- #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.