# If $C$ is a component of $Y$ and a component of $Z$, is it a component of $Y\cup Z$?

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.

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

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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.

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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:

Lemma

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 $Y\cup Z$.

Proof:

$C$ 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 $Y \cup Z$.

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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