# Complete-Not Compact (TIFR 2013 problem 23)

Question:

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.