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At this page:

you can find an example of a compact but not sequentially compact space. My question is: how to prove the existence of "$r \in I$ such that its binary expansion has its $k$-th digit $0$ iff $k$ is odd, and $1$ otherwise"? intuitively it is clear, but in practice do we need AC (or, at least, the countable AC)?

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Please try to make the question self contained (at least in notation), is usually causing me some problems when loading up. – Asaf Karagila Oct 13 '11 at 16:17
The example in that page is not completely formal: for one thing, it does not say what to do with numbers that have dual binary expansion... – Arturo Magidin Oct 13 '11 at 16:20
There seems to be a typo in that page. They actually want $r$ to be a number such that the $n_k$th digit is $0$ if $k$ is odd and $1$ otherwise. But the arguments below work fine to ensure the existence of such a number. – Nate Eldredge Oct 13 '11 at 17:20
up vote 2 down vote accepted

The existence of the real number $r\in[0,1]=I$ such that its binary expansion has $k$th digit $0$ if $k$ is odd and $1$ if $k$ is even does not depend on the Axiom of Choice in any way whatsoever. This is nothing more than the number $$r = \sum_{i=0}^{\infty}\frac{1}{2^{2i+1}},$$ which is a convergent series of real numbers, being a series of positive terms that is bounded above by $$\sum_{i=1}^{\infty}\frac{1}{2^i} = 1.$$

Where exactly do you believe that this requires the Axiom of Choice? If your real numbers are defined as equivalence classes of Cauchy sequences, then $r$ is the equivalence class of the sequence of its partial sums, which can be defined using induction/recursion theorem. If your real numbers are defined as Dedekind cuts, then $r$ is the cut determined by the union of the cuts determined by the partial sums, which again can be defined using recursion/induction.

(P.S: $$\begin{align*} r &= \sum_{i=0}^{\infty}\frac{1}{2^{2i+1}} = \frac{1}{2}\sum_{i=0}^{\infty}\frac{1}{2^{2i}}\\ &= \frac{1}{2}\sum_{i=0}^{\infty}\left(\frac{1}{4}\right)^i\\ &=\frac{1}{2}\left(\frac{1}{1-\frac{1}{4}}\right)\\ &= \frac{1}{2}\left(\frac{4}{3}\right)\\ &=\frac{2}{3}. \end{align*}$$ and, luckily, $\frac{2}{3}$ exists, even without the Axiom of Choice...)

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Indeed, this number is more conveniently written as $r=2/3$. – Nate Eldredge Oct 13 '11 at 17:18

Firstly, the definitions of compactness without the axiom of choice are not equivalent as you may want them to.

Secondly, Tychonoff's theorem is in fact equivalent to the axiom of choice. Indeed in this case where we take products of $[0,1]$ we can relax the demand for weaker choice principles (for example ultrafilter lemma, maybe less for this specific case).

However to define $r$ as $r(n)=0\iff n=2k+1$ you do not need the axiom of choice. Consider $P(\omega)$ under the equivalence relation:

$$A\sim B\iff A=B\lor \left(A=\omega\setminus B\land\left(|A|<\omega\lor |B|<\omega\right)\right)$$

That is two sets are equivalent if one is finite and the other one is its complement, or if they are equal. It is quite simple to see that this is an equivalence relation and every equivalence class is either a singleton of a pair.

We can also take $\mathcal A$ to be all the subsets which are only equivalent to themselves; and the finite sets when the equivalence class is a pair.

Define the bijection $f\colon\mathcal A\to[0,1]$ as: $$f(A)=\sum_{n\in A} 2^{-n}$$

The number $r$ which you seek to define is exactly $f(\{2k\mid k\in\omega\})$. All this, without any choice.

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