[proofplan]
We prove the equivalence by expanding the two definitions and comparing their quantifiers. The sup norm condition controls $|f_n(x)-f_m(x)|$ for every $x \in E$ because each pointwise absolute value is bounded above by the supremum. Conversely, a uniform pointwise Cauchy estimate bounds the whole supremum; to preserve the strict inequality in the norm condition, we apply the uniform Cauchy hypothesis with $\varepsilon/2$.
[/proofplan]
[step:Record the sup norm expression for pairwise differences]
Let $m,n \in \mathbb{N}$. Define the difference function
\begin{align*}
h_{m,n}:E &\to \mathbb{R}
\end{align*}
by $h_{m,n}(x)=f_n(x)-f_m(x)$ for every $x \in E$. Since $f_n$ and $f_m$ are bounded, the function $h_{m,n}$ is bounded. Therefore its sup norm is finite and satisfies
\begin{align*}
\|f_n-f_m\|_\infty = \|h_{m,n}\|_\infty = \sup_{x \in E}|f_n(x)-f_m(x)|.
\end{align*}
For every $x \in E$, the defining property of the supremum gives
\begin{align*}
|f_n(x)-f_m(x)| \leq \|f_n-f_m\|_\infty.
\end{align*}
[/step]
[step:Derive uniform Cauchyness from the sup norm Cauchy condition]
Assume that $(f_n)_{n=1}^{\infty}$ is Cauchy with respect to $\|\cdot\|_\infty$. Let $\varepsilon>0$. By the sup norm Cauchy condition, there exists $N \in \mathbb{N}$ such that, for all $m,n \geq N$,
\begin{align*}
\|f_n-f_m\|_\infty < \varepsilon.
\end{align*}
Fix $m,n \geq N$ and $x \in E$. By the pointwise bound from the previous step,
\begin{align*}
|f_n(x)-f_m(x)| \leq \|f_n-f_m\|_\infty < \varepsilon.
\end{align*}
Thus, for every $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that for all $m,n \geq N$ and all $x \in E$,
\begin{align*}
|f_n(x)-f_m(x)| < \varepsilon.
\end{align*}
This is exactly uniform Cauchyness of $(f_n)_{n=1}^{\infty}$ on $E$.
[/step]
[step:Use an epsilon adjustment to bound the supremum from uniform Cauchyness]
Assume that $(f_n)_{n=1}^{\infty}$ is uniformly Cauchy on $E$. Let $\varepsilon>0$. Since $\varepsilon/2>0$, uniform Cauchyness gives $N \in \mathbb{N}$ such that, for all $m,n \geq N$ and all $x \in E$,
\begin{align*}
|f_n(x)-f_m(x)| < \frac{\varepsilon}{2}.
\end{align*}
Fix $m,n \geq N$. The number $\varepsilon/2$ is an upper bound for the set
\begin{align*}
\{|f_n(x)-f_m(x)| : x \in E\}.
\end{align*}
Therefore,
\begin{align*}
\|f_n-f_m\|_\infty = \sup_{x \in E}|f_n(x)-f_m(x)| \leq \frac{\varepsilon}{2} < \varepsilon.
\end{align*}
Hence for every $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that, for all $m,n \geq N$,
\begin{align*}
\|f_n-f_m\|_\infty < \varepsilon.
\end{align*}
Thus $(f_n)_{n=1}^{\infty}$ is Cauchy with respect to the sup norm.
[guided]
We assume that $(f_n)_{n=1}^{\infty}$ is uniformly Cauchy on $E$, and we want to prove the Cauchy condition for the sup norm. The target is a strict estimate
\begin{align*}
\|f_n-f_m\|_\infty < \varepsilon.
\end{align*}
The minor point is that a strict pointwise bound by $\varepsilon$ only implies that the supremum is at most $\varepsilon$, not necessarily strictly less than $\varepsilon$. We avoid this by asking the uniform Cauchy condition for the smaller tolerance $\varepsilon/2$.
Let $\varepsilon>0$. Since $\varepsilon/2>0$, uniform Cauchyness gives an index $N \in \mathbb{N}$ such that, whenever $m,n \geq N$ and $x \in E$,
\begin{align*}
|f_n(x)-f_m(x)| < \frac{\varepsilon}{2}.
\end{align*}
Now fix $m,n \geq N$. The displayed inequality holds for every $x \in E$, so $\varepsilon/2$ is an upper bound for the set of all pointwise differences
\begin{align*}
\{|f_n(x)-f_m(x)| : x \in E\}.
\end{align*}
By the definition of the supremum as the least upper bound, the supremum of this set is bounded above by $\varepsilon/2$. Using the definition of the sup norm, we obtain
\begin{align*}
\|f_n-f_m\|_\infty = \sup_{x \in E}|f_n(x)-f_m(x)| \leq \frac{\varepsilon}{2}.
\end{align*}
Since $\varepsilon/2<\varepsilon$, this gives
\begin{align*}
\|f_n-f_m\|_\infty < \varepsilon.
\end{align*}
The index $N$ works for all $m,n \geq N$, so the sequence $(f_n)_{n=1}^{\infty}$ is Cauchy with respect to the sup norm.
[/guided]
[/step]
[step:Combine the two implications]
The second step proves that sup norm Cauchyness implies uniform Cauchyness on $E$. The third step proves that uniform Cauchyness on $E$ implies sup norm Cauchyness. Therefore the two conditions are equivalent.
[/step]
