[proofplan] We prove the two implications directly from the definitions. If the sequence is uniformly Cauchy, then for each fixed point $x \in E$ the pointwise sequence $(f_n(x))$ is Cauchy in the [complete metric space](/page/Complete%20Metric%20Space) $Y$, so it has a unique limit; these pointwise limits define the candidate function $f:E \to Y$. The uniform Cauchy estimate is then combined with pointwise convergence and an $\varepsilon/2$ argument to prove [uniform convergence](/page/Uniform%20Convergence) to $f$. Conversely, uniform convergence to a function $f$ implies uniform Cauchyness by applying the triangle inequality through the common limiting value $f(x)$. [/proofplan]

[step:Construct the pointwise limit from the uniform Cauchy hypothesis]Assume that $(f_n)_{n=1}^{\infty}$ is uniformly Cauchy on $E$. Fix $x \in E$. For every $\varepsilon>0$, the uniform Cauchy hypothesis gives $N \in \mathbb{N}$ such that for all $m,n \ge N$ and all $z \in E$, \begin{align*} d_Y(f_n(z),f_m(z))<\varepsilon. \end{align*} Applying this with $z=x$ shows that the sequence $(f_n(x))_{n=1}^{\infty}$ is Cauchy in $Y$. Since $(Y,d_Y)$ is complete, there exists a point $y_x \in Y$ such that $f_n(x) \to y_x$ in $Y$. The limit is unique because [metric spaces are Hausdorff](/theorems/290). Therefore there is a unique function \begin{align*} f:E &\to Y \end{align*} defined by the rule \begin{align*} f(x)=y_x=\lim_{n \to \infty} f_n(x) \end{align*} for every $x \in E$. If $E=\varnothing$, the same definition is interpreted as the unique function from $\varnothing$ to $Y$, and all later estimates over $x \in E$ are vacuous.[/step]

[guided]We start from the uniform Cauchy assumption and first build the only possible candidate for the uniform limit. Fix an arbitrary point $x \in E$. The uniform Cauchy condition says that for every $\varepsilon>0$ there is one index $N \in \mathbb{N}$ such that, whenever $m,n \ge N$, the estimate \begin{align*} d_Y(f_n(z),f_m(z))<\varepsilon \end{align*} holds simultaneously for every $z \in E$. In particular, it holds at the fixed point $z=x$. Hence, for every $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $m,n \ge N$, \begin{align*} d_Y(f_n(x),f_m(x))<\varepsilon. \end{align*} This is exactly the statement that the pointwise sequence $(f_n(x))_{n=1}^{\infty}$ is Cauchy in the [metric space](/page/Metric%20Space) $(Y,d_Y)$. The completeness of $Y$ is used precisely here: every [Cauchy sequence](/page/Cauchy%20Sequence) in $Y$ converges to a point of $Y$. Therefore there exists $y_x \in Y$ such that \begin{align*} \lim_{n \to \infty} f_n(x)=y_x. \end{align*} Because limits in a metric space are unique, the point $y_x$ is determined uniquely by $x$. This allows us to define a function \begin{align*} f:E &\to Y \end{align*} by assigning to each $x \in E$ the pointwise limit \begin{align*} f(x)=y_x=\lim_{n \to \infty} f_n(x). \end{align*} This is the natural candidate for the uniform limit: any uniform limit must also be the pointwise limit at every point. If $E=\varnothing$, there are no points $x$ at which a choice has to be made. The function $f:\varnothing \to Y$ is the unique empty function, and every universal statement over $x \in \varnothing$ is vacuously true. Thus the argument covers the empty-domain case as well.[/guided]

[step:Upgrade the pointwise limit to uniform convergence] We prove that $(f_n)_{n=1}^{\infty}$ converges uniformly to the function $f:E \to Y$ constructed above. Let $\varepsilon>0$. Since $(f_n)_{n=1}^{\infty}$ is uniformly Cauchy on $E$, there exists $N \in \mathbb{N}$ such that, for all $m,n \ge N$ and all $x \in E$, \begin{align*} d_Y(f_n(x),f_m(x))<\frac{\varepsilon}{2}. \end{align*} Fix $n \ge N$ and $x \in E$. Since $f_m(x) \to f(x)$ in $Y$ as $m \to \infty$, there exists $M \in \mathbb{N}$ such that $M \ge N$ and \begin{align*} d_Y(f_M(x),f(x))<\frac{\varepsilon}{2}. \end{align*} The triangle inequality in $(Y,d_Y)$ gives \begin{align*} d_Y(f_n(x),f(x)) \le d_Y(f_n(x),f_M(x))+d_Y(f_M(x),f(x)). \end{align*} By the choice of $N$ and $M$, \begin{align*} d_Y(f_n(x),f(x))<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{align*} This estimate holds for every $n \ge N$ and every $x \in E$, so $(f_n)_{n=1}^{\infty}$ converges uniformly to $f$ on $E$. [/step]

[step:Derive uniform Cauchyness from uniform convergence] Conversely, assume that there exists a function $f:E \to Y$ such that $(f_n)_{n=1}^{\infty}$ converges uniformly to $f$ on $E$. Let $\varepsilon>0$. By uniform convergence, there exists $N \in \mathbb{N}$ such that, for all $k \ge N$ and all $x \in E$, \begin{align*} d_Y(f_k(x),f(x))<\frac{\varepsilon}{2}. \end{align*} Let $m,n \ge N$ and let $x \in E$. Applying the triangle inequality in $(Y,d_Y)$ through the point $f(x)$ gives \begin{align*} d_Y(f_n(x),f_m(x)) \le d_Y(f_n(x),f(x))+d_Y(f(x),f_m(x)). \end{align*} Using the symmetry of the metric for the second term and the choice of $N$, we obtain \begin{align*} d_Y(f_n(x),f_m(x))<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{align*} Thus for every $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $m,n \ge N$ and all $x \in E$, \begin{align*} d_Y(f_n(x),f_m(x))<\varepsilon. \end{align*} Therefore $(f_n)_{n=1}^{\infty}$ is uniformly Cauchy on $E$. [/step]

[step:Conclude the equivalence] The first two steps show that every [uniformly Cauchy sequence](/page/Uniformly%20Cauchy%20Sequence) of functions $f_n:E \to Y$ converges uniformly to a function $f:E \to Y$, using completeness of $(Y,d_Y)$. The third step shows that every uniformly convergent sequence of functions is uniformly Cauchy. Hence the two stated conditions are equivalent. [/step]