Is a countable product of compact intervals in $\mathbf R$ compact (without using the AC)?

Question

Let $\{I_n=[a_n,b_n]\}_{n\in\mathbf N}$ be a countable collection of closed, bounded intervals in $\mathbf R$. Is the infinite Cartesian product $$\prod_{n=1}^\infty I_n$$ compact without using the Axiom of Choice?

Answer 1

See Herrlich's book The Axiom of Choice [1], Theorem 3.13:

(ZF) $[0,1]^\mathbb N$ is compact.

Of course your question is about a general product of closed intervals, however note that the functions $f_n\colon[a_n,b_n]\to[0,1]$ defined by $$f_n(x)=\frac{x-a_n}{b_n-a_n}$$ are homeomorphisms, and therefore we can transfer the product of closed intervals to a product of $[0,1]$.

It should also be noted that this is no longer true for general uncountable products, indeed even $[0,1]^\mathbb R$ may fail to be compact (see [1, Section 4.8: theorem 4.70 and E 13]).

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Bibliography:

1. Herrlich, H. Axiom of Choice. Lecture Notes in Mathematics, Springer, 2006.