Let the metric space $X = [0,1) \cup (2,3]$ with $d(x,y) = |x-y|$. Prove that $[0,1)$ is a closed subset. I know that it is a subset, because the set contains every limit point in the metric, but isn't $(2,3]$ closed as well? Wouldnt this be a violation of a closed set being closed if the complement is open?

  • $\begingroup$ They're both closed and open. $\endgroup$ – Daniel Fischer Oct 7 '13 at 23:38
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    $\begingroup$ A set can be both open and closed. $\endgroup$ – Pedro Tamaroff Oct 7 '13 at 23:38
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    $\begingroup$ "Clopen" is a term that is actually used sometimes. $\endgroup$ – MartianInvader Oct 7 '13 at 23:45
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    $\begingroup$ @MartianInvader But it is awful! =O $\endgroup$ – Pedro Tamaroff Oct 7 '13 at 23:50

The terminology is perhaps unfortunate. In topology, open and closed are not antonyms. A set can be

  1. open and not closed,
  2. closed and not open,
  3. open and closed, or
  4. neither open nor closed.

The empty set and the whole space are always both open and closed. You may have noticed that no other set in $\Bbb R$, $\Bbb R^2$, or $\Bbb R^3$ is both open and closed: this is because those spaces are "connected".

Edit: The last sentence takes things very badly out of historical context. Take with a grain of salt.


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