Let $(X, \mathfrak{T})$ be a topological space

Let $V, U \in \mathfrak{T}$, and suppose that $V \subseteq U$

Then when it is true that $V \subseteq \overline V \subseteq U$, where $\overline V$ is the closure of $V$?

(Properties of the underlying topology for example)

This doesn't seem to be true in general, consider the particular point topology

$(X, \mathfrak{T}_p)$

Let $V, U \in \mathfrak{T}_p$, $V \subseteq U$ , then $p \in V, U$, but $\overline V = X$

  • 4
    $\begingroup$ This is clearly not true in general indeed (the trivial example $U=V$ shows so). Now, what kind of condition/statement would you expect? Tautologically speaking, $\bar{V}\subseteq U$ is necessary and sufficient... $\endgroup$ – Clement C. Jun 11 '16 at 20:21
  • $\begingroup$ Well, it's true when open=closed, at least. But otherwise, it won't hold in general. $\endgroup$ – Hayden Jun 11 '16 at 20:27
  • $\begingroup$ One of the cases that this is going to be true is when $U$ is a closed set. $\endgroup$ – Sina Jun 11 '16 at 20:41
  • $\begingroup$ Another case when it's true is when you're in a metric space and diam($U$) < diam($V$). $\endgroup$ – Benjamin Gadoua Jun 11 '16 at 21:36

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