# 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$\subseteq$X:C$\subseteq$A and C is open in X}
By the books definition, a neighborhood is open.

## The Attempt at a Solution

Let C'$\subseteq$A be a neighborhood of p contained in A
Then p$\in$$\bigcup${C$\subseteq$X:C$\subseteq$A and C is open in X}
p$\in$Int A

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

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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$\in$A,
There exists an open set C'$\subseteq$A s.t. p$\in$C'
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!