- #1

sa1988

- 222

- 23

## Homework Statement

Let ##A## be a subspace of a topological space ##X##, and let ##B\subset A##.

Show that ##B## is closed if and only if there exists a closed subset ##C \subset X## such that ##B = C \cap A##

## Homework Equations

## The Attempt at a Solution

So I've started by just drawing the two cases in which ##B## is said to exist. It doesn't really help with much but it's a way for me to see clearly what I'm supposed to be working with.

I believe the right hand image demonstrates the 'only if' part of the statement, where we have ##B \subset A \subset X## which is open by definition. (Since the complement of ##A## is ##X## which is open because it's a topological space, thus ##A## is closed, and then the same applies to show that ##B## is open since the complement of ##B## is closed).

So ##B## can only possibly be closed in the other case, shown in the left hand image, which is the 'if' part. From this we can see that ##\exists## neighbourhoods ##N## of ##n \in B## where ##N \subset B## and ##N \subset C##, so ##N \subset B \cap C ##

But if ##N \subset B \cap C ##, this means ##B## contains elements with neighbourhoods that go beyond its boundaries into the

*closed*set ##C##.

Hence ##B## must be closed too, in the case where ##B = A \cap C##, and only this case.

Is this proof heading in the right direction? I fear I may have been a bit too hand-wavy with the 'only if' part.

Thanks.