Boundedness leading to pointwise convergence implying uniform convergence? Consider a sequence of functions $\{f_n\}$ on some closed interval $I \subset \mathbb{R}$. Let's assume that $f_n$ is bounded on $I$ by $M \in \mathbb{R}$ for each $n \in \mathbb{N}$. If $\{f_n\}$ converges pointwise to some function $f$ on $I$ does it necessarily follow that $\{f_n\}$ converges uniformly to $f$ on $I$?
Intuitively I think this is the case, which I believe is supported by the following argument. As $\{f_n\}$ is pointwise convergent on $I$ we can construct a set for each $\epsilon>0$: $$K_\epsilon =\{K(x) \in \mathbb{N}\ : x \in I\ \text{and}\ \forall n>K(x)\  ,|f_n(x)-f(x)|<\epsilon\}$$ Due to the fact that $\{f_n\}$ is bounded on a closed interval, and is pointwise convergent to $f$, I believe that $max (K_\epsilon)$ exists for each $\epsilon >0$. Then for $\epsilon > 0$ we have $n>max(K_\epsilon )$ implies $$|f_n(x)-f(x)|< \epsilon \ \ \forall x \in I$$ and so by definition $\{f_n\}$ is uniformly convergent on $I$. My only issue is that I'm not able to prove that the maximum of the set exists, I just intuitively feel that it does under the imposed conditions. If anyone could confirm or explain where I've gone wrong with my reasoning I'd be much obliged.

The answer is no. Take $f_n(x)=x^n$ on $[0,1]$. $|f_n(x)|\le1$ for all $x\in[0,1]$ and all $n\in\mathbb{N}$. $f_n$ converges pointwise to the function $f(x)=0$ if $0\le x<1$, $f(1)=1$, but the convergence is not uniform. Even more, no subsequence converges uniformly.
