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Fix a closed set $V$ in a topo space $X$. I've already proved the converse. These are equivalent conditions for any open set in $V$ to be everywhere dense in $V$.

What I'm trying to prove:

If for all open $U_1, U_2$ in $X$ such that $U_i \cap V \neq \varnothing \implies V \cap U_1 \cap U_2 \neq \varnothing$, then for any open $U$ in $V$, we have $\overline{U} = V$, where I'm assuming the closure is taken in $V$.

Attempt:

Suppose that for some open subset of $V$, $U = V \cap U'$ ($U'$ open in $X$), $\overline{U} \neq V$. Then we'd have that $V \setminus \overline{U} \neq \varnothing$, but $V \setminus \overline{U} = V \cap (X \setminus \overline{U}_{\text{ in }X})$, and I got lost somewhere in the symbols :)

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I think a direct proof is easier here. Suppose $U$ is a nonempty open subset of $V$, and $U=V\cap U'$ for some $U'$ open in $X$. To prove $\overline{U}=V$ it suffices to show every point in $V$ is a limit point of $U$. To that effect, let $x\in V$ and $U''$ an open set containing $x$. Then $U' \cap V \neq \varnothing$ (since $U$ is nonempty), and $U'' \cap V \neq \varnothing$ (since $x \in U'' \cap V$). Then by hypothesis, $V \cap U' \cap U'' \neq \varnothing$, i.e., $U \cap U'' \neq \varnothing$. Hence $x$ is a limit point of $U$.

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  • $\begingroup$ Thanks! Your rating does not reflect well your genius. $\endgroup$ – Shine On You Crazy Diamond Dec 31 '15 at 4:49
  • $\begingroup$ Glad I could help. I'm a relatively new member here. $\endgroup$ – kccu Dec 31 '15 at 4:51
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Of course we assume $U\neq \varnothing$.

Seeking a contradiction, suppose that $\bar U\neq V$. Then there exists a proper closed subset $C$ of $V$ with $U\subseteq C\subseteq V$. Note that $U^\prime=V\setminus C$ is open in V and that $U^\prime\neq\varnothing$. That is, $U\cap V\neq\varnothing$ and $U^\prime\cap V\neq\varnothing$. It follows that $U^\prime\cap U\neq\varnothing$, a contradiction since $U^\prime=V\setminus C\subseteq V\setminus U$. Hence $\bar U=V$.

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