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

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?

• And what have you tried? What do you think? – t.b. Aug 4 '12 at 11:01

See Herrlich's book The Axiom of Choice , 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]).

### Bibliography:

1. Herrlich, H. Axiom of Choice. Lecture Notes in Mathematics, Springer, 2006.
• Thank you for the reference :D – gifty Aug 4 '12 at 11:11
• I never thought much about compactness in the absence of AC, but I wonder if the different notions of compactness (for metric spaces, say) are even equivalent without AC? And if not, which notion of compactness is referred to here? – Harald Hanche-Olsen Aug 4 '12 at 12:22
• @Harald: It's the topological definition (every open cover has a finite subcover) that is meant here, which is the strongest notion of compactness among the standard ones. For metric spaces topological compactness implies a) total boundedness and completeness b) countable compactness and c) every sequence has an accumulation points, each of which in turn implies sequential compactness. The converses are wrong without choice but with countable choice you can show that sequential compactness implies compactness, so these are equivalent. See also this. – t.b. Aug 4 '12 at 14:14
• @HaraldHanche-Olsen: Herrlich has a nice treatment of this question in chapter 3 (in which the theorem appears). The concepts do split up, and if some of them are equivalent then the axiom of choice holds. Talking only about metric spaces then indeed there are non-equivalent forms of compactness, but their equivalence is not necessarily equivalent to the axiom of choice anymore. – Asaf Karagila Aug 4 '12 at 14:15
• heh, more or less simultaneous... – t.b. Aug 4 '12 at 14:16