Proposition 2.27

Proposition 2.27: Let $$A \subseteq X$$, and $$\mathcal{U}(A)= \{ U : U \text{ is open and } U \subseteq A\}$$. Show that $$Int(A)= \bigcup \mathcal{U}(A)$$, that is, $$int(A)$$ is the union of all open subsets of $$A$$.

Ian_mi
Let $$x \in \text{Int}(A)$$. Then there exists an open set $$G$$ such that $$x \in G \subseteq A$$. Therefore $$x \in \bigcup \{G : G \subseteq A \text{ and } G \text{ is open}\}$$, hence $$\text{Int}(A) \subseteq \bigcup \{G : G \subseteq A \text{ and } G \text{ is open}\}$$. Let $$x \in \bigcup \{G : G \subseteq A \text{ and } G \text{ is open}\}$$. Then there exists an open set $$G$$ such that $$x \in G$$ and $$G \subseteq A$$. Therefore $$x \in \text{Int}(A)$$ so $$\bigcup \{G : G \subseteq A \text{ and } G \text{ is open}\} \subseteq \text{Int}(A)$$. Thus $$\bigcup \{G : G \subseteq A \text{ and } G \text{ is open}\} = \text{Int}(A)$$.

71.237.205.56 01:49, March 23, 2011 (UTC)Ian_mi