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

[Hodgey8806]

Homework Statement

Let A be a topological space and let A\subseteqX 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\subseteqA and C is open in X}

The Attempt at a Solution

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

 

[micromass]

Sounds good!

 

[Hodgey8806]

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

Let C'\subseteqA be open and non-empty
Let p\inC
Thus p\in\bigcup{C\subseteqX:X\subseteqA and C is open in X}
Thus p\inInt A

 

[micromass]

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'\subseteqA be open and non-empty
Let p\inC
Thus p\in\bigcup{C\subseteqX:X\subseteqA and C is open in X}
Thus p\inInt A

 

[Hodgey8806]

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

Do I still need to clean up?

 

[micromass]

Ah ok. I guess it's ok now!

 

[Hodgey8806]

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!