# Thread: Y is a subspace of a X which is a topology

1. ## Y is a subspace of a X which is a topology

Let Y be a subspace of X. If A is closed in Y and Y is closed in X, then A is closed in X.

First are there 2 cases to consider? (1) A is a proper subset of Y (2) $A=Y\cap U$ or does this not matter?

Anyways, all I have is this:

Since A is closed in Y, $Y-A$ is open in Y. Likewise, $X-Y$ is open in X.

I am not sure what to do next.

2. ## Re: Y is a subspace of a X which is a topology

Originally Posted by dwsmith
Let Y be a subspace of X. If A is closed in Y and Y is closed in X, then A is closed in X.

First are there 2 cases to consider? (1) A is a proper subset of Y (2) $A=Y\cap U$ or does this not matter?

Anyways, all I have is this:

Since A is closed in Y, $Y-A$ is open in Y. Likewise, $X-Y$ is open in X.

I am not sure what to do next.
By definition that $A$ is closed in $Y$ you have that $A=Y\cap C$ for some closed subset $C$ of $X$....so.

3. ## Re: Y is a subspace of a X which is a topology

Originally Posted by Drexel28
By definition that $A$ is closed in $Y$ you have that $A=Y\cap C$ for some closed subset $C$ of $X$....so.
I don't know. All I can think of is to do $Y-(Y\cap A)$ then but I don't see how that will help.

4. ## Re: Y is a subspace of a X which is a topology

Ok, I think I have it.

$A=Y\cap U$ where U is some closed set in X. Now, A is the intersection of two closed sets in X so A is closed in X.

5. ## Re: Y is a subspace of a X which is a topology

Originally Posted by dwsmith
Ok, I think I have it.

$A=Y\cap U$ where U is some closed set in X. Now, A is the intersection of two closed sets in X so A is closed in X.
Correct.