In my Real Analysis textbook, it proved the Bolzano-Weierstrass theorem which states that every bounded sequence in $\mathbb{R}$ has a convergent subsequence. The author proved this theorem by splitting the interval in which the bounded sequence sits into two, and taking the half with infinitely many terms. He continued to split the interval with infinitely many terms into two. He then used the nested intervals theorem to conclude that the intersection of all the intervals is a singleton, and constructed a sequence that converges to that point by taking an element from every "half interval" (half in the sense that it was obtained by cutting a larger interval in half).
As an exercise, it asks me to prove the same theorem in the following way:
Let $A = \{x \in \mathbb{R} \, | \, \exists \, \text{infinitely many} \, a_n\,\text{'s such that} \, x < a_n\}$. Show that $\{a_n\}$ has a subsequence converging to $\sup A$.
Since this also proves the Bolzano-Weierstrass theorem, I thought I could apply the same procedure, but either it is a dead-end, or I got confused because I failed.
Best Answer
Since the sequence $(a_n)$ is bounded it is clear that $A$ is bounded. Hence $\sup A$ exists. Now for each $k \in \mathbb{N}$, by the definition of supremum, there is $x_k \in A$ such that $x_k >\sup A - \frac{1}{k}$. Hence there is an infinite family $J_k \subset \mathbb{N}$ such that $a_n >x$, for all $n \in J_k$. If $a_n> \sup A$ for infinitely many $n \in J_k$ (there is infinite set $I_k \subset J_k$ such that this happens) you have that $(a_n)_{n \in I_k}$ converges to $\sup A$. If not there is $\epsilon>0$ such that $a_n>\sup A +\epsilon$ for all $n \in I_k$ contradicting the definition of supremum because in this way $\sup A +\epsilon \in A$. So if some $k \in \mathbb{N}$ we have that $a_n> \sup A$ for infinitely many $n \in J_k$ we know that there is a subsequence converging to $\sup A$. Suppose this is not the case for all $k \in \mathbb{N}$. Hence, for each $k \in \mathbb{N}$, there is an infinite family (again denoted by $J_k$) such that $\sup A \geq a_n >x_k > \sup A -\frac{1}{k}$ for all $n \in J_k$. Now, choose $n_1 \in J_1$. Choose $n_2 \in J_2 \setminus \{n_1\}$. And for the $k-th$ case choose $n_k \in J_k \setminus \{n_1, \cdots, n_{k-1}\}$ (This is always possible since the sets are infinite). Now notice the subsequence $a_{n_k}$ converges to $\sup A$ because for all $k\in \mathbb{N}$ $$ \sup A \geq a_{n_k} > \sup A -\frac{1}{k}. $$