$f: \mathbb{Q} \rightarrow \mathbb{R} \ \ \lim _{q \rightarrow t, \ q\in \mathbb{Q}} f(q) =g$ There's a continuous function $f: \mathbb{Q} \rightarrow \mathbb{R}$
Prove that  $ \exists t \in \mathbb{R} \setminus \mathbb{Q} \ \ \exists g \in \mathbb{R} :\ \ \lim _{q \rightarrow t, \ q\in \mathbb{Q}}f(q)=g$.
Prove that there are $2^{\aleph _0}$ such numbers $t$.
I know that if a function is continuous on rational points, then it's continuous on whole $ \mathbb{R}$, but that isn't relevant to the problem, is it?
I would appreciate all your help.
Thanks.
 A: Here is my attempt; I hope that there is a simpler solution. (But if you have already seen the facts I mention below, it is not that complicated.)
Let us define
$$\omega(f,x)=\limsup\limits_{\substack{t\to x\\t\in\mathbb Q}} f(t)-\liminf\limits_{\substack{t\to x\\t\in\mathbb Q}} f(t).$$
(This make sense for $x\in\mathbb R$ too; even though the function $f$ is only defined for rationals.)
The continuity on $\mathbb Q$ implies that $\omega(f,x)=0$ if $x\in\mathbb Q$.


The set  $\{x\in\mathbb R; \omega(f,x)<\varepsilon\}$ is open in $\mathbb R$ for each $\varepsilon>0$.

If we define, for example, $$g(x)=\liminf\limits_{\substack{t\to x\\t\in\mathbb Q}} f(t),$$
then 
$$\omega(f,x)=\limsup\limits_{t\to x} g(t)-\liminf\limits_{t\to x} g(t)=\omega(g,x).$$
The fact that the function $\omega(g,x)$ is upper semicontinuous and the above set is open is often used in the proof that the set of continuity points of any function is $G_\delta$.
See, for example:


*

*Kaczor, Nowak: Problems in Mathematical Analysis II; Problem 1.4.20

*Set of points of continuity are $G_{\delta}$

Hence $M=\{x\in\mathbb R; \omega(f,x)=0\}$ is a $G_\delta$ set containing rationals. If we show that $|M|=2^{\aleph_0}$, then we have $2^{\aleph_0}$ points in $M\setminus\mathbb Q$ with the required properties.

So it suffices to show that:

Every dense $G_\delta$ subset of reals is uncountable.

By a result of Mazurkiewicz (mentioned, e.g., in this answer) such subset can be endowed with a metric which makes it into complete metric space.
If we have complete metric space with no isolated points then it is uncountable by Cantor-Bendixson theorem.
Some arguments showing that such sets must be uncountable have been given also here:


*

*How to show that $\mathbb{Q}$ is not $G_\delta$

*Does a dense $G_\delta$ subset of a complete metric space without isolated points contain a perfect set?
A: Let $\Bbb P=\Bbb R\setminus\Bbb Q$, and for $x\in\Bbb R$ let $B(x,\epsilon)=(x-\epsilon,x+\epsilon)$. Suppose that the first result is false. Then for each $x\in\Bbb P$ there is an $n(x)\in\Bbb N$ such that for each $\epsilon>0$ there are $p,q\in B(x,\epsilon)\cap\Bbb Q$ with $|f(p)-f(q)|\ge 2^{-n}$. For $k\in\Bbb N$ let $A_k=\{x\in\Bbb R\setminus\Bbb Q:n(x)=k\}$; by the Baire category theorem there are an $m\in\Bbb N$ and an open interval $(a,b)$ in $\Bbb R$ such that $A_m$ is dense in $(a,b)\cap\Bbb P$ and hence in $(a,b)$. Fix $p\in(a,b)\cap\Bbb Q$. Since $f$ is continuous, there is an $\epsilon>0$ such that $|f(q)-f(r)|<2^{-m}$ for all $q,r\in B(p,\epsilon)\cap\Bbb Q$, and it follows that $A_m\cap(a,b)\cap B(p,\epsilon)=\varnothing$, which is a contradiction.
In fact this shows that the sets $A_k$ are nowhere dense in $\Bbb P$. They are also closed in $\Bbb P$. To see this, suppose that $x\in\operatorname{cl}_{\Bbb P}A_k$, and let $\epsilon>0$ be arbitrary; then there is a $y\in A_k\cap B(x,\epsilon/2)$, so there are $p,q\in B(y,\epsilon/2)\cap\Bbb Q\subseteq B(x,\epsilon)\cap\Bbb Q$ with $|f(p)-f(q)|\ge 2^{-k}$, and the claim follows. For $k\in\Bbb N$ let $U_k=\Bbb P\setminus A_k$; $U_k$ is a dense open subset of $\Bbb P$, so $G=\bigcap_{k\in\Bbb N}U_k$ is a dense $G_\delta$-set in $\Bbb P$, and clearly $f$ has a continuous extension to each $x\in G$. Thus, to finish the problem we need only show that $|G|=2^\omega$; this is done in this answer to an earlier question, also noted by Martin Sleziak in his answer.
A: HINT: 

Question 2, because $R$ has $2^{\aleph_0}$ numbers, and $Q$ has countable numbers, so $R\setminus Q$ still has $2^{\aleph_0}$ numbers...


ADDED: Suppose that $|R\setminus Q|=\kappa<2^{\aleph_0}$, then $|R|=|R\setminus Q \cup Q|=\kappa+\aleph_0=\kappa<2^{\aleph_0}$, which leads a contradiction.
