This is the first half of proving Bolzano-Weierstrass theorem

Question

Just making sure I'm on the right track so far

Every bounded infinite set of real numbers has at least one accumulation point

Pf: Let S be a bounded set. Since S is bounded, there are real numbers $x$ and $y$ such that $S \subset [x,y]$

If $x_1$ is the midpoint then at least one of the sets $[x,x_1], [x_1, y]$ must contain an infinite set of members of S.

Choose either of these sets and rename it $[a_1,b_1]$

If $x_2$ is the midpoint of this set then one of the sets $[a_1,x_2] , [x_2, b_1]$ contains an infinite set of members of S

Again, we choose one and call it $[a_2,b_2]$, if $x_3$ is the midpoint of this set then,

$[a_2,x_3], [x_3,b_2]$ contains an infinite set of members of S

Choose one and call it $[a_3, b_3]$

We continue to do this until we obtain for each positive integer n, a closed interval $[a_n,b_n]$

We know three things now:

1. Since for each $[a_n,b_n]$ we take the midpoint of the set $[a_{n-1},b_{n-1}]$, we know that $b_n-a_n = 2^{-n}(y-x)$

2. Since we continually assume that $[a_i,b_i] _{i< n}$ contains infinite points, $[a_n,b_n]$ contains infinite points

3. Since our set gets smaller and smaller as we approach n , we can see that $[a_n,b_n] \subset [x,y]$ and we know that

Answer 1

Yes this is exactly correct. You know $[a_{n+1},b_{n+1}]\subseteq [a_n,b_n] $ for every $n$ and that $a_n-b_n \rightarrow 0$ as $n\rightarrow \infty$. You can define your point to be $x=\cap_{n\in \mathbb{N}}[a_n,b_n]$. Now you have to argue that $x$ actually is an accumulation point which you can do since your sets are closed.