Convergence of sequence of functions implies convergence of the max of the sequence of functions to the max of the collection of functions Claim:
Let $f_1,..., f_N:X\longrightarrow \mathbb{R}^+$ be a set of bounded functions and suppose that for each $i=1,...,N$, there exists a sequence of functions $\{g_{n,i}\}_{n\in\mathbb{N}}$ such that $g_{n,i}\downarrow f_i$ pointwise. Define the sequence $\{g_n\}_{n\in\mathbb{N}}$ of functions by:
\begin{equation*}
g_n:= \max_{i\le N} g_{n,i} \qquad \forall n\in\mathbb{N}
\end{equation*}
Then $g_{n}\downarrow \max_{i\le N} f_i$ pointwise.
Is this claim true?? It seems intuitively true...but I couldn't show it with an $\epsilon$, $\delta$ proof!
 A: That the sequence $\left(g_n\left(x\right)\right)_{n\geqslant 1}$ is non-increasing follows from the fact that $g_{n+1,i}(x)\leqslant g_{n,i}(x)$ for each $i$ hence $\max_{1\leqslant i\leqslant N}  g_{n+1,i}(x)\leqslant\max_{1\leqslant i\leqslant N}  g_{n,i}(x)$. 
Now, for the pointwise convergence: fix a positive $\varepsilon$ and $x\in X$ . For any $i\in\left\{1,\dots,N\right\}$, there exists $n_i$ such that if $n\geqslant n_i$, then $ -\varepsilon  \leqslant g_{n,i}(x) -f_i(x) \leqslant \varepsilon$. Let $n_0:=\max_{1\leqslant i\leqslant N}n_i$. Then for any $n\geqslant n_0$, 
$$ f_i(x)-\varepsilon  \leqslant g_{n,i}(x)  \leqslant f_i(x)+\varepsilon.$$
Taking the $\max$ over $i\in \left\{1,\dots,N\right\}$, we get 
$$\max_{1\leqslant i\leqslant N} f_i(x)  -\varepsilon  \leqslant \max_{1\leqslant i\leqslant N} g_{n,i}(x)  \leqslant \max_{1\leqslant i\leqslant N} f_i(x)+\varepsilon$$
hence for all $n\geqslant n_0$,
$$\left\lvert g_n(x)- \max_{1\leqslant i\leqslant N} f_i(x)\right\rvert\leqslant\varepsilon .$$
