The author of my intro analysis text has an exercise to give a proof of Bolzano-Weierstrass using axiom of completeness directly.

Let $(a_n)$ be a bounded sequence, and define the set $$S=\{x \in \mathbb{R} : x<a_n \text{ for infinitely many terms } a_n\}.$$

Show that there exists a subsequence $a_{n_k}$ converging to $s = \sup S$.

I feel I am close. I know that for any $\epsilon > 0$, there must be infinitely many $a_n$ such that $\sup S - \epsilon < a_n < \sup S + \epsilon$. (If there were only finitely many $a_n$ in that interval, then $\sup S + \frac{\epsilon}{2} \in S$, contradicting $\sup S$ as an upper bound.) However, I don't know how to pinpoint a single subsequence $(a_{n_k})$ such that all such elements with $k \geq \text{ some } N$ are in this interval for all $\epsilon$.

  • $\begingroup$ I'd change the title: AOC usually means the Axiom of Choice. $\endgroup$ – YoTengoUnLCD Apr 11 '16 at 19:47
  • 1
    $\begingroup$ Choose $n_1$ that works for $\epsilon=1$ [If there are infinitely many, then of course there is at least one]. Then choose $n_2 > n_1$ that works for $\epsilon=1/2$, and so on. $\endgroup$ – GEdgar Apr 11 '16 at 20:02

Since $(a_n)$ is bounded, $S$ is nonempty and bounded above. So by AoC there exists an upper bound $s=\sup S$.

Consider $s-\frac1k$ and $s+\frac1k$ where k is an arbitrary but fixed natural number. Since any number smaller than $s$ is not an upper bound of $S$, so $\exists (s'\in S)(s-\frac1k < s')$. Observe the property which forms $S$, by transtivity of $<$, $s-\frac1k$ also has the property: $s-\frac1k < a_n$ for infinitely many terms $a_n$. Apply this similar reasoning on $s+\frac1k$ we can see that $s+\frac1k \notin S$, so there are none or only finitely many terms $a_n$ satisfying $1+\frac1k < a_n$, this is the same as saying there are infinite many terms $a_n$ satisfying $a_n \ge 1+\frac1k$. Combine these two parts we get: for all $k\in N$, there are infinitely many terms of $a_n$ satisfying $s-\frac1k < a_n \le s+\frac1k$.

The last statement gave us a hint of how to build a subsequence of $(a_n)$. For every different $k\in N$, we can pick a term from infinitely many terms that satisfy that inequality. For example we can pick $a_{n_1}$ from $\{a_n : s-1<a_n<s+1\}$. After we picked $a_{n_k}$, we only need to satisfy $n_{k+1}>n_k$ to make this is indeed a subsequence, no repeatition or backward happened. (This is always can be done because for every pick we have infinitely many terms in hand)

Then we need to check whether this subsequence $(a_{n_k})$ converges to something. By intuition this should be $\sup S$. To satisfy the inequality $|a_{n_k}-s|<\epsilon$ for every $\epsilon >0$, choose $K>\frac1\epsilon$. If $k\ge K$ then $\frac1k < \epsilon$ which implies $$s-\epsilon < s-\frac1k < a_{n_k} \le s+\frac1k < s+\epsilon$$

So the B-W theorem has been proved using AoC.

  • $\begingroup$ Thanks for good explanation. $\endgroup$ – Silent Jan 21 '17 at 11:20
  • $\begingroup$ "so there are none or only finitely many terms $a_n$ satisfying $1+\frac{1}{k}<a_n$, this is the same as saying there are infinite many terms $a_n$ satisfying $1+\frac{1}{k}\leq a_n$" I can't related these two statement same. now it's too late but can you give me any logic to convince it @Tuoyu.Z $\endgroup$ – emonhossain May 20 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.