Uniform convergence and continuity again

Question

There is a good chance I am over complicating this.

If $(f_n)$ is a sequence of real functions converging uniformly on $[0,1]$ to a function $f$, and if $f_n$ continuous at $x_n \in [0,1]$ with $x_n \to x$, does it follow that $f$ is continuous at $x$?

So I won't write it all out but my attempted proof essentially goes:

$|f(x)-f(y)| \le |f(x)-f_n(x)|+ |f_n(x)-f_n(x_n)| + |f_n(x_n)-f_n(y)| +|f_n(y)-f(y)|$ by triangle inequality.

The first and last terms can be made arbitrarily small by convergence of $f_n$ to $f$, and in particular uniform convergence for that last term.

Now I tread with trepidation, since I made a similar mistake on a different problem. If I can bound the second term I can certainly bound the third since I can choose how close $y$ need be to $x$. But I'm not sure I can use the continuity of $f_n$ at $x_n$ alone to show the second term can be made arbitrarily small since any $\delta$ will depend on $n$, and of course $n$ increases as $x_n \to x$. I assume I need to use the uniform convergence of $f_n$, but I am not sure how. How should I proceed? Alternatively the proposition could be false in which please let me know without spoiling the counterexample.

Answer 1

The proposition is false.

Define, for all $n$,

$$f_n(0)=1, \quad f_n(x)=0 ~\forall x \in (0,1];$$
$$x_n=\frac{1}{n}.$$