# Question about metric spaces and uniform continuity

I'm attempting to prove the following statement:

"Suppose that $$f:X \to Y$$ is a uniformly continuous function between metric spaces. Suppose that $$\{x_n\}$$ and $$\{x'_n\}$$ are two sequences in $$X$$ such that $$d_X(x_n,x'_n) \to 0$$ in $$\mathbb{R}$$. Show that $$d_Y(f(x_n),f(x'_n)) \to 0$$.

Since $$d_X(x_n,x'_n) \to 0$$, what could I say about these sequences? Also, where would the uniform continuity come in? Any help would be appreciated!

• Write out the definition of uniform continuity and think about how it might be applied when $d_X(x_n,x_n’)$ is sufficiently small. – Nap D. Lover Feb 26 '20 at 0:52
• My definition of uniform continuity is $\forall \epsilon>0$, $\exists \delta >0$ such that when $d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\epsilon$. – MATH-LORD Feb 26 '20 at 0:56
• Could I say that since $d_X(x_n,x'_n) \to 0$, then $d_X(x,y)<\delta$ for all $\delta$? (since $\delta>0$) – MATH-LORD Feb 26 '20 at 0:59

Your definition of uniform continuity: $$\forall \epsilon > 0$$, $$\exists \delta > 0$$ such that $$d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\epsilon$$.

$$d_X(x_n, x_{n}^{\prime}) \to 0$$ means $$\forall \epsilon_1 > 0$$, $$\exists N_1 > 0$$ such that $$n>N_1 \implies d_X(x_n, x_{n}^{\prime}) < \epsilon_1$$.

You need to show $$d_Y(f(x_n), f(x_{n}^{\prime})) \to 0$$ that is you need to show $$\forall \epsilon_2 > 0$$, $$\exists N_2 > 0$$ such that $$n>N_2 \implies d_Y(f(x_n), f(x_{n}^{\prime})) < \epsilon_2$$.

So let $$\epsilon_2 > 0$$ be given.

Then for $$\epsilon = \epsilon_2$$ in the definition of uniform continuity, there is a $$\delta_{\epsilon_2} > 0$$ such that $$d_X(x,y)<\delta_{\epsilon_2} \implies d_Y(f(x),f(y))<\epsilon_2$$.

Now for $$\epsilon_1 = \delta_{\epsilon_2}$$ in the second statement, there is an $$N_{\delta} > 0$$ such that $$n>N_{\delta} \implies d_X(x_n, x_{n}^{\prime}) < \delta_{\epsilon_2}$$.

Thus, choosing $$N_2 = N_{\delta}$$ in the third statement gives you that desired chain of implications $$n>N_{\delta} \implies d_X(x_n, x_{n}^{\prime}) < \delta_{\epsilon_2} \implies d_Y(f(x_n), f(x_{n}^{\prime})) < \epsilon_2$$.

That is, given any $$\epsilon_2 > 0$$ you manage to find the corresponding $$N_2 > 0$$ so that the third statement holds.