Bolzano -Weierstrass Theorem and uniform Continuity

Question

The following problem has hints, but I am unable presently to use it.

Suppose $f$ is uniformly continuous on $(a,b]$, and let $\{x_n\}$ be any fixed sequence in $(a,b]$ converging to $a$. Show that the sequence $\{f(x_n)\}$ has a convergent subsequence (Hint: Use the Bolzano Weierstrass Theorem)

Now let $L$ be the limit of the convergent subsequence of $\{f(x_n)\}$. Prove, using the uniform continuity of $f$ on $(a,b]$, that $$\lim_{x\to a^+}f(x)=L.$$

Answer 1

Hint: Bolzano Weierstrass says that every bounded real sequence has a converging subsequence. If $x_n \to a$ and $f$ is uniformly continuous. Is $(f(x_n))_{n\geq 1} \subset \mathbb{R}$ bounded?

Answer 2

$f$ is uniformly continuous on $(a,b]$ and is therefore bounded. $x(n)$ is a fixed sequence in $(a,b]$ converging to a so $f(x(n))$ is a bounded sequence. By Bolzano Weierstrass $f(x(n))$ has a convergent subsequence. Let $L$ be the limit of the convergent subsequence, $x(n_k)\to x$ so therefore $$f(x(n_k))\to f(x)$$ as $f$ is uniformly continuous. So as a sequence can only have one limit, $$f(x(n))\to L.$$

Answer 3

Since $f$ is uniformly continuous on $(a,b]$, it is bounded. Moreover, $\lim\limits_{x\rightarrow a}f(x)$ exists, say $c$. So we can define \begin{align} g(x)=\begin{cases}f(x), \,\,\,x\in (a,b]\\ c,\,\,\,x=c.\end{cases}\end{align} Then $[a,b]$ is a closed and bounded interval by the intermediate value theorem. Then applying Bolzano-Weierstrass theorem on $g([a,b])$ yields the result.