Converging sequence of functions on a compact set.

Question

I have this yes/no question regarding uniform convergence of real functions:
"If a sequence of functions defined on a compact set converges uniformly, then the uniform limit is a bounded function."

The answer seems to me to be no, since we don't have any boundedness/continuity requirements on the sequence, reason, but I can't put my finger on it.

I begin by considering a compact set $C$, the sequence $f_n:C\rightarrow \mathbb{R}$ and the limit $f:C\rightarrow \mathbb{R}$. Since $f_n\rightarrow f$ uniformly, $\lim (\sup_{x\in C}|f_n(x)-f(x)|)=0$.
Standard arguments give that $|f_n(x)-f(x)|<1,\forall n\geq n_0,x\in C$.
Now I can't put a bound on $f$ using $f_n$, that's why I suspect the answer is no.

So, to show the answer is no, I need to find a counterexample, for say $C=[0,1]$. Is this right? Is there a simple counterexample? I'm only saying there is a counterexample because an attempt at proof seems to lead nowhere…

Thanks in advance for any replies.

Answer 1

If you take $(f_n)$ to be a constant sequence, such that the common value for $f_n$ is an unbouded function, then obviously $(f_n)$ converges uniformly but the limit is unbounded.