Proof: Discreteness of Topological Groups

[pivoxa15]

Homework Statement

Prove: a topological group is discrete if the singleton containing the identity is an open set.

The statement is in here http://en.wikipedia.org/wiki/Discrete_group

The Attempt at a Solution

Is that because if you multiply the identity with any element in the group, you get a new element with nothing surrounding it because it's like you can also multiply the area around the identity to the new position. In other words mapping open sets to open set?

f is cts => open sets are mapped to open sest in a topological group.

 

[StatusX]

It's not true that continuous functions map open sets to open sets. For example, the map f(x)=x^2 maps the open set (-1,1) to the non-open set [0,1). Rather, the inverse image of an open set under a continuous map is an open set. So look at the preimage of the identity under certain maps, say, f_g:G->G, where f_g(h)=gh.

 

[pivoxa15]

It's not true that continuous functions map open sets to open sets. For example, the map f(x)=x^2 maps the open set (-1,1) to the non-open set [0,1). Rather, the inverse image of an open set under a continuous map is an open set. So look at the preimage of the identity under certain maps, say, f_g:G->G, where f_g(h)=gh.

I realized that after posting. I may have remembered under some circumstances, open sets are mapped to open sets. What is this circumstance?

 

[morphism]

Maps that do that are called 'open maps'. Just like continuous maps don't have to be open, open maps don't have to be continuous either.

 

[pivoxa15]

Maps that do that are called 'open maps'. Just like continuous maps don't have to be open, open maps don't have to be continuous either.

Under what circumstances are open maps continuous and vice versa?

 

[pivoxa15]

So look at the preimage of the identity under certain maps, say, f_g:G->G, where f_g(h)=gh.

But do all f_g have to be continuous maps? It isn't implied from the axioms for a topological group.

 

[morphism]

Multiplication maps are always continuous. This follows from the fact that multiplication is jointly continuous in a topological group.

 

[pivoxa15]

Multiplication maps are always continuous. This follows from the fact that multiplication is jointly continuous in a topological group.

RIght, I found a proof of it using product topology and component mapping, pi which is open.