I am trying an NBHM question of analysis.I am trying to disprove the following statement:
Every bounded continuous real valued function on $\mathbb Q^2$ can be extended to a bounded continuous function on $\mathbb R^2$.
My idea was to construct a function having bounded oscillation at a point and continuous elsewhere.Now if the point of oscillation is outside $\mathbb Q^2$,then we are done.So,I first try to construct a function in two variables that has bounded oscillation at $0$,then I can translate the oscillation to a point not belonging to $\mathbb Q^2$.My idea is to use the function $f(x)=\cos(\frac{1}{x}),x\neq 0$.My claim is that the surface obtained when I rotate the graph of $f$ about $y$-axis gives me the surface in $\mathbb R^3$,the function corresponding to which is having bounded oscillation at $0$.So,basically I am looking for a $3D$ version of $\cos(\frac{1}{x})$,but I cannot find the explicit formula to write the function in the form $f(x,y)$,will $\cos(\frac{1}{x^2+y^2})$ work?
Pick any irrational number $r$ and define $$F : (\mathbb R \setminus \{r\}) \times \mathbb R \to \mathbb R, F(x,y) = \cos\frac{1}{x-r} .$$ This is a bounded continuous function. Let $f : \mathbb Q^2 \to \mathbb R$ denote its restrction to $\mathbb Q^2$. It is again a bounded continuous function. It does not have a continuous extension to $\mathbb R^2$. If it had one, then $\lim_{x \to r}f(x,0) = \lim_{x \to r}\cos\frac{1}{x-r} $ must exist, and this is not true.