# Oddity about a closed set

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?

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

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".