Let ${ f_n\left(x\right):[0,1] \to \mathbb{R} }$ uniformly convergent to function $f(x)$ which is bounded on the interval $[0,1]$. Prove the following: ${ \lim_{n\to\infty} \sup_{[0,1]} f_n\left(x\right) = \sup_{[0,1]} f\left(x\right) }$.
I tried to show that $f$ is bounded and therefore it has a supremum $S$ for some $x_0 \in [0,1]$. In addition, for every $\epsilon > 0$ there exist $n_0$ such that for every $n > n_0$ and for every $x \in [0,1]$, $|f_n(x) - f(x)| < \epsilon$. I tried to put the $x_0$ instead of $x$, and replace the $f(x_0)$ with $S$, but it doesn't helped much...
How can I use the boundedness wisely?
i) For $x_n\in [0,1]$, assume that $\sup\ f = \lim\ f(x_n)$
Assume that $|\sup\ f-f(x_n)|<\varepsilon$ for some $n$. By pointwise convergence, $i>i_0$ implies $|f_i(x_n)-f(x_n)|<\varepsilon$ so that $$\sup\ f_i\geq f_i(x_n)\geq \sup\ f -2\varepsilon$$ for all $i>i_0$
Hence $\lim\ \sup\ f_i\geq \sup\ f$
ii) $|\lim\ \sup\ f_i - \sup\ f_i|<\varepsilon$ for all $i>i_0$.
Further $| f_i(x_i)- \lim\ \sup\ f_i |<2\varepsilon$ for all $i>i_0$
By uniform convergence, when $i$ is large, then $|f(x_i)-f_i(x_i)| < \varepsilon$ so that $|f(x_i)-\lim\ \sup\ f_i |<3 \varepsilon$ for large $i$.