighbourhood of all of its point. Does it not imply the equivalence between the two definitions? Commented Jul 24 at 17:04
• @loyiideruu There exists a space $X$ and a point $x\in X$ such that for every neighbourhood $U$ of $x$ there is a connected neighbourhood $V$ of $x$ with $V\subseteq U$, but for some $U$ there is no open connected neighbourhood $V\subseteq U$. See this answer. Commented Jul 24 at 17:15
• @loyiideruu I have updated my answer to account for your definition of "neighborhood". Commented Jul 24 at 17:47