Set of oscillation lower then $\epsilon$ is a $G_{\delta}$ set

Question

I have the following problem:

Let $(X,d)$ be a metric space and $f:X\rightarrow \mathbb{R}$ be some function. Let

$$\omega(f)(x,r) = sup\{d(f(x_0),f(y_0))\, |\, x_0,y_0\in B(x,r)\}$$ and $$osc(f)(x)=\lim_{r\rightarrow 0}\omega(f)(x,r).$$ Show that {$x\in X\,|\, osc(f)(x)<\epsilon$} is an open set, with the hint, that this set is equal to $$\cup_{r>0}\{x\,|\,\omega(f)(x,r)<\epsilon\}.$$

How can i show that these two sets are equal and why is the set $\cup_{r>0}\{x\,|\,\omega(f)(x,r)<\epsilon\}$ open? I would argue, that because $\omega(f)(x,r)$ is the greatest distance of the function of two points inside an open ball, this set must be open too, but is this sufficient?

Lastly I have to show, that the points of continuity of $f$ form a $G_\delta$-set. This is pretty obvious, because $f$ is continuous iff $osc(f)(x)=0$. With the above, we can see that {$x\,|\,osc(f)=0$}$=\cap_{n\geq 1}\cup_{r>0}${$x\,|\,\omega(f)(x,r)<\frac{1}{n}$}. Does this mean, that there exists a function on $\mathbb{R}$ that is continuous only in the rationals?

Answer 1

Fix $r>0$ and let $x$ be an element such that $\omega(f)(x,r)<\epsilon$.

By definition, $$\omega(f)(x,r) = \sup\{d(f(x_0),f(y_0))\, |\, x_0,y_0\in B(x,r)\}$$

Now, if $y$ is such that $d(x,y)<r/2$ we have that $B(y,r/2)\subset B(x,r)$, because if $d(y,x_0)<r/2$ by triangle inequality $d(x,x_0)\leq d(x,y)+d(y,x_0)<r$.

Thus, it follows immediately that $$\omega(f)(y,r/2) = \sup\{d(f(x_0),f(y_0))\, |\, x_0,y_0\in B(y,r/2)\}\leq \omega(f)(x,r)$$

and so $\omega(f)(y,r/2)<\epsilon$. This shows $\displaystyle\bigcup_{r>0}\{x\,|\,\omega(f)(x,r)<\epsilon\}$ is open, and so $\{x\in X\,|\, \operatorname{osc}(f)(x)<\epsilon\}$ is open, as desired.

With respect to the last question, proving that the set of continuity points is $G_\delta$ precisely shows that there cannot exist a function only continuous at $\mathbb{Q}$, as rationals are not a $G_\delta$ set.