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Thread: Topology: Closure

  1. #1
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    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)
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  2. #2
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    Quote Originally Posted by horowitz View Post
    For sets A and B in a topological space X, the following hold:
    (2) Int(A)∩ Int(B) =Int (A∩B)
    If $\displaystyle x\in \text{int}(A) \cap \text{int}(B)$ then $\displaystyle x\in \text{int}(A) \text{ and }x\in \text{int}(B)$.
    This means there are open subsets $\displaystyle U,V$ so $\displaystyle x\in U\subseteq A \text{ and }x\in V\subseteq B$.
    Thus, $\displaystyle x\in U\cap V \subseteq A\cap B$, remember that $\displaystyle U\cap V$ is open subset.
    Thus, $\displaystyle x\in \text{int}(A\cap B)$.

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

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