2
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – jawheele May 29 at 6:06
7
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Tri Nguyen May 29 at 6:06
  • 1
    $\begingroup$ 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. $\endgroup$ – Lord Shark the Unknown May 29 at 6:15
  • $\begingroup$ Ok that makes sense. Thank you so much. I was just confused on the openness of neighborhoods and the openness requirements of connected component (and how they are both open and closed in the relative topology). $\endgroup$ – Tri Nguyen May 29 at 6:18
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.