# TIFR 2013 problem 23 | Complete-Not Compact

Try this problem 23 from TIFR 2013 named - Complete not compact.

Question: TIFR 2013 problem 23

True/False?

Let $X$ be complete metric space such that distance between any two points is less than 1. Then $X$ is compact.

Hint:

What happens if you take discrete space?

Discussion:

Discrete metric space as we know it doesn't satisfy the distance < 1 condition. But we can make slight changes to serve our purpose.

In $X$ define $d(x,y)=\frac{1}{2}$ if $x\ne y$. Otherwise, $d(x,x)=0$.

$d$ is indeed a metric, and it gives the same discrete topology on $X$. Namely, every set is open because every singleton is open. And therefore every set is closed.

We want $X$ to be complete. If $x_n$ is a sequence in $X$ which is Cauchy, then taking $\epsilon=\frac{1}{4}$ in the definition of Cauchy sequence, we conclude that the sequence is eventually constant.

Since the tail of the sequence is constant, the sequence converges (to that constant).

This shows that $X$ is indeed Complete.

We don't want $X$ to be compact. Not all $X$ will serve that purpose, for example a finite set is always compact. We take a particular $X=\mathbb{R}$.

Since singleton sets are open, if we cover $X$ by all singleton sets, then that cover has no finite subcover. Hence $X$ is not compact.

Therefore the given statement is False.