[proofplan] We prove both implications by unwinding the quantifiers in the definition of [uniform continuity](/page/Uniform%20Continuity). If $f$ is not uniformly continuous, the negated definition supplies one fixed separation size $\varepsilon_0>0$ that cannot be controlled at any scale $\delta>0$; choosing $\delta=1/n$ produces the required sequences. Conversely, if such sequences exist and $f$ were uniformly continuous, then the scale supplied for $\varepsilon_0$ would eventually contain all pairs $(x_n,y_n)$, contradicting their uniform separation in $Y$. [/proofplan] [step:Negate uniform continuity and choose pairs at scale $1/n$] Assume first that $f$ is not uniformly continuous. By definition, uniform continuity of $f:X\to Y$ means that for every $\varepsilon>0$ there exists $\delta>0$ such that, for all $x,y\in X$, \begin{align*} d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\varepsilon. \end{align*} Negating this statement gives a number $\varepsilon_0>0$ such that for every $\delta>0$ there exist points $x,y\in X$ satisfying \begin{align*} d_X(x,y)<\delta \end{align*} and \begin{align*} d_Y(f(x),f(y))\geq \varepsilon_0. \end{align*} For each $n\in\mathbb N$, apply this negated statement with $\delta=1/n$. Choose points $x_n,y_n\in X$ such that \begin{align*} d_X(x_n,y_n)<\frac{1}{n} \end{align*} and \begin{align*} d_Y(f(x_n),f(y_n))\geq \varepsilon_0. \end{align*} Thus $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ are sequences in $X$ with the required fixed separation in $Y$. [/step] [step:Show the constructed pairs converge together in $X$] It remains in this direction to prove that $d_X(x_n,y_n)\to 0$. Let $\eta>0$ be given. Choose $N\in\mathbb N$ such that $1/N<\eta$. Then for every $n\geq N$, \begin{align*} 0\leq d_X(x_n,y_n)<\frac{1}{n}\leq \frac{1}{N}<\eta. \end{align*} Therefore $\lim_{n\to\infty} d_X(x_n,y_n)=0$. [guided] We have already built the sequences so that the $Y$-distance never falls below the same positive number $\varepsilon_0$. The remaining point is to check that the $X$-distance actually tends to $0$, rather than merely being small in some informal sense. Let $\eta>0$ be arbitrary. Since the real sequence $(1/n)_{n=1}^{\infty}$ converges to $0$, there exists $N\in\mathbb N$ such that $1/N<\eta$. For every $n\geq N$, the construction gives \begin{align*} d_X(x_n,y_n)<\frac{1}{n}. \end{align*} Because $n\geq N$, we also have $1/n\leq 1/N$. Combining these inequalities with nonnegativity of the metric $d_X$ gives \begin{align*} 0\leq d_X(x_n,y_n)<\frac{1}{n}\leq \frac{1}{N}<\eta. \end{align*} This is exactly the definition of convergence of the real sequence $(d_X(x_n,y_n))_{n=1}^{\infty}$ to $0$. Hence \begin{align*} \lim_{n\to\infty} d_X(x_n,y_n)=0. \end{align*} [/guided] [/step] [step:Use the sequences to contradict uniform continuity] Conversely, suppose there exist $\varepsilon_0>0$ and sequences $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ in $X$ such that \begin{align*} \lim_{n\to\infty} d_X(x_n,y_n)=0 \end{align*} and, for every $n\in\mathbb N$, \begin{align*} d_Y(f(x_n),f(y_n))\geq \varepsilon_0. \end{align*} Assume, for contradiction, that $f$ is uniformly continuous. Applying uniform continuity with the positive number $\varepsilon_0$ gives a number $\delta>0$ such that, for all $x,y\in X$, \begin{align*} d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\varepsilon_0. \end{align*} Since $d_X(x_n,y_n)\to 0$, there exists $N\in\mathbb N$ such that \begin{align*} d_X(x_N,y_N)<\delta. \end{align*} The implication above, applied with $x=x_N$ and $y=y_N$, yields \begin{align*} d_Y(f(x_N),f(y_N))<\varepsilon_0. \end{align*} This contradicts the assumed inequality \begin{align*} d_Y(f(x_N),f(y_N))\geq \varepsilon_0. \end{align*} Therefore $f$ is not uniformly continuous. [/step]