Why set of integer under indiscrete topology is compact? > Why set of integer under indiscrete topology is compact? After looking this question I got surprised as In rudin I read in chapter 2 that compact space is independent of metric space chosen. As metric space is special case of topology above should hold. But where is I am making mistake .How to show that above is compact? Any Help will be appreciated

Compactness is independent of the metric chosen _among those that define the same topology_ (commonly termed _equivalent metrics_ ).

For instance, the closed interval $[0,1]$ is compact in the topology induced by the usual metric and all equivalent metrics. It is _not_ compact under the _discrete metric_ $\delta(x,y)=1$ if $x\
e y$ and $\delta(x,x)=0$: this metric induces a different topology, namely the discrete topology.

Any set endowed with the indiscrete topology is compact (for lack of infinite open covers), so this says nothing about compactness under different topologies.