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Let $X$ be a topological space, and $C, U \subset X$. Let $\overline{A}(X)$ denote closure of $A$ in $X$. Does it hold that

$(1) \quad \overline{C \cap U}(U) = \overline{C}(X) \cap U$?

Background

It can be shown that

$(2) \quad \overline{C \cap U}(U) = \overline{C \cap U}(X) \cap U$,

When $C \subset U$, $(1)$ is shown to hold by $(2)$. Otherwise, since closure is increasing, from $(2)$ we have

$(3) \quad \overline{C \cap U}(U) \subset \overline{C}(X) \cap U$.

So the question could also be whether the superset relation holds.

Edit: $(1)$ also holds when $U$ is open in $X$; see my answer below.

kaba
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3 Answers3

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No: take $C=\mathbb{Q},U=\mathbb{R}\setminus\mathbb{Q}$, and $X=\mathbb{R}$. Then

$$\overline{C\cap U}(U)=\overline{\varnothing}(U)=\varnothing$$

while

$$\overline{C}(X)\cap U=\mathbb{R}\cap(\mathbb{R}\setminus \mathbb{Q})=\mathbb{R}\setminus\mathbb{Q}.$$

Balloon
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Denoting closure of $A$ in (sub)space $B$ by $\operatorname{cl}_B(A)$, we can say

If $A \subseteq X$ is a subspace of the space $X$ and $B \subseteq A$ then $$\operatorname{cl}_A(B) = \operatorname{cl}_X(B) \cap A$$

but as Balloon's example shows, we cannot allow arbritrary $B \subseteq X$, $B$ has to be a subset of $A$ to talk about its closure in $A$, after all.

Henno Brandsma
  • 242,131
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Balloon provided an answer to the general question. I will show that the result holds when $U \cap D$ is open in any closed $D \subset X$ such that $C \setminus U \subset D$. This holds, for example, when $U$ is open in $X$ ($D = X$). The following holds without restrictions:

$$ \begin{align} \overline{C}(X) \cap U & = \overline{(C \cap U) \cup (C \setminus U)}(X) \cap U \\ & = (\overline{C \cap U}(X) \cup \overline{C \setminus U}(X)) \cap U \\ & = (\overline{C \cap U}(X) \cap U) \cup (\overline{C \setminus U}(X) \cap U) \\ & = \overline{C \cap U}(U) \cup (\overline{C \setminus U}(X) \cap U) \\ & = \overline{C \cap U}(U) \cup (\overline{C \setminus U}(D) \cap (U \cap D)). \end{align} $$

When $V$ is open in $Y$, we have the additional property that for any $A \subset Y$, we have $A \cap V = \emptyset \iff \overline{A}(Y) \cap V = \emptyset$. We have $(C \setminus U) \cap (U \cap D) = \emptyset$. Hence $\overline{C \setminus U}(D) \cap (U \cap D) = \emptyset$, and so

$$\overline{C}(X) \cap U = \overline{C \cap U}(U).$$

kaba
  • 2,035