Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a topological space, $Y$ and $Z$ subspaces of $X$. Let $C$ be a connected subset of $Y\cap Z$ such that $C$ is a component of $Y$ and a component of $Z$. Does it follow that $C$ is a component of $Y\cup Z$?

Intuitively, I would say yes, but I don't know how to prove it.

In case further assumptions are necessary, you can go as far as: $X$ is a compact metric space, $Y$ is open, $Z$ (and therefore $C$) is closed, and $C=Y\cap Z$.

Any help is much appreciated.

share|cite|improve this question
up vote 10 down vote accepted

In general, no. Let $X=[0,1]$, $Y=\mathbb{Q}\cap[0,1]$, $Z=([0,1]\setminus\mathbb{Q})\cup\{0\}$, and $C=\{0\}$.

share|cite|improve this answer
Thank you very much! – Stefan Walter Feb 14 '11 at 19:26
@Stefan: Happy to help. However, I'm a bit surprised by the acceptance, because I didn't answer the question with the hypotheses that $Y$ is open and $Z$ is closed in a compact metric space. – Jonas Meyer Feb 14 '11 at 19:29
Oh yes, I overlooked that. I'll temporarily unaccept the question. But do you think those hypotheses can make the statement true? I wouldn't want to just make finding a counterexample a tedious task. – Stefan Walter Feb 14 '11 at 19:49
@Stefan: I don't know. A counterexample would have to not be locally connected, because a component of an open subspace of a locally connected space is open, and if $C$ is clopen then it will be a component in $X$. But I haven't really gotten anywhere on a counterexample or proof. – Jonas Meyer Feb 14 '11 at 19:54

I don't have a definitive answer, but here is a counterexample that seems to have everything but compactness of X.

In the unit square take the subspace $$ X = \bigcup \{ \, [ (1/n, 0), (0, 1) ] \mid n \in {\mathbb N} \, \} \cup \{ (0, 0) \}, $$ where $[ \cdot, \cdot ]$ denotes a line segment. Because $X \setminus \{ (0, 0) \}$ is path-connected and dense, $X$ is connected.

However, if we take the open subset $Y = X \setminus \{ (0, 1) \}$ and the closed subset $Z = \{ (0, 0), (0, 1) \}$, we find that $C = Y \cap Z = \{ (0, 0) \}$ is a component of both. For $Z$ this is obvious. For $Y$ note that any subset containing $(0, 0)$ and some other point can be separated along a segment $[ (0, 1), (\frac{2}{2n+1}, 0) ]$.

Other salient features of this space are:

  • $X$ is locally compact, except at $(0, 0)$ and $(0, 1)$
  • $Z$ is compact, as it would have to be if $X$ were compact
  • Z is locally connected
  • Y is locally connected, except at $(0, 0)$

The most obvious ways of compactifying this counterexample fail. Simply taking the closure in the unit square produces a space where $C$ no longer is a component of $X$. Since $X$ is not locally compact, the one-point compactification is not Hausdorff. It might work as a counterexample for compact $T_1$ spaces but I have not verified this.

share|cite|improve this answer

I think I can now give a positive result, using some of the extra assumptions ($X$ compact Hausdorff, $Y$ open, $Z$ closed). It will be convenient to use the following:


Let $X$ be a Hausdorff space and $C \subset X$ have a compact neighbourhood $K$. Then $C$ is a component of $X$ if and only if $C$ is a component of $K$

Proof of `only if':

If $C$ is not a component of $K$, then $C$ is not connected, or there is a connected subset of $K$ that is a proper superset of $C$. Either way, $C$ is not a component of $X$.

Proof of `if':

Assume $C$ is a component of $K$ and let $B$ be the boundary of $K$ in $X$. Clearly $C$ is connected, so we need to prove that no proper superset of $C$ is connected.

Let us consider $K$ as a subspace. Since $K$ is a compact Hausdorff space, $C$ is a quasicomponent. (for a proof see this answer) Because $C \cap B = \emptyset$, this means that for every $b \in B$ there is a clopen neighbourhood $U_b$ disjoint from $C$. These neighbourhoods form a cover of $B$, that by compactness has a finite subcover. Let $U$ be the union of this subcover. Being a finite union of clopen sets, $U$ is clopen and so is its complement.

Because none of the $U_b$ intersect $C$, and because $B \subset U$, we have $C \subset K \setminus U \subset K \setminus B \subset K$. Since $K$ is closed in $X$ and $K \setminus B$ is open in $X$, $K \setminus U$ is clopen in $X$ too. We may conclude that any connected superset of $C$ must be a subset of $K \setminus U$, therefore a subset of $K$, therefore by assumption equal to $C$.

Wihout too much trouble we can now prove:

Let $X$ be a compact Hausdorff space, $Y$ an open subspace and $Z$ a closed subspace. Let $C$ be a connected subset of $Y \cap Z$ such that $C$ is a component of $Y$ and a component of $Z$. Then $C$ is a component of $Y\cup Z$.


$C$ is a component of, therefore closed in $Z$, which is closed in $X$, so $C$ is closed in $X$. $Y$ is open in $X$, so $X \setminus Y$ is closed in $X$.

$X$ is normal, so $C$ and $X \setminus Y$ have disjoint neighbourhoods $U$ and $V$. If we take $K = \operatorname{Cl} U$ then $$ C \subset U \subset K \subset X \setminus V \subset Y \subset Y \cup Z $$ and $K$ is compact.

Starting from the fact that $C$ is a component of $Y$, we now apply the lemma one way to find that $C$ is a component of $K$, then the other way to find it is a component of $Y \cup Z$.

share|cite|improve this answer
I think the last proposition can be simplified to: If $X$ is a compact normal space and $Y$ is an open subset and $C$ is closed and is a connected component of $Y$, then it is a connected component of $X$. – Stefan Hamcke Feb 8 '14 at 23:13
Well, yes. It is phrased in this rather complicated way just to follow the wording of the original question as closely as possible. The lemma is the really interesting bit, as far as I am concerned. – Niels Diepeveen Feb 9 '14 at 1:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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