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.

Some Useful Links

• TIFR 2014 Problem 22
• Fundamental Theorem of Abelian Groups - Video