In a topology, can one always choose a connected neighborhood of a point?

Question

I am trying to show something that requires me finding a connected neighborhood of a point. This seems true, but I don't know why I am doubting myself:

Let X be a topological space. If $x \in X$, then there exists an open neighborhood $U$ of $x$ that is connected.

Can I just pick any open neighborhood and then if it's not connected, then decompose it into disconnected components and pick the one that $x$ is in. Then if it's still not connected then I can repeat. But I am afraid what if this procedure doesn't terminate? Is the statement above true in general?

From my book, the definition of a neighborhood is: $U$ is a neighborhood of $x \in X$ if there exists an open set $V$ such that $x \in V \subset U \subset X$.

Answer 1

No. Consider $\Bbb Q$ in its usual topology. Every subset of $\Bbb Q$ is totally disconnected, and no point has a connected neighbourhood.

Answer 2

Your question has a negative answer. However, there are some special cases in which the answer is trivially "yes". Assume that $X$ has only finitely many connected components. These are closed and therefore also open (because their number is finite). Then the component of any point $x$ is a connected open neighborhood of $x$.

But there is a more interesting situation. In your question you write "Can I just pick any open neighborhood and then if it's not connected, then decompose it ...". This indicates that you hope that each open neighborhood $U$ of any $x$ contains a connected open neighborhood $V$ of $x$. This is of course not true in general, but spaces having this property received an own name: These are the locally connected spaces. For example, open subspaces of $\mathbb R^n$ are locally connected.