TITLE: $f: \mathbb{Q} \rightarrow \mathbb{R} \ \ \lim _{q \rightarrow t, \ q\in \mathbb{Q}} f(q) =g$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.