# Is $\{ \infty \}$ open or closed in $\overline{ \mathbb R}$?

Is $\{ \infty \}$ open or closed in $\overline{ \mathbb R}$ ? Here $\overline{ \mathbb R} = \mathbb R \cup \{ \infty \} \cup \{ -\infty \}$.

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I would prefer the notation $+ \infty$, however it is a closed point. $\overline{\mathbb R}$ is homeomorphic to a closed compact interval. – Andrea May 18 '12 at 17:28
@Andrea Thank you. – Misaj May 18 '12 at 17:30
Although 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. – user12014 May 18 '12 at 19:33

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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+1: "Your question seems to assume it's either open or closed, but most sets are neither open nor closed." There's an old saying. A set is not like a door. It can be open, closed, both or neither. – Grumpy Parsnip May 18 '12 at 17:49
@JimConant I liked your "A set is not like a door..." – nikita2 Dec 7 '12 at 4:47

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.

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