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

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 $$x \in V \subset U \subset X$$.

• Your suggested procedure fails not in that it doesn't terminate (it terminates after one step-- the connected component containing $x$ is connected by definition), but in that the connected components of a space need not contain a nonempty open set. Indeed, in Lord Shark's example, the connected component of $\mathbb{Q}$ containing $x$ is just $\{x\}$, which is not open. – jawheele May 29 at 6:06

No. Consider $$\Bbb Q$$ in its usual topology. Every subset of $$\Bbb Q$$ is totally disconnected, and no point has a connected neighbourhood.
• By the usual topology, do you mean the topology inherited from $\mathbb{R}$? In that case, given a neighborhood, how can I show that it's disconnected? What is the decomposition? Thank you for your help. I am just learning about connectedness today. – Tri Nguyen May 29 at 6:06
• One can think of it as the subspace topology inherited from $\Bbb R$ or as a metric space itself. If $a$ and $b$ are rationals with $a<b$, then there is an irrational $\xi$ between $a$ and $b$. Then $\Bbb Q$ is the disjoint union of open sets $\{t\in\Bbb Q:t<\xi\}$ and $\{t\in\Bbb Q:t>\xi\}$. Thus no subset of $\Bbb Q$ containing $a$ and $b$ can be connected. – Lord Shark the Unknown May 29 at 6:15
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.