I am regarding the real numbers with the topology that is induced by the euclidean metric. and now i have a set given by $$M:=\{f:\mathbb{R} \rightarrow \mathbb{R}; \forall n \in \mathbb{N}: f(2n) \in (0,1)\} .$$ the topology on $ \mathbb{R}^{\mathbb{R}}$ is supposed to be the product topology and i should calculate the interior of this set.

my idea was, that the interior of this set is empty as there are infinite conditions on the function. Am i right?

  • 3
    $\begingroup$ You probably mean $\Bbb R^{\Bbb N}$, but otherwise I think you're correct. $\endgroup$
    – Lord_Farin
    Apr 27 '13 at 22:31

A basic open in $\mathbb{R}^{\mathbb{R}}$ (with the product topology) consists of functions whose values are restricted to intervals at finitely many points, i.e. $U=\{f|f(x_i)\in(a_i,b_i), i=1,..,n\}$ for some choice of $n, x_i, a_i,b_i$. The set $M$ you describe contains no open set.


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