Prove the sequence of functions converges and the limit function is continuous Suppose $F :=$ {$f_n: \Bbb R \rightarrow \Bbb R, n=1,2,3,...$} is an equicontinuous family. If the sequence $f_n(q)$ converges for each $q \in \Bbb Q$, show that $f_n(r)$ converges for each $r \in \Bbb R$, and the limit function is continuous. 
I have proved that if for each $q \in \Bbb Q$, the set {$f_n(q): \forall n$} is bounded, then there is a subsequence $f_{n_k}$ which converges for each $q \in \Bbb Q$. I'm not sure if this helps to prove the above statement. 
Could someone provide a proof please? Thanks.
 A: $f_n(r)$ converges
$F$ is equicontinuous, so :
$$\forall \epsilon >0, \exists d, \forall n, |x-y| \leq d \Rightarrow |f_n(x)-f_n(y)| \leq \epsilon$$
Since $f_n(q)$ converges for all $q \in \mathbb{Q}$, $\exists m_q,M_q$ such that $m \leq f_n(q) \leq M$. Because $F$ is equicontinous and $\mathbb{Q}$ dense in $\mathbb{R}$, then :
$$\forall r \in \mathbb{R}, \forall \epsilon >0, \exists q \in \mathbb{Q},\forall n, |f_n(r)-f_n(q)| \leq \epsilon $$
Hence : $m_q-\epsilon \leq f_n(r) \leq M_q+\epsilon$ so $f_n(r)$ is bounded. Then we have a subsequence $f_{n_k}(r)$ who converges, we call $f(r)$ that limit.
So now we have :
$$|f_n(r)-f(r)| \leq |f_n(r)-f_n(q)|+|f_n(q)-f_{n_k}(q)|+|f_{n_k}(q)-f_{n_k}(r)|+|f_{n_k}(r)-f(r)|$$
The term number :


*

*is less than $\epsilon$ because $F$ is equicontinuous

*is less than $\epsilon$ for $n,k$ big enough because $f_n(q)$ converges to $f(q)$.

*is less than $\epsilon$ for the same reason as 1.

*is less then $\epsilon$ for $k$ big enough because $f_{n_k}(r)$ converges to $f(r)$.


Finally, for $n$ big enough : $|f_n(r)-f(r)| \leq 4 \epsilon$.
$f_n$ converges pointwise to $f$.
$f$ is continuous
Let's set $\epsilon > 0$ and suppose $|x-y|<d$ such that $\forall n, |f_n(x)-f_n(y)| \leq \epsilon /3$ (this is possible because $F$ is equicontinuous). Then :
$$|f(x)-f(y)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|$$
Now : $\exists N, \forall n \geq N, |f(x)-f_n(x)| \leq \epsilon /3,|f_n(y)-f(y)| \epsilon /3$ because of the pointwise convergence we have proven.
Finally : 
$$|f(x)-f(y)| \leq \epsilon$$
That concludes the proof.
