Solution check: Uniform continuity

Question

While reading Bartle's book, i came across the following:
Problem:
Prove that if $f$ is uniformly continuous on a bounded subset $A$ of $mathbb{R},$ then $f$ is bounded on $A$.
Attempt:
Given $A subset mathbb{R}$ such that $A$ is bounded and $f: A rightarrow mathbb{R}$ an uniformly continuous function, we will prove that $f$ is bounded.
Given $(x_{n})$ a sequence of points in $A$ one can obtain a convergent subsequence $(x_{nk})$, by using the Bolzano-Weierstrass theorem. Since $f$ is uniformly continuous and $(x_{nk})$ is a Cauchy sequence (since it is convergent), $(f(x_{nk}))$ is also Cauchy.
Finally, since every Cauchy sequence is bounded, there exists $a$ and $b$ in $mathbb{R}$ such that: $$a leqslant fleft(x_{n k}right) leqslant b$$
Therefore, $f$ is bounded.
Questions:
1. Is the solution correct?
2. If the solution is indeed correct, is it well written?
3. If the solution is incorrect, can someone gently me explain why and provide a solution?
Thanks in advance, Lucas!

Falcon · Accepted Answer

It seems that you have the main idea but I disagree with your conclusion, you cannot conclude that $|f(x)| le a$ for all $x in A$ with what you wrote. Let us try by contradiction: suppose that $f$ is not bounded, $i.e.$ it exists $(x_n)_n subset A$ such that
$$forall K > 0, exists N_K in mathbb N~text{ s.t. }~ forall n ge N_K: |f(x_n)| ge K.$$
As you wrote above we can find a subsequence $(x_{n_k})_k$ such that $(|f(x_{n_k})|)_k$ converges to a certain $L in mathbb R^+$. Since the function $mathbb N to mathbb N: k mapsto n_k$ is strictly increasing, we find that $k le n_k$ (it is a well known result about strictly increasing function from $mathbb N$ to $mathbb N$, try to show it). Moreover, for $K = L + 1$ (for example), there is $N_K$ such that for all $k ge N_K$, $|f(x_{n_k})| > K$ which contradicts the fact that $(|f(x_{n_k})|)_k$ converges to $L$.

Solution check: Uniform continuity

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