Theorem 2.37

Theorem 2.37: For any subset $$A$$ of a topological space $$X$$, $$Int(A),Bd(A),Ext(A)$$ partition the space into three disjoint sets. That is, no two of $$Int(A),Bd(A),Ext(A)$$ have a nonempty intersection (they are pairwise disjoint) and $$X= Int(A)\cup Bd(A)\cup Ext(A)$$.

Olmana
We've seen partitions before; $$A$$ and $$A^c$$ form a partition of $$X$$.

Let's restate some definitions in a way that makes their relationships clear:

Call $$x$$ an interior point of $$A$$ if there exists an open set $$U$$ containing $$x$$ such that $$U \cap A^c = \emptyset$$.

Call $$x$$ an exterior point of $$A$$ if there exists an open set $$U$$ containing $$x$$ such that $$U \cap A = \emptyset$$.

Call $$x$$ an boundary point of $$A$$ if every open set $$U$$ containing $$x$$ contains a point in $$A$$ and a point in $$A^c$$.

Now the definition of $$Bd(A)$$ is the logical negation of the definition of $$Int(A) \cup Ext(A)$$; therefore:

$$(Bd(A))^c = Int(A) \cup Ext(A)$$

In other words, $$Bd(A)$$ and $$Int(A) \cup Ext(A)$$ form a partition of $$X$$.

From the definitions, interior points are in $$A$$ and exterior points are in $$A^c$$. Since $$A$$ and $$A^c$$ are disjoint, $$Int(A)$$ and $$Ext(A)$$ must be disjoint. In other words,

$$Int(A)$$ and $$Ext(A)$$ form a partition of $$Int(A) \cup Ext(A)$$

Since $$Bd(A)$$ and $$Int(A) \cup Ext(A)$$ form a partition of $$X$$, and $$Int(A)$$ and $$Ext(A)$$ form a partition of $$Int(A) \cup Ext(A)$$, the three sets together, $$Bd(A)$$, $$Int(A)$$ and $$Ext(A)$$, form a partition of $$X$$.

Olomana 17:50, March 20, 2011 (UTC)

Ian_mi
By propositions 2.33b, 2.33d, 2.33g, and 2.36, $$\text{Int}(A)^c = \text{Cl}(A^c) =\text{Int}(A^c) \cup \text{Bd}(A^c) = \text{Ext}(A) \cup \text{Bd}(A)$$. Therefore $$\text{Int}(A)$$ and $$\text{Ext}(A) \cup \text{Bd}(A)$$ are disjoint and $$X = \text{Int}(A) \cup \text{Int}(A)^c = \text{Int}(A) \cup \text{Ext}(A) \cup \text{Bd}(A)$$. $$\text{Ext}(A) = \text{Int}(A^c)$$ and $$\text{Bd}(A) = \text{Bd}(A^c)$$ so $$\text{Ext}(A)$$ and $$\text{Bd}(A)$$ are also disjoint. This proves the theorem.

Ian mi 03:45, March 24, 2011 (UTC)Ian_mi