1. ## Topology: Closure

For sets A and B in a topological space X, the following hold:
(1) Cl(X-A)=X-Int(A)
(2) Int(A)∩ Int(B) =Int (A∩B)

2. Originally Posted by horowitz
For sets A and B in a topological space X, the following hold:
(2) Int(A)∩ Int(B) =Int (A∩B)
If $x\in \text{int}(A) \cap \text{int}(B)$ then $x\in \text{int}(A) \text{ and }x\in \text{int}(B)$.
This means there are open subsets $U,V$ so $x\in U\subseteq A \text{ and }x\in V\subseteq B$.
Thus, $x\in U\cap V \subseteq A\cap B$, remember that $U\cap V$ is open subset.
Thus, $x\in \text{int}(A\cap B)$.

If $x\in \text{int}(A\cap B)$ then $x\in U\subseteq A\cap B$ for some open subset $U$.
But then $x\in U\subseteq A \text{ and }x\in U\subseteq B$.
Therefore, $x\in \text{int}(A) \text{ and }x\in \text{int}(B) \implies x\in \text{int}(A)\cap \text{int}(B)$.

Thus, $\text{int}(A\cap B) = \text{int}(A)\cap \text{int}(B)$.

3. Originally Posted by horowitz
(1) Cl(X-A)=X-Int(A)
Let $x\in \overline{X-A}$ then it means for any open set $U$ with $x\in U$ we have that $U\cap (X-A)\not = \emptyset$. But then there is not open set $U$ with $x\in U$ so that $U\subseteq A$ (otherwise the intersection would be empty). Therefore, $x$ is not and interior point of $A$, thus, $x\in X - \text{int}(A)$. Conversely, let $x\in X - \text{int}(A)$, this means $x\in X$ and $x\not \in \text{int}(A)$. But if $x\not \in \text{int}(A)$ it means for any open set $U$ so that $x\in U$ we have that $U\not \subseteq A$. Thus, for any open set $U$ with $x\in U$ we see that $U\cap X - A\not = \emptyset$ which means $x$ lies on the closure of $X-A$.

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### prove that in a topological space X - int(A) =cl(X-A)

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