Proposition 2.33.c

Proposition 2.33.c: $$Int(A) = A \setminus Bd(A)$$.

Ian_mi
By proposition 2.33b, $$\text{Int}(A) \cup \text{Bd}(A) = \text{Cl}(A) = A \cup A' = A \cup (A' \cap A^c)$$ so $$\text{Int}(A) \setminus \text{Bd}(A) = (A \setminus \text{Bd}(A)) \cup ((A' \cap A^c) \setminus \text{Bd}(A))$$ so it suffices to show that $$\text{Int}(A) \setminus \text{Bd}(A) = \text{Int}(A)$$ and $$(A' \cap A^c) \setminus \text{Bd}(A) = \emptyset$$.

Let $$x \in \text{Int}(A)$$. Then there exists an open set $$G$$ containing $$x$$ such that $$G \subseteq A$$. Therefore $$G$$ contains no point of $$A^c$$ so $$x$$ is not a boundary point of $$A$$. Thus $$\text{Int}(A) \setminus \text{Bd}(A) = \text{Int}(A)$$.

Let $$x \in A' \cap A^c$$. Then every open set containing $$x$$ contains a point of $$A$$ since $$x$$ is a limit point of $$A$$ and a point of $$A^c$$ since $$x \in A^c$$. Therefore $$x$$ is a boundary point of $$A$$ so $$A' \cap A^c \setminus \text{Bd}(A) = \emptyset$$. This proves the proposition.

Ian mi 03:19, March 24, 2011 (UTC)Ian_mi

Comments
Shouldn't this proposition read $$Int(A) = Cl(A)\setminus Bd(A)$$ ?

baruch_shahi 00:26, March 26, 2011 (UTC -5)

Maybe, but it is true as well.

Ian mi 14:54, March 26, 2011 (UTC)Ian_mi