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prove that the closure of the intersection of A and B is the subset of the intersection of the closure of A and the closure of B.

my proof: let x be in cl(A intersects B), then x is in the intersection of A and B and x is in the set of the limit of the intersection of A and B. case i: x is in cl(A intersects B), then x is in A and x is in B, then x is in the closure of A and x is in the closure of B. Therefore, x is in the intersection of the closure of A and the closure of B.

For the second part, I'm not sure how to do it. Please help! Thank you.

I'm using the Topology without tears book. (3.2 #2) It's online.

user110969
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2 Answers2

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We know that A $\cap$ B is a subset of A, and A $\cap$ B is a subset of B. Also, A is a subset of cl(A) and B is a subset of cl(B) (since cl(X) is the smallest closed set containing the set X). Thus, we can conclude that A $\cap$ B is a subset of each of cl(A) and cl(B). Therefore, cl(A $\cap$ B) is a subset of cl(A) $\cap$ cl(B).

Rogue
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The easiest way to do this, I think, is to use the fact that $\newcommand{\cl}{\operatorname{Cl}}$$\cl$ is "increasing": If $X \subseteq Y$ then $\cl X \subseteq \cl Y$. Apply this to $A\cap B$, $A$, and $B$.

dfeuer
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  • How about: A ⊆ clA, B ⊆ clB. So, A∩B ⊆ cl A ∩ cl B. Therefore,Cl(A ∩ B) ⊆ cl A ∩ cl B – user110969 Dec 09 '13 at 09:35
  • Not quite. The last step there doesn't work. Start with $A \cap B\subseteq A$ and $A\cap B \subseteq B$. – dfeuer Dec 09 '13 at 09:43
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    @dfeuer: For sure the "last step" works. $\text{cl}(A \cap B)$ is the smallest closed set that contains $A \cap B$. Thus $A \cap B \subseteq \text{cl}(A) \cap \text{cl} (B) \implies \text{cl}(A \cap B) \subseteq \text{cl}(A) \cap \text{cl}(B)$, since $\text{cl}(A) \cap \text{cl}(B)$ is closed. – dohmatob Nov 27 '16 at 23:43