If you have a locally compact Hausdorff space, then every point has a
local basis of compact neighbourhoods. I.e. for any point there is a
compact set which contains an open set which also contains the point. Now,
if every compact is finite, than there are finite open sets. Take such a
set E, which contains p. As the space is Hausdorff for each x in E which
is not p one can find an open set U_x which contains p and does not
contain x. Then the intersection of E and all U_x is just {p}, and as an
intersection of finitely many open sets it should be open. I.e. the
topology is discrete.

I want to remark that even for a Hausdorff topological group, unless the
unity (and thus all other points) have a countable neighbourhood basis,
the statement does not hold. I.e. one can construct a Hausdorff
topological group in which only finite sets are compact, but it is not a
discrete.

> Dear All
> please help me with the proof of this point.
> the group $G$ is discrete if the Haar measure $\mu$ is discrete.
> In the proof of this point the writer has said if $G$ is not
> discrete then it contains a compact set that is infinite. Is
> there any one who can help me with the above line.
> $G$ is also a locally compact Haussdorff space.
> Thanks
> Fatima
>
>
>