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I was looking at this answer, because I need to answer the same question:

Show that $F\subset Y$ is closed in $Y$ iff $F=Y\;\cap\;H$ where $H\subset X$ is closed in $X$.

The second answer (given by @egreg) states:

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

But this would only be the case if $A \subset Y $, wouldn't it? So wouldn't that be incorrect?

I really like the format of his proof but I can't figure out how to correct it if it's incorrect.

Could somebody shed some insight?

Community
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Lorkus
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  • As pointed out in the answer of @egreg, the question that you have linked has an incorrect statement. You have simply copied that incorrect statement. You should look more carefully at the answer of egreg, who supplies the corrected statement. – Lee Mosher Sep 25 '16 at 15:40
  • @LeeMosher I edited the statement, but that doesn't really change my question. In the case that $Y \subset A$, $x$ could be an element of $A$. So $x \notin A$ wouldn't hold, right? – Lorkus Sep 25 '16 at 16:04

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$ F\subset Y $ closed in $Y$ $\iff$ $Y-F$ is open in $Y$ $\iff$ $\exists$ $O$ open in $X$; $ O \cap Y$ = $Y-F$ $\iff$ $\exists$ $H= X-O$ closed in $X$; $ H\cap Y$= $F$

outlaw
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