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

>

>

>