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Let $T$ be a set and $H\subset T$. I want to show that $$\overline{H}=T \setminus ({T\setminus H)}^{\mathrm{o}}.$$ I am able to understand this intuitively, from drawing a picture and generally just thinking about it, however I am unable to formulate a proof into words. I assume a standard proof would go about showing $\overline{H} \subset T \setminus ({T\setminus H)}^{\mathrm{o}}$ and the other way round, however I am unsure how the proper argument would be formulated using the concepts of neighbourhoods of points and so on. Help would be appreciated thanks.

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  • $\begingroup$ $T$ is not just a set, but a topological space. You could use that a point $p$ belongs to $\overline{H}$ if and only is every neighborhood of $p$ intersects $H$. I don't know which definition of $\overline{H}$ you use, but this is one of the equivalent definitions (you may need to prove that, if you use a different definition), and then using this definition should make the proof easy, as you suggest above. $\endgroup$ – Mirko May 4 '17 at 1:41
  • $\begingroup$ Proposition is false if T isn't the space. For example, let T = H = (0,1) subset R. $\endgroup$ – William Elliot May 4 '17 at 3:44
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Use int T\H = T - cl H.

Proof is a logical tour de force:
x in int T\H iff some open U nhood x with U subset T\H
iff not (for all open U nhood x, not U subset T\H)
iff not (for all open U nhood x, some y in U with y not in T\H)
iff not (for all open U nhood x, some y in U cap H)
iff not x in cl H.

int = interior; cl = closure.

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