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I have the following problem:

Let $(X,\tau)$ be a topological space and $(Y,\tau_y)$ be its subspace, that is $Y\subset X$ and $\tau_y=\{\;A\;\cap\;Y\mid A\in\tau\;\}$. Let also $F\subset Y$ and $H\subset X$, where $H$ is closed in $X$. Show that $F=Y\;\cap\;H$ if and only if $F$ is closed in $Y$.

I solved this problem by assuming that if $F$ is closed then $F=Y\;\cap\;H$, where $H$ is closed in $X$, but I got stuck proving this in the other direction, that is showing that if $F=Y\;\cap\;H$ then $F$ is closed in $Y$.

I tried to do something like this:

Assumption: $F=Y\;\cap\;H$, where $H$ is closed in $X$. If $H$ is closed in $X$ then $X\;\setminus\;H$ is open in $X$ and then $X\;\setminus\;H \in \tau$ which means that $B = (X\;\setminus\;H)\;\cap\;Y\in \tau_y$.

How should I proceed now? Should I investigate the set $B$ now more or what?

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For any set $A \subseteq X$ we have $A = (A \cap H) \sqcup (A \cap (X \setminus H))$. If $F = Y \cap H$ then in particular, we have

$$ Y = (Y \cap H) \sqcup (Y \cap (X \setminus H)) = F \sqcup (Y \cap (X \setminus H)) $$

so the complement of $F$ in $Y$ is $Y \cap (X \setminus H)$ which is, by definition, an open subset of $Y$ (as $X \setminus H$ is an open subset of $X$).

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  • $\begingroup$ Thank you for your help! =) $\endgroup$ – jjepsuomi Jan 25 '16 at 12:44
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The statement of the problem is somewhat imprecise and it should be

A subset $F$ of $Y$ is closed in $Y$ if and only if $F=Y\cap H$, for some closed subset $H$ of $X$

($\Rightarrow$) Suppose $F$ is closed in $Y$; then $Y\setminus F$ is open in $Y$, so $Y\setminus F=Y\cap A$, for some open subset $A$ of $X$. Set $H=X\setminus A$, which is closed. Let's show that $F=Y\cap H$.

Let $x\in F$; then $x\notin Y\setminus F$, so $x\notin Y\cap A$ and therefore $x\notin A$. So $x\in H$.

Let $x\in Y\cap H$; then $x\in H$, so $x\notin A$ and therefore $x\notin Y\setminus F$. Hence $x\in F$.

($\Leftarrow$) Let $F=Y\cap H$, for $H$ closed in $X$. Then, easily, $Y\setminus F=Y\cap(X\setminus H)$, so $Y\setminus F$ is open in $Y$.

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