How to get started on proving that the limit of a set A exists in its Accumulation Set

Let $X$ and $Y$ be metric spaces. If $Y$ is complete, $A\subset X$, $f:A\rightarrow Y$ is a uniformly continuous function. Prove

(1) If $x_0\in \bar{A}\backslash A$, $\{x_n\}_{n=1}^{+\infty}\subset A$, $\lim_{n\rightarrow+\infty} x_n=x_0$. Then the limit \begin{equation*} \lim_{n\rightarrow\infty}f(x_n) \end{equation*} exists.

Since $A \subset \bar{A}$ but how can $x_0$ exist in A as given that $x_0$ is part of the set that is whats not in A but in A. So from the definitions that means $x_0 \in A'$ which are the accumulation points of A. But idk where to go from here

Solution 1:

$f(x_1), f(x_2) ,\ldots$ is a sequence in $Y$ and you want to show it converges. You are given that $Y$ is complete. This is a big hint that you need to show that the sequence $f(x_1), f(x_2), \ldots$ is Cauchy.

You know that $x_1, x_2, \ldots$ is Cauchy. See if you can use an important property of $f$ to establish that $f(x_1), f(x_2), \ldots$ is Cauchy.

Solution 2:

1. $(X, d) $ be a metric space and $Y\subset X$ . Let, $\{y_n\} \subseteq Y $ be a sequence.Then $\{y_n\}$is cauchy sequence in $(Y, d_Y) $ iff $\{y_n\} $ is cauchy sequence in $(X, d) $

2. Every convergent in a metric space is cauchy.

3. Uniformly continuous function maps cauchy sequence to cauchy sequence.

4. A cauchy sequence in a complete metric space is convergent.

Proof:

$\{x_n\}_{n=1}^{+\infty}\subset A$ and $\{x_n\} \to x_0 $ in $(X, d) $

Hence, $\{x_n\}$ is cauchy sequence in $(A, d_A) $

$f:A\to X$ uniformly continuous. Hence $\{f(x_n) \}$ is cauchy sequence in $(Y, d') $.

Due to completeness of $(Y, d') $, the sequence $\{f(x_n) \}$ converges. i.e \begin{equation*} \lim_{n\rightarrow\infty}f(x_n) \text {exists} \end{equation*}