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?
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They're both closed and open. – Daniel Fischer Oct 07 '13 at 23:38
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1A set can be both open and closed. – Pedro Oct 07 '13 at 23:38
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1"Clopen" is a term that is actually used sometimes. – MartianInvader Oct 07 '13 at 23:45
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1@MartianInvader But it is awful! =O – Pedro Oct 07 '13 at 23:50
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The terminology is perhaps unfortunate. In topology, open and closed are not antonyms. A set can be
- open and not closed,
- closed and not open,
- open and closed, or
- 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.
dfeuer
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