Let $(E,τ)$ be a topological space. Let $A∈τ$. Let $B⊆E$ . Prove that $A∩cl(B) ⊆ cl(A∩B)$. Conclude:

  1. $A∩B=∅ ⇒ A∩ cl(B)=∅$,
  2. If $cl(B)=E$ then $cl(A∩B)=cl(A)$.

Let $F \subseteq E$ denote a closed set.

Let $x \in A \cap cl(B) \implies x \in A \displaystyle \bigcap \bigg(\bigcap \{F \subset E : B \subseteq F, F : \text{closed} \}\bigg) \\ \implies x \in A \wedge x \in \bigg(\bigcap \{F \subset E : B \subseteq F, F : \text{closed} \}\bigg) \\ \implies x\in cl(A \cap cl(B) ) \displaystyle \overset ? = cl(A \cap B) $

Not sure if that last step is valid.

For the sought conclusions:

  1. $A \cap B = \emptyset, \text{BWOC: assume}\,\, A \cap cl(B) \not=\emptyset \Rightarrow \exists x \in A \cap cl(B) \implies x \in cl(A\cap B) $ $\rightarrow\leftarrow\,\, A \cap B = \emptyset \implies cl(A \cap B) = cl(\emptyset) = \emptyset $

  2. $cl(B) = E \implies (\forall U \in \tau, B\cap U \not= \emptyset \Rightarrow B \cap A \not= \emptyset ), \\ cl(A \cap B) \subseteq cl(A) \cap cl(B) = cl(A) \cap E = cl(A) \Rightarrow cl(A\cap B) \subseteq cl(A), \\ \text{ From the above}\,\,: (A = A \cap E \subseteq cl(A\cap E))\Rightarrow cl(A) \subseteq cl(A \cap E) \subseteq cl(A) $

  • 3
    $\begingroup$ I think more people would read and understand your arguments if you wrote them in standard english. $\endgroup$ – Forever Mozart Apr 12 '16 at 4:27

Your last step in the first part of the proof isn't quite right. You didn't quite show that $x \in cl(A\cap cl(B))$ or that this set is equivalent to $cl(A\cap B)$. The key observation here is that $x$ is contained in any closed subset $F \supseteq B$ and $x$ is any closed subset $H \supseteq A$. Notice however that for any closed subsets of this form we have that $A \cap B \subseteq H \cap F$ and $H \cap F$ is closed hence $x$ is an element of any closed subset containing $A \cap B$. The second claim follows from the fact that

$$ A \cap cl(B) \;\; \subseteq \;\; cl(A\cap B) \;\; =\;\; cl(\emptyset) \;\; =\;\; \emptyset. $$

A similar claim can be made for the case $cl(B) = E$.


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