1. ## prove or disprove

For any topological space $(X,\tau )$ and for any group $A\subseteq X$

$int(cl(A))=cl(int(A))$

Thanks

2. Originally Posted by sinichko
For any topological space $(X,\tau )$ and for any group $A\subseteq X$

$int(cl(A))=cl(int(A))$

Thanks
Think about reals and the rationals. What is the closure? What is the interior?

3. Originally Posted by sinichko
For any topological space $(X,\tau )$ and for any group $A\subseteq X$

$int(cl(A))=cl(int(A))$

Thanks
Originally Posted by Focus
Think about reals and the rationals. What is the closure? What is the interior?
Alternatively:

Think about it like this. If $K=\left(\overline{A}\right)^{\circ}=\overline{A^{\ circ}}$ then we have that $\overline{K}=\overline{\overline{A^{\circ}}}=\over line{A^{\circ}}=K$ and so $K$ is closed. But, $K^{\circ}=\left(\left(\overline{A}\right)^{\circ}\ right)^{\circ}=\left(\overline{A}\right)^{\circ}=K$ and so $K$ is open. So, in a connected space $X$, $K=X$. But, $X=\left(\overline{A}\right)^{\circ}\subseteq\overl ine{A}\subseteq X$ and so $A$ must be dense in $X$. So more examples would be be $\mathbb{S}^1\subseteq\mathbb{R}^2$, $\text{SO}(n)\subseteq\text{GL}^+(n,\mathbb{R})$, etc.

The above also shows (like Focus) pointed out that your subgroup cannot have empty interior.

Also, note the above applies equally well to non-subgroup subsets.