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How can a finite set E in R² not be open? Isn't every point of E an interior point of E (since the point will be contained in its own neighbourhood)? And if every point in E is an interior point, then every point in E is also interior in R², since E is a finite subset of R². Then, E should be open in R².

Parul
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    You might want to look up the definition of "neighborhood". A neighborhood has to be a union of open disks. – J126 Sep 20 '13 at 18:21
  • A point $x$ is in the interior of $E$ if you can find a small enough ball centered at $x$ which lies entirely in E. For a non-empty finite set, you cannot find such balls. –  Sep 20 '13 at 18:25
  • Your tag "metric spaces" implies that you are using a metric topology. If this is so, then please make it explicit. What is open and what is closed is determined by the topology alone. – Fly by Night Sep 20 '13 at 18:32
  • Actually, I must disagree with @Joe here: A neightbourhood of a point must contain an open disk centered at that point, but that does not make it a union of open disks. For a simple counterexample, take a closed disk of positive radius centered at the point in question. – Harald Hanche-Olsen Sep 20 '13 at 18:46
  • Set $E$ is open in $E$ for the reason you state in your first question. If we just say $E$ is open, however, we would mean $E$ is open in $\mathbb R^2$. That is false in this case, so your assertion "every point in $E$ is also interior in $\mathbb R^2$" is false. – GEdgar Sep 20 '13 at 18:52

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A finite set is never open, because it cannot contain any open balls (so therefore can't be a union of them).

Matt Rigby
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    Some triviality: A non-empty finite set... –  Sep 20 '13 at 18:26
  • @Parul : In case you don't understand Ustun's comment, the empty set is finite, open, and closed. This may seem trivial, but it means a finite set in your problem actually can be open. – Stefan Smith Sep 21 '13 at 03:13