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I need to show that if $X$ is a topological space, and $Y$ a subspace of $X$. Then $E$ is closed in the subspace topology on $Y$ iff $E = Y \cap F$, where $F$ is closed in $X$.

I have the following proof, and need to know if this is complete:

Suppose first $E$ is closed in $Y$. Then $Y - E$ is open so that $Y - E = Y \cap U$, where $U$ is open in $X$. Then $Y \cap (X - U) = Y \cap X - Y \cap U = Y - (Y - E) = E$. Hence $E$ is the intersection of a closed set in $X$ with $Y$.

Conversly, suppose $E = Y \cap F$, with $F$ closed in $X$. Now $X - F$ is open so $(X - F) \cap Y$ is open in $Y$. We get that $Y \cap (X - F) = Y - (Y \cap F) = Y - E.$ Hence $E$ is a closed set in Y.

My questions concerns the following set equality: $Y \cap (X - U) = Y \cap X - Y \cap U$. Is this true? And can someone please provide proof.

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The property you are asking about is true.

Write $A-B$ as $A \cap B^c$. It should then be easy to show, using elementary properties of $\cup$ and $\cap$, that $$ Y \cap (X-U) = Y \cap U^c$$ and $$ Y \cap X - Y \cap U = Y \cap U^c $$

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