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I would like to prove that a subset $A$ of a topological space $X$ is locally closed if and only if there is an open set $U$ and a closed set $F$ such that $A = U \cap F$.

My definition of $A$ being locally closed is that for all $x \in A$ there exists a neighbourhood $V_x$ of $x$, such that $A\cap V_x$ is closed in $V_x$ with regards to the subspace topology.

I am having some troubles proving this equivalence, because I do not really understand the definitions in the first place. As far as I know, a set $M$ is said to be closed in $V_x$, if it is a complement of an element of the subspace topology, i.e. if there exists an open set $U \subset X$, such that $M = (V_x \cap U)^C$. Is this definition correct? I am pretty sure it is, but when I use this then I reach some contradictions so I am pretty confused at the moment.

Any help would be greatly appreciated!

Jack4t3
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  • Use that $F$ is closed in a subspace $Y$ of $X$ iff there is some closed $F'$ of $X$ such that $F' \cap Y =F$. – Henno Brandsma Oct 15 '17 at 21:02
  • @HennoBrandsma how can I proof that by using the definition involving a complement of an open set in $Y$? – Jack4t3 Oct 15 '17 at 21:10
  • $Y \setminus F = O' \cap Y$ for some open $O'$ and then use $X\setminus O'$ as $F$. – Henno Brandsma Oct 15 '17 at 21:12
  • @HennoBrandsma Oh, I get it now. The whole time I was using complements in $X$ and not in $Y$, thanks a lot! – Jack4t3 Oct 15 '17 at 21:13
  • Clearly, if $A = U\cap F$, then $A$ is locally closed. But the other direction is false, I guess. – amsmath Oct 15 '17 at 21:13
  • Where did you see this definition of locally closed? It seems weird ! – Red shoes Oct 15 '17 at 22:41
  • Following up on the comment by Henno Brandsma, you can also say that if $F\subset Y\subset X$ then $ F$ is closed in $Y$ iff $F=Y\cap Cl_X(F).$ – DanielWainfleet Oct 16 '17 at 04:29
  • @Redshoes : I don't know where the OP came across the above definition, but here is one important context in which they come up:in algebraic geometry, quasiprojective varieties are locally closed with respect to Zariski topology. – user676464327 Mar 09 '20 at 02:47

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For each $x\in A$, pick $U_x$ open such that $A\cap U_x$ is closed in $U_x$. Let $(a_i)$ be a net in $A$ convergent to some $a\in U_x$. Then by definition, eventually $a_i\in U_x$. Then we can just take the final subnet of $(a_i)$ to witness that $a\in \overline{A\cap U_x}$ and hence $a\in A\cap U_x$. This implies that $A\cap U_x=\overline A\cap U_x$.

It follows that $A=(\bigcup_x{U_x})\cap \overline{A}$. The other direction is easy.

tomasz
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