[proofplan]
Fix a rational tolerance $\varepsilon>0$ and use the finite-net continuity hypothesis to obtain a scale $\rho$ and an admissible radius $\alpha$ around every $\rho$-net centre. Choose a rational $\delta>0$ small enough that moving from $x$ to $x'$ by less than $\delta$ keeps $x'$ inside the same $\alpha$-ball around a $\rho$-net centre chosen near $x$. The two finite-net continuity estimates then compare both $f(x)$ and $f(x')$ to $f(c)$, and the triangle inequality in $Y$ gives the desired $\varepsilon$-bound.
[/proofplan]
[step:Choose a rational modulus scale from the finite-net continuity hypothesis]
Fix a rational number $\varepsilon>0$. By the assumed finite-net continuity condition, there exist rational numbers $\rho>0$ and $\alpha>0$ such that $2\rho<\alpha$ and, for every centre $c$ occurring in $\operatorname{Net}_X(\rho)$ and every point $z\in X$,
\begin{align*}
d_X(z,c)<\alpha \implies d_Y(f(z),f(c))<\varepsilon/2.
\end{align*}
Since $2\rho<\alpha$, the rational number $\alpha-\rho$ is positive. Define the rational number $\delta>0$ by
\begin{align*}
\delta := \frac{\alpha-\rho}{2}.
\end{align*}
Then $\delta+\rho=(\alpha+\rho)/2<\alpha$, because $\rho<\alpha$.
[/step]
[step:Use the $\rho$-net to place nearby points inside one admissible centre ball]
Let $x,x'\in X$ satisfy $d_X(x,x')<\delta$. By the defining covering property of the net operation, there exists a centre $c\in X$ occurring in the finite list $\operatorname{Net}_X(\rho)$ such that
\begin{align*}
d_X(x,c)<\rho.
\end{align*}
Because $2\rho<\alpha$, we have $\rho<\alpha$, and therefore
\begin{align*}
d_X(x,c)<\alpha.
\end{align*}
By the triangle inequality in the [metric space](/page/Metric%20Space) $(X,d_X)$,
\begin{align*}
d_X(x',c)\leq d_X(x',x)+d_X(x,c)<\delta+\rho<\alpha.
\end{align*}
Thus both $x$ and $x'$ lie inside the $\alpha$-ball around the same centre $c$.
[guided]
We now explain why the net centre is chosen near $x$, rather than separately choosing centres near both $x$ and $x'$. The hypothesis compares values of $f$ only to values at net centres. To compare $f(x)$ with $f(x')$, it is useful to compare both of them to a single value $f(c)$.
Since $\operatorname{Net}_X(\rho)$ is a $\rho$-net for $X$, applied to the point $x\in X$ it gives a centre $c\in X$ occurring in the finite list $\operatorname{Net}_X(\rho)$ such that
\begin{align*}
d_X(x,c)<\rho.
\end{align*}
The finite-net continuity hypothesis is available inside radius $\alpha$ around this centre. We first verify that $x$ lies in that radius. From $2\rho<\alpha$ we get $\rho<\alpha$, hence
\begin{align*}
d_X(x,c)<\rho<\alpha.
\end{align*}
Next we verify that $x'$ also lies in the same $\alpha$-ball around $c$. The triangle inequality in $(X,d_X)$ gives
\begin{align*}
d_X(x',c)\leq d_X(x',x)+d_X(x,c).
\end{align*}
Using the assumptions $d_X(x,x')<\delta$ and $d_X(x,c)<\rho$, and using symmetry of the metric to identify $d_X(x',x)=d_X(x,x')$, we obtain
\begin{align*}
d_X(x',c)<\delta+\rho.
\end{align*}
By the definition $\delta=(\alpha-\rho)/2$,
\begin{align*}
\delta+\rho=\frac{\alpha+\rho}{2}.
\end{align*}
Since $\rho<\alpha$, this quantity is strictly less than $\alpha$. Therefore
\begin{align*}
d_X(x',c)<\alpha.
\end{align*}
Thus the same centre $c$ is close enough to both $x$ and $x'$ for the finite-net continuity estimate to apply twice.
[/guided]
[/step]
[step:Compare both function values to the same net-centre value]
The centre $c$ occurs in $\operatorname{Net}_X(\rho)$, and we have proved both $d_X(x,c)<\alpha$ and $d_X(x',c)<\alpha$. Applying the finite-net continuity hypothesis first with $z=x$ and then with $z=x'$, we obtain
\begin{align*}
d_Y(f(x),f(c))<\varepsilon/2
\end{align*}
and
\begin{align*}
d_Y(f(x'),f(c))<\varepsilon/2.
\end{align*}
Using symmetry of the metric $d_Y$, the [second inequality](/theorems/2136) gives
\begin{align*}
d_Y(f(c),f(x'))<\varepsilon/2.
\end{align*}
By the triangle inequality in the metric space $(Y,d_Y)$,
\begin{align*}
d_Y(f(x),f(x'))\leq d_Y(f(x),f(c))+d_Y(f(c),f(x'))<\varepsilon/2+\varepsilon/2=\varepsilon.
\end{align*}
Therefore, for this rational $\varepsilon>0$, the rational number $\delta=(\alpha-\rho)/2>0$ satisfies
\begin{align*}
d_X(x,x')<\delta \implies d_Y(f(x),f(x'))<\varepsilon
\end{align*}
for all $x,x'\in X$. Since $\varepsilon>0$ was arbitrary, $f$ is uniformly continuous.
[/step]
