Uniformly convergence and bounded

Question

I know if {$f_n$} is a sequence of bounded function converge uniformly to f on E, then f is bounded on E. I wonder if it is true for pointwise convergence. So here is my supposed proof:

if $f$ is not bounded above, $\forall c$, there $\exists x_c$ st $f(x_c)>c$,

Since $f_n$ converges to f pointwise, for this $x_c$, pick $\epsilon=1$, there $\exists N$, for $\forall n \ge N$, $f_n(x_c)>f(x_c)-1>c-1$.

Thus $\forall n\ge N$, $f_n$ is not bounded above. We have contradiction.

Thus f is bounded above. By the same way f is bounded below.

This proof seems correct to me and thanks for your time.

Answer 1

Thanks to the comments above. I have found the error in my proof:

$\forall c$, $f_n$ is not bounded above by $c-1$ for all $n\ge N(c)$. When we change $c$, the corresponding $N(c)$ will change, so we can not show $f_n$ is not bounded above by any $c-1$.

However, if convergence is uniform, we will be able to find a common $N$ that is independent of $c$ and get the result we want.