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

## 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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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.

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!

That works.

Excellent! Thank you for the help!

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 :)

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!