- #1

Hodgey8806

- 145

- 3

## Homework Statement

I tried to state the whole thing, but here it is again:

Prove: A point is in the boundary of if and only if every neighborhood of it contains both a point of A and a point of X\A.

## Homework Equations

Boundary def: X\(Int A [itex]\cup[/itex] Ext A)

Def of Int A: [itex]\bigcup[/itex]{C[itex]\subseteq[/itex]X:C[itex]\subseteq[/itex]A and C is open in X}

Def of Ext A: X\[itex]\bar{A}[/itex] where [itex]\bar{A}[/itex] means closed (including boundary)

Def of [itex]\bar{A}[/itex]: [itex]\bigcap[/itex]{B[itex]\subseteq[/itex]X:B[itex]\supseteq[/itex]A and B is closed in X}

## The Attempt at a Solution

Please bare with me as this is a little messy. I used a bunch of set theory rules:

Let p[itex]\in[/itex][itex]\delta[/itex]A

p[itex]\in[/itex]X\(Int A [itex]\cup[/itex] Ext A)

1) p[itex]\in[/itex](X\Int A) [itex]\wedge[/itex] 2)p[itex]\in[/itex](X\Ext A)

Thus breaking each one down separately:

1)p[itex]\in[/itex]X\[itex]\bigcup[/itex]{C[itex]\subseteq[/itex]X:C[itex]\subseteq[/itex]A and C is open in X}

p[itex]\in[/itex][itex]\bigcap[/itex]{X\C:C[itex]\subseteq[/itex]A and C is open in X}[itex]\subseteq[/itex]=X\A--(I realize now this could have been stated without these set properties.)

2)p[itex]\in[/itex]X\(X\[itex]\bar{A}[/itex])

p[itex]\in[/itex][itex]\bar{A}[/itex][itex]\supseteq[/itex]A

Thus every neighborhood of p contains both a point of A and a point of X\A.

Q.E.D.