Is $ \{ \infty \} $ open or closed in $ \overline{ \mathbb R}$ ? Here $\overline{ \mathbb R} = \mathbb R \cup \{ \infty \} \cup \{ -\infty \} $.
Is $ \{ \infty \} $ open or closed in $ \overline{ \mathbb R}$?
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0Although the answers given are more insightful, notice also that you should know the answer before you can prove it: $\{\infty\}$ is finite, hence compact, hence closed. – 2012-05-18
2 Answers
It's closed in the usual topology.
Your question seems to assume it's either open or closed, but most sets are neither open nor closed.
We want to define such things as $\lim\limits_{x\to3} f(x) = +\infty$. That has to mean that for every open set $A$ containing $+\infty$, there is some open set $B$ containing $3$ small enough so that if $x\in B$ and $x\ne 3$, then $f(x)\in A$. It is from such considerations as that that the definitions of open and closed subsets of $\overline{\mathbb{R}}$ are derived.
A set is closed precisely if its complement is open. The set $(-a,a)$ is open. The union of every set of open sets is open. The union of all sets of the form $(-a,a)$ is all of $\mathbb{R}$, and clearly excludes both $+\infty$ and $-\infty$. The set $\{-\infty\}\cup(-\infty,b)$ is open. So we have $ \big(\{-\infty\}\cup(-\infty,b)\big)\cup\big( \bigcup_{a>0} (-a,a) \big) $ is open. Its complement is $\{+\infty\}$. So that set is closed.
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0@JimConant I liked your "A set is not like a door..." – 2012-12-07
Another way to look at it is to consider the complement $\{-\infty\} \cup \mathbb{R}$ which is a neighborhood of all of its elements, i.e. it is open. Hence its complement, $\{\infty\}$, is closed.