Question about Bolzano-Weierstrass theorem for sequences, functions and sets

Question

There are three formulations of the Bolzano-Weierstrass theorem:

• Every infinite and bounded subset $A \subset \mathbb{R}$ has an accumulation point.
• Every bounded sequence $x_n \in \mathbb{R}^{\mathbb{N}}$ has convergent subsequence.
• Continuous function $f: [a,b] \to \mathbb{R}$ on bounded and closed interval $[a,b]$ is bounded and has minimum and maximum.

My question is: How is the first formulation connected to the third? There is obviously a connection between the first and second because the definition of the accumulation point of the sequence is that $M$ is the accumulation point if the sequence has a subsequence that converges to $M$. Still, I do not see what that has to do with the minimum and maximum of function? What is the point here?

Edit: What I have realized is that the Borel-Lebesgue theorem implies the first formulation here.

Answer 1

$(i)$, $(ii)$ and the Borel-Lebesgue theorem are equivalent, and also equivalent to the statement that the compact subsets of $\mathbb R$ are exactly its closed and bounded subsets.

$(iii)$ is usually thought of as very useful consequence of this fact, with the following proof :

$(i)\Rightarrow (iii)$ Let $f:[a,b]\to \mathbb R$ be a continuous function and $s = \sup_{x\in [a,b]}f(x) \in (-\infty,+\infty]$. There is a sequence $(x_n) $ of numbers in $[a,b]$ such that $\lim_{n\to +\infty} f(x_n) = s$. By $(i)$, it has a convergent subsequence $x_{\varphi(n)}\to x$. Since $f$ is continuous, we have :$$s = \lim_{n\to +\infty} f(x_n) = f(x)$$ Therefore $s<+\infty$ and $f$ reaches its maximum. The proof is identical for the minimum.

However, $(iii)$ can also be read as the statement that segments in $\mathbb R$ are pseudo-compact spaces. It can be proven that compact topological spaces are also pseudo-compact. The converse is not true in general, but holds for metric spaces (such as $\mathbb R$) (see this MathSE post)