# Show If a point A is in the interior, then it has a neighborhood contained in A.

## Homework Statement

Let A be a topological space and let A$\subseteq$X be any subset.
Show: If a point A is in the interior, then it has a neighborhood contained in A.

## Homework Equations

Neighborhoods are defined to be open in my book.
Int(A) = $\bigcup${C$\subseteq$A and C is open in X}

## The Attempt at a Solution

Let p$\in$Int(A).
Then p$\in$$\bigcup${C$\subseteq$X:C$\subseteq$A and C is open in X}
So, $\exists$ an open set C' s.t. p$\in$C' and C$\subseteq$A
Q.E.D.

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Sounds good!

Thank you! Could you check my work backwards now?
I can say:

Let C'$\subseteq$A be open and non-empty
Let p$\in$C
Thus p$\in$$\bigcup${C$\subseteq$X:X$\subseteq$A and C is open in X}
Thus p$\in$Int A

That's correct. But you got to clean up your presentation. What's C' for example??

Thank you! Could you check my work backwards now?
I can say:

Let C'$\subseteq$A be open and non-empty
Let p$\in$C
Thus p$\in$$\bigcup${C$\subseteq$X:X$\subseteq$A and C is open in X}
Thus p$\in$Int A

C' was meant to be one set that has that element and
I meant to type let p$\in$C'
and I meant to write
p∈⋃{C⊆X:C⊆A and C is open in X}

Do I still need to clean up?

Ah ok. I guess it's ok now!

Thank you very much! If you have time, would you mind helping me with my other question? It's about the equivalence of a bounded set with a closed ball. I just want to check my proof one direction. Thanks again!