Uniform continuity preserves uniqueness of convergent sequences?

Question

If $f: (a, b) \to \Bbb R$ is uniformly continuous, $\{x_n\}$ and $\{x'_n\}$ are sequences in $(a, b)$ with $x_n \to b$, $x'_n \to b$, $f(x_n) \to y$, and $f(x'_n) \to \overline{y}$, prove $y = y'$.

So by the definition of convergent sequence, we know that for all $\delta > 0$, there exists an $N$ such that $d(x_n, b) < \delta$ for all $n > N$, and similarly a $N'$ with $x'_n$, so we can just take the max of $N$ and $N'$.

I'm not sure how to use uniform continuity here. We can't say that if $n > N$, then $d(x_n, b) < \delta \implies d(f(x_n), f(b)) < \epsilon$ by uniform confinuity since $b$ is not in the domain of the function, so $f(b)$ doesn't make sense.

What is the correct approach I should take? Is there a way to use Cauchy sequences here?

Answer 1

Hint: Define a sequence by $(\tilde{x}_n)_{n \in \mathbb{N}} := (x_1,x_1',x_2,x_2',\ldots)$.

• Show that $(f(\tilde{x}_n))_{n \in \mathbb{N}}$ is a Cauchy sequence; hence, convergent.
• Choosing suitable subsequences, deduce that the limit equals $y$ and $y'$; hence, $y=y'$.