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!

  • $\begingroup$ 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. $\endgroup$ – Nap D. Lover Feb 26 '20 at 0:52
  • $\begingroup$ My definition of uniform continuity is $\forall \epsilon>0$, $\exists \delta >0$ such that when $d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\epsilon$. $\endgroup$ – MATH-LORD Feb 26 '20 at 0:56
  • $\begingroup$ Could I say that since $d_X(x_n,x'_n) \to 0$, then $d_X(x,y)<\delta$ for all $\delta$? (since $\delta>0$) $\endgroup$ – 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.

Note: Please feel free to edit for sake of readability.


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