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Define $G_{1}:=\{f_{n}(x):=\sum\limits_{i=1}^{n}\frac{x^{i}}{i!}|n \in \mathbb N\}\subseteq C([0,1])$

The background to this question is that I want to show that $G_{1}$ is relatively compact. It is clear I need to use Arzela-Ascoli but in order to do that I need to prove equicontinuity first, and this is where I am struggling.

I know that $(f_{n})_{n}$ converges uniformly on $[0,1]$ to some $f$ and $f_{n}(x)\leq \exp(x)$. From this I know that for any $\epsilon > 0$ there exists $N \in \mathbb N$ so that $\vert\vert f_{n}-f\vert\vert_{\infty}<\epsilon$ for all $n \geq N$, where $\vert\vert f_{n}-f\vert\vert_{\infty}=\sup\limits_{x\in [0,1]}\vert f_{n}(x)-f(x)\vert$ but how does this help me show continuity let alone equicontinuity.

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Let $\varepsilon>0$ be given.

Since the limit $f(x)$ is a uniformly continuous function on $[0,1]$, there exists $\delta_0>0$ such that $|x-y|\le \delta_0 \implies |f(x)-f(y)|\le\varepsilon/3$.

Since $f_n\to f$ uniformly, we can find $N$ such that $\|f-f_n\|_\infty<\varepsilon/3$ for all $n\ge N$. This implies that $$\begin{align} |f_n(x)-f_n(y)| &\le |f_n(x)-f(x)|+|f(x)-f(y)|+|f(y)-f_n(y)| \\ &\le \varepsilon/3 + \varepsilon/3 + \varepsilon/3 \\ &= \varepsilon \end{align}$$ whenever $|x-y|\le \delta_0$ and $n\ge N$.

For the case $i<N$, each $f_i$ is uniformly continuous so we can find $\delta_i>0$ such that $|f_i(x)-f_i(y)|<\varepsilon/3$. For each such $i$ the above calculation checks out whenever $|x-y|\le \delta_i$.

Lastly, we take $\delta = \min\{\delta_0,\delta_1,\dots,\delta_{n-1}\}$ to get equicontinuity.

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