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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

2 Answers 2

up vote 3 down vote accepted

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..." –  Nikita Evseev 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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