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
On
As mentioned in the comments, if by indiscrete topology you mean the trivial topology $\{0, X\}$, then it's not metrizable. It's not Hausdorff, for example. This said, any subset of real numbers is compact in this topology because we have only two open sets in this topology!
Secondly, there are many metric topologies on $\mathbb{R}$ that are compact, yet $\mathbb{R}$ is not compact with its usual topology. I'm sure Rudin hasn't written anything like 'compactness is independent of the metric chosen'. You misinterpreted it.
On the other hand, compactness is an intrinsic property. Namely, it's not like openness or closedness that depends on the subspace topology. I think this is what Rudin has written.
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 $x\ne y$ 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.