Prove: If p has a neighborhood contained in A, then p is in the interior of A.

[Hodgey8806]

Homework Statement

If p has a neighborhood contained in A, then p is in the interior of A.

Homework Equations

Int A = \bigcup{C\subseteqX:C\subseteqA and C is open in X}
By the books definition, a neighborhood is open.

The Attempt at a Solution

Let C'\subseteqA be a neighborhood of p contained in A
Then p\in\bigcup{C\subseteqX:C\subseteqA and C is open in X}
p\inInt A

Seems too simple, but I'm going through the book's definition of "neighborhood of p"--and open subset of X.

 

[voko]

You are thinking along the right lines. But C' as you chose it leads you nowhere. You have not proved that p is in any of C.

 

[Hodgey8806]

Should it state something like:
Let p\inA,
There exists an open set C'\subseteqA s.t. p\inC'
Thus it's in the union of all open sets.

Would that fix the issue? Thanks!

 

[voko]

That works.

 

[Hodgey8806]

Excellent! Thank you for the help!

 

[voko]

You just need to put things in a logical sequence without any gaps. First there is a neighborhood V. It contains some open set C'. Which must necessarily be one of all those open sets. So p is in the union. Q. e. d. If you are not explicit about every small detail, your professor might become unhappy :)

 

[Hodgey8806]

Lol, my professor is a stickler--but she is an excellent teacher! So she's only a stickler because she has covered the material. So hopefully I catch these things before a test---though we have a quiz coming up!

If you have time, do you mind helping me with my other post? I'm not quite sure I can stop where I did--though it makes sense to me!