How to Prove the Closure of a Subspace H is a Topological Group in G"

[NeroKid]

Homework Statement

Let H be a subspace of G. show that if H is also a subgroup of G then the closure of H is a topological group
G is a topological group

Homework Equations

The Attempt at a Solution

let closure of H is a subspace of G then the map of the operation of G restricted to cl(H) is continuous
I have the A2 axiom satisfied for cl(H) but I can't prove the the A1 + A3 +A4

 

[Klockan3]

A1-4 doesn't tell us anything, please write them down. It is hard to help you if we don't know which definition of a topological group you use.

 

[NeroKid]

A1: for all x y in G xRy is in G
A2 : for all x,y ,z in G xR(yRz) = (xRy)Rz
A3 : for all x in G there exists a 0 in G such that xR0=0Rx=x
A4: xRx^-1 = 0 = x^-1 R x

 

[micromass]

Hi NeroKid!

You know that

$$f:G\times G\rightarrow G:(x,y)\rightarrow x*y$$

is continuous. And since H is a subgroup, you know that

$$f(H\times H)\subseteq H$$

Now, what can you say about

$$f(cl(H)\times cl(H))$$

 

[NeroKid]

tks i get it know ;))