# Is empty set with usual metric space $(X=\mathbb{R},d)$ bounded?

Here R is set of real numbers and $d$ is distance metric i.e $|x-y|$. I wanted to find if empty set under given metric space is compact and we know by Heine Borel theorem that a set is compact iff it is closed and bounded. Empty set is closed because its complementary set i.e. set of real numbers $\mathbb{R}$ is open.Hence i just need to prove if the empty set is bounded.

The problem i am facing is , how can we define $|x-y|$ on empty set , to find $\exists m$ , $m \in \mathbb{R}$ , such that $|x-y|\leqslant m$.

• Write out the definition carefully with quantifiers. – Ted Shifrin Sep 15 '18 at 18:13
• A set $E$ is bounded iff there exists $m>0$ such that, for all $x,y \in E$ we have $d(x,y) \le m$. So what do you get when $E = \varnothing$? – GEdgar Sep 15 '18 at 18:15
• It is vacuously bounded. – Eduardo Longa Sep 15 '18 at 18:15
• @GEdgar I cannot have any pair $x,y$ such that$d(x,y)<=m$ , hence i cannot say it is bounded or not bounded. – user585269 Sep 15 '18 at 18:20
• Let $E = \emptyset$, and $m > 0$. Can you find a pair $x,y \in E$ such that $d(x,y) > m$? – Theoretical Economist Sep 15 '18 at 18:23

This is answered in comments already, but here's an honest-to-god StackExchange answer. Your question can be asked more succinctly as: "Is $\{\}$ bounded when considered as a subset of $\mathbb{R}$ with its standard metric?".

The answer is Yes. In fact, the empty set is bounded when thought of as a subset of any metric space. To see why, you have to write down a definition of boundedness. It should probably come out as this:

$X$ is bounded with respect to the metric $d$ if there exists a number $M \in \mathbb{R}$ such that, $d(a,b) < M$ for all $a, b \in X$.

Side note: I would have got equivalent definitions if I'd said $M \in \mathbb{N}$ or asked that $d(a, b) \le M$. You can check you understand the definition above by checking you understand why these definitions would give the same thing.

I claim that $\{\}$ is bounded with respect to any $d$. To prove this, I can set $M$ to be any number. After all, I just need to prove that if you choose $a, b \in \{\}$ then $d(a,b) < M$. But you can't choose find such $a$ or $b$ because the empty set is... empty! As such, the inequality is never actually checked.

As you correctly wrote in the question, you can now use the Heine-Borel theorem to show that the empty set is compact as a subset of $\mathbb{R}$. In fact, the Heine-Borel theorem is quite heavy machinery to use to prove the compactness of the empty set. If you think for a minute about the definition of compactness based on finite subsets of open covers, you should be able to prove directly that the empty set is compact as a subset of any topological space.

• To the proposer: If $S$ is any statement, phrase, clause, formula, etc., and if $X=\emptyset$ then the sentence $\forall a\in X\;(S)$ is true. Because its negation is $\exists a\in X\;(\neg S),$ which cannot be true unless some $a$ exists that belongs to $X.$ – DanielWainfleet Sep 16 '18 at 1:01