My version of Bolzano-Weierstrass theorem states that:
"A bounded sequence in $\Bbb R$ always have a convergent subsequence."
Till now, I only knew one or two standard proofs that are always cited in books.
But suddenly when I , out of just sheer curiosity tried proving it, I found that there is probably a much shorter and concise way of proving it. Surprisingly enough, till now, never did I encounter this way of proving.
I will state that hereby:
Let $(x_n)$ be a bounded sequence. This means, $X=(x_n)$ and there exists a real number $M$ such that $|x_n|\leq M,\forall n\in\Bbb N.$
This means, that the set $X$ or the sequence is bounded above. As the Set $X\neq \emptyset,$ we can conclude by the Axiom of Completeness that $s=\sup X\in\Bbb R$ exists.
We claim, that $s$ is a limit point of $X$. In other words, the claim implies, that there exists a convergent subsequence of the sequence $(x_n).$
For every $\frac{1}{n}$ (,where $n\in\Bbb N$) we choose, an $a_n\in X$ such that, $s-\frac 1n\lt a_n\leq s\lt s+\frac{1}{n},$ and the existence of such an $a_n$ is justified, as $s$ is a supremum. Continuing like this, we obtain a subsequence $(a_n)$ of $(x_n)(=X).$
We claim $\lim a_n=s.$
For any $\epsilon\gt 0$ there exists an $n_0\in\Bbb N$ such that $\frac{1}{n_0}\lt\epsilon$ (By Archimedean Property). Now, we have, $\exists n_0\in\Bbb N$ such that $$s-\epsilon\lt s-\frac {1}{n_0}\lt a_{n_0}\leq s\lt s+\frac{1}{n_0}\lt s+\epsilon,$$ and if $n\geq n_0$ then, $s-\epsilon\lt s-\frac{1}{n_0}\leq s-\frac{1}{n}\lt a_n\lt s+\frac 1n\leq s+\frac{1}{n_0}\lt s+\epsilon.$ This proves that $\lim a_n=s.$
Again, this establishes our initial claim as well as $(a_n)$ is a convergent subsequence of $(x_n).$
I don't understand whether this really is an alternative proof or not? I haven't been able to find any errors till now, so I want to know if this a valid argument?
Best Answer
One flaw that I see is that you didn't show that the $\{a_n\}$ are all distinct elements of $X$. You say you can find points $a_n$ in the set arbitrarily close to the supremum, but when the supremum is isolated you just end up with the same point repeated.
For example, consider the set $X = \{\pm1,\pm\tfrac{1}{2},\pm\tfrac{1}{3},\cdots\}$ The supremum $s$ is the point $1$. But following the procedure in your argument, the subsequence you obtain is $a_1, 1, 1, \cdots$, where there is a wide choice for $a_1$ but all subsequent points have to be $1$. That is not a (proper) subsequence.
The fix is to divide $X$ by half repeatedly and every time you do so one of the sub-division will always be infinite. The subdivisions get smaller and smaller (because $X$ is bounded) and being infinite always contain one element. Choosing one element of these divisions (there is always a fresh choice because the set is infinite) produces a convergent subsequence which may or may not converge to the supremum.