Let $(f_n)$ be a sequence of continuous real valued functions defined on a metric space $(X,d)$, and suppose that $(f_n)$ converges uniformly to a function $f$. Prove that
$$\lim_{n\to\infty}f_n(x_n)=f(x)$$
for every sequence $(x_n)$ in $X$ with $x_n\to x$.
These are what I know that might be useful in the proof:
Since $(f_n)$ converges uniformly to $f$, for every $\epsilon>0$, there exists $N$ such that $n\geq N\implies|f_n(x)-f(x)|<\epsilon$ for all $x\in X$.
$f$ is continuous.
Since every sequence $(x_n)$ in $X$ with $x_n\to x$, we have for every $r>0$, there exists $N$ such that $n\geq N\implies d(x_n,x)<r$.
My questions are :
From point 1 above, since it applies for all $x\in X$, can we easily conclude $\lim_{n\to\infty}f_n(x_n)=f(x)$ since $x_n\in X$ and $x\in X$?
And $x_n$ already converges to $x$ as given in the question, so there is really nothing to prove?
I don't quite understand how to relate the convergence of $x_n$ and of $f_n$ in order to write the proof. Any ideas?
It is quite intuitive that the statement to be proven is correct, but I'm just wondering how to write it down by using those basic definitions of convergence of sequence and function and to relate them.
I have a very rough idea like this: Since $x_n\to x$ then $f_n(x_n)\to f_n(x)$ because $f_n$ is continuous and $f_n(x)\to f(x)$, combining the two we got $f_n(x_n)\to f(x)$. I am really not sure whether it is correct?
Thanks for the help!