Interior, closure, open set in topological space

**zadir** · March 1st 2011, 12:22 PM

$T=\mathbb R^2 \: A=\mathbb Q^+ \times \mathbb R^+$

$\overline{A}=?$ /closure of A/
$Int A=?$
In $(A, \tau_A)$ is $A$ open? $(A, \tau_A)$ /topological space/

Thank you in advance!

**girdav** · March 1st 2011, 12:57 PM

You can show that if you have two topological spaces $(X,\mathcal T_1)$ and $(Y,\mathcal T_2)$ and if you consider the product space $(X\times Y,\mathcal T\otimes \mathcal U)$ you will have for any $A\subset X$ and $B\subset Y$ that $\overline{A\times B}=\overline{A}\times \overline B$ and $\mathrm{Int}(A\times B)=\mathrm{Int}(A)\times \mathrm{Int}(B)$ .

**zadir** · March 2nd 2011, 12:44 PM

Originally Posted by girdav

You can show that if you have two topological spaces $(X,\mathcal T_1)$ and $(Y,\mathcal T_2)$ and if you consider the product space $(X\times Y,\mathcal T\otimes \mathcal U)$ you will have for any $A\subset X$ and $B\subset Y$ that $\overline{A\times B}=\overline{A}\times \overline B$ and $\mathrm{Int}(A\times B)=\mathrm{Int}(A)\times \mathrm{Int}(B)$ .

Thank you for your help.
To tell the truth I don't know how to continue, how to use that in the problem.

**Drexel28** · March 2nd 2011, 06:46 PM

Originally Posted by zadir

Thank you for your help.
To tell the truth I don't know how to continue, how to use that in the problem.

He's saying that if you consider giving $\mathbb{R}^2$ the usual topology induced by the usual norm then the induced topology is the same as if $\mathbb{R}^2$ the product topology (when both copies of the reals are endowed with the usual topology). That said, for the product topology it's easy to prove things like $\text{cl}_{\mathbb{R}\times\mathbb{R}}\left(A\time s B\right)=\text{cl}_{\mathbb{R}}(A)\times\text{cl}_ {\mathbb{R}}(B)$ . So now since $\mathbb{Q},\mathbb{R}$ are both dense in $\mathbb{R}$ you can conclude that $\mathbb{Q}\times\mathbb{R}$ is dense in $\mathbb{R}^2$ . etc.

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