Proposition 2.33.b

Proposition 2.33.b: $$Cl(A) = Int(A) \cup Bd(A)$$.

Ian_mi
Let $$x \in \text{Bd}(A)$$. If $$x \in A$$ then $$x \in \text{Cl}(A)$$ since $$A \subseteq \text{Cl}(A)$$. Otherwise every open set containing $$x$$ contains a point of $$A$$ which is distinct from $$x$$ since $$x \in A^c$$. Therefore $$x$$ is a limit point of $$A$$ so $$x \in \text{Cl}(A)$$. Thus $$\text{Bd}(A) \subseteq \text{Cl}(A)$$. $$\text{Int}(A) \subseteq A \subseteq \text{Cl}(A)$$ so $$\text{Int}(A) \cup \text{Bd}(A) \subseteq \text{Cl}(A)$$.

Let $$x \in A$$. If $$x \notin \text{Int}(A)$$ then every open set containing $$x$$ contains a point of $$A^c$$. Since $$x \in A$$ every open set containing $$x$$ also contains a point of $$A$$, hence $$x \in \text{Bd}(A)$$. Thus $$A \subseteq \text{Int}(A) \cup \text{Bd}(A)$$. Let $$x \in A^c$$ be a limit point of $$A$$. Then every open set containing $$x$$ contains a point of $$A$$ since $$x$$ is a limit point of $$A$$ and of $$A^c$$ since $$x \in A^c$$. Thus $$x$$ is a boundary point of $$A$$ so $$A^c \cap A' \subseteq \text{Bd}(A)$$. Hence $$\text{Cl}(A) = A \cup A' = A \cup ((A \cup A^c) \cap A') = A \cup (A \cap A') \cup (A^c \cap A') = A \cup (A^c \cap A') \subseteq \text{Int}(A) \cup \text{Bd}(A)$$.

Ian mi 02:36, March 24, 2011 (UTC)Ian_mi