From 'Treatise on Analysis (vol 2)':

(12.2.3) Let $L$ be a subset of a topological space $E$. Then the following properties are equivalent: (a) For each $x\in L$ there exists a neighborhood $V$ of $x$ in $E$ such that $L\cap V$ is closed in $V$;
(b) $L$ is an open subset of the subspace $\bar{L}$ (the closure of $L$ in $E$);
(c) $L$ is the intersection of an open subset and a closed subset of $E$.

The book proves this with b $\Rightarrow$ c, c $\Rightarrow$ a, a $\Rightarrow$ b. The last one however (a to b) is what I'm stuck on. I am following their approach exactly for this particular proposition.

Their proof for a $\Rightarrow$ b goes:

For each $x \in L$, we have $V \cap L = V \cap \bar{L}$, because $V \cap L$ is closed in $V$; this shows that in the subspace $\bar{L}$ the point $x$ is an interior point of $L$, and therefore $L$ is open in $\bar{L}$.

So the part I'm stuck on is showing that $\bar{L} \cap V = L \cap V$. I wasn't sure whether the closure in that expression was w.r.t. subspace $V$ or the space $E$. But assuming either leaves me stuck.

Hints are more welcome, so that some work is left for me to learn from.



The closure operator of $V$ goes as $L\mapsto \bar L\cap V$, no? Then the property of (a) says that $L\cap V$ is closed in $V$.

  • $\begingroup$ This would work but I would first have to prove that $Cl_v(L\cap V) = \bar{L} \cap V$. For the definition of $Cl_E(L)$, I've been using the intersection of all closed sets in $E$ containing $L$ $\endgroup$ – Shine On You Crazy Diamond Sep 17 '12 at 23:58

Let $x\in V\cap \bar L$. We want to show that $x\in V\cap L$.

We assume that $x\notin V\cap L$. Since $V\setminus (V\cap L)$ is open in $V$, we have $x\in O\cap V$, where $O$ is open in $E$ and $O\cap V\subset V\setminus (V\cap L)=V\setminus L$. Taking the intersection with $L$ on both sides, we get $O\cap V\cap L\subset V\setminus L \cap L$ hence $O\cap V\cap L$ is empty, which contradicts the fact that $x\in V\cap \bar L$.

The other inclusion is obvious.

  • $\begingroup$ I wrote the proof of the inclusion $V\cap \overline{L}\subset V\cap L$ by contradiction, assuming their is an element in the latter but not in the former. Then I use the definition of trace topology (topology on a subset) and the hypothesis. By the way, I have rewritten a bit the proof, since it was not very clear. $\endgroup$ – Davide Giraudo Sep 18 '12 at 15:11
  • $\begingroup$ I'll go through it in a few days and re-adjust my votes (got to travel home). Any idea how to patch my below proof? It seemed like it should work, then doesn't quite. $\endgroup$ – Shine On You Crazy Diamond Sep 18 '12 at 16:11

$\overline{L}$ stands for closure of L below: To show $\overline{L}\cap V=L\cap V$:

Let $x\in\overline{L}\cap V$. Then $x\in \overline{L}\cap \overline{V}=\overline{L\cap V }=\overline{K\cap V }$ for some closed set $K$ ( as $L\cap V$ is closed in $V$) $=\overline{K}\cap\overline{V}=K\cap\overline{V}$. Hence, as $x\in V$, $x\in K\cap V = L\cap V$.


We can use the fact that neighborhoods of points in the boundary $\partial L$ of a subset $L$ of a topological space $X$ have a non-empty intersection with $L$.

If $L \cap V$ is closed in $V$, then $\bar{L} \cap V \subseteq L \cap V$.

Proof: Let $x\in \bar{L}\cap V$. Assume $x \notin L$. Then $x$ must be in $\partial L$. Since $L \cap V$ is closed in $V$, $V \setminus (V \cap L)$ = $O \cap V$ for some open set $O$ of $X$. But since $O$ contains $x$ which is in the boundary of $L$, $O \cap L$ is nonempty. So $O \cap V \cap L$ = $(V \setminus L) \cap L = \emptyset$


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