Uniform continuity in Metric spaces

Question

I would like to prove the following:

Proposition: Let $(X, d)$ and $(Y, \rho)$ be metric spaces and $f : X\to Y$ a function. Assume that for any two sequences $(x_n), (y_n)$ in $X$ the condition $d(x_n,y_n) \to 0$ implies $\rho(f(x_n), f(y_n)) \to 0$. Then $f$ is uniformly continuous.

My proof: Suppose for a contradiction that $f$ is not uniformly continuous. That means there exists an epsilon so that for $\delta = \frac{1}{n}$ where n is arbitrary natural number, there exists two sequences $(x_n),(y_n)$ in X so that $d(x_n,y_n)< \frac{1}{n}$ and $\rho(f(x_n),f(y_n) \geq \epsilon$. By the squeeze theorem, since $\frac{1}{n} \rightarrow 0$ we have $d(x_n,y_n) \rightarrow 0$. Since $(x_n),(y_n)$ are contained in X for every $n \in \mathbb{N}$ $\rho(f(x_n),f(y_n)) \geq \epsilon$ it follows that $\rho(f(x_n),f(y_n)$ does not converge to 0.

May someone tell me a better way to justify the last sentence please.

Answer 1

The correct statement is presumably this:

Let $(X, d)$ and $(Y, \rho)$ be metric spaces and $f : X\to Y$ a function. Assume that for any two sequences $(x_n), (y_n)$ in $X$ the condition $d(x_n,y_n) \to 0$ implies $\rho(f(x_n), f(y_n)) \to 0$. Then $f$ is uniformly continuous.

Now your proof is entirely correct. If $\rho(f(x_n), f(y_n)) \to 0$ then for large enough $n\in\mathbb{N}$ it will hold $\rho(f(x_n), f(y_n)) < \varepsilon$ which contradicts $\rho(f(x_n), f(y_n)) \ge \varepsilon, \forall n\in\mathbb{N}$.