# $S=\operatorname{cl}_S(A), T = \operatorname{cl}_T(B) \implies \operatorname{cl}_{S\cup T}(A \cup B) = S \cup T$

Let $$X$$ be a topological space with subsets $$A \subseteq S \subseteq X$$ and $$B \subseteq T \subseteq X$$ such that $$S=\operatorname{cl}_S(A), T = \operatorname{cl}_T(B)$$. Prove that $$\operatorname{cl}_{S\cup T}(A \cup B) = S \cup T$$.

Note: all subsets have subspace topologies.

Attempt: Let $$x \in S \cup T$$. WLOG, $$x \in S$$. Thus since $$S = \operatorname{cl}_S(A)$$, we know that $$V \cap A \neq \emptyset$$ for all (open) neighborhoods $$V$$ of $$x$$ in $$S$$.

Suppose to the contrary there is a neighborhood $$W$$ of $$x$$ in $$S \cup T$$ with $$W \cap (A \cup B) = \emptyset$$. Then, by possibly shrinking $$W$$, we may assume that $$W$$ is open in $$S \cup T$$. This means that $$W = (S \cup T) \cap G = (S \cap G) \cup(T \cap G)$$ for some subset $$G$$ open in $$X$$. Now, also $$W \cap A = \emptyset$$ and thus $$(S \cap G) \cap A= \emptyset$$. But $$S \cap G$$ is open in $$S$$ and contains $$x$$. Thus $$S \cap G$$ is a neighborhood of $$x$$ in $$S$$, contradicting what we wrote at the beginning of the proof.

Thus, every neighborhood $$W$$ of $$x$$ in $$S \cup T$$ has $$W \cap (A \cup B) \neq \emptyset$$ and it follows that $$x \in \operatorname{cl}_{S \cup T}(A \cap B)$$.

Is this correct?

• "By possibly shrinking" is unacceptably vauge. – William Elliot Feb 12 '20 at 21:09
• I mean the following with that: You know by definition of neighborhood that $W$ contains an open set containing $x$. Replace $W$ by this open set. That being said, is my proof correct? @WilliamElliot – user745578 Feb 12 '20 at 21:20

Direct Proof.
A is dense within S and B is dense within T
implies A $$\cup$$ B is dense within S $$\cup$$ T.
Proof.
Assume x in S $$\cup$$ T.
Let V be open within S $$\cup$$ T.
Exists X-open U with V = U $$\cap$$ (S $$\cup$$ T).
Case x in S: x in S-open U $$\cap$$ S.
. . As A is dense with in S, exists a in A $$\cap$$ U $$\cap$$ S.
Case x in T: Likewise exists b in B $$\cap$$ U $$\cap$$ T.
As every not empty (S $$\cup$$ T)-open set intersects A $$\cup$$ B,
A $$\cup$$ B is dense within S $$\cup$$ T.

• This answer might benefit from some Latex. – user745578 Feb 12 '20 at 21:21

For a subspace $$S$$ we have a formula for the closure in a subspace: if $$A \subseteq S \subseteq X$$ then $$\operatorname{cl}_S(A)=\operatorname{cl}(A) \cap S$$

where the unmarked closure is taken in the whole space $$X$$.

So $$\operatorname{cl}_{S \cup T}(A \cup B)=(S \cup T) \cap \operatorname{cl}(A \cup B) = (S \cup T) \cap (\operatorname{cl}(A) \cup \operatorname{cl}(B))$$

and the right hand side already contains the sets (it's a union of 4 intersections, two of which are:) $$\operatorname{cl}(A) \cap S = \operatorname{cl}_S(A)=S$$ and $$\operatorname{cl}(B) \cap T = \operatorname{cl}_T(B)=T$$ and so

$$\operatorname{cl}_{S \cup T}(A \cup B)=S \cup T$$

follows trivially.

• Thanks! I'm well aware of this fact, but actually I needed the result for a countable collection of subsets, and not a finite one. Your proof does not seem to generalise to the infinite case while mine does (I think, the problem is that countable union of closure is not closure of the union). Could you tell me if my proof is correct please? – user745578 Feb 13 '20 at 6:15