1

Here's Lemma 9.2 in Topology by James R. Munkres, 2nd edition:

Given a collection $\mathscr{B}$ of non-empty sets (not necessarily disjoint), there exists a function $$c \colon \mathscr{B} \to \bigcup_{B \in \mathscr{B}} B$$ such that $$c(B) \in B$$ for each $B \in \mathscr{B}$.

The function $c$ is called a choice function for the collection $\mathscr{B}$.

This lemma has been derived using the Axiom of Choice, which is as follows:

Given a collection $\mathscr{A}$ of disjoint non-empty sets, there exists a set $C$ consisting of exactly one element from each element of $\mathscr{A}$; that is, a set $C$ such that $C$ is contained in the union of the elements of $\mathscr{A}$, and for each $A \in \mathscr{A}$, the set $C \cap A$ contains a single element.

The set $C$ can be thought of as having been obtained by choosing one element from each of the sets in $\mathscr{A}$.

Now let $\mathscr{B}$ be the collection of all the non-empty subsets of the set $\{0,1\}^\omega$ of all the infinite sequences of $0$s and $1$s.

Can we explicitly find a choice function for this collection without using the Axiom of Choice?

I know that this is possible for each of the collections of all the non-empty subsets of $\mathbb{N}$, $\mathbb{Z}$, and $\mathbb{Q}$, respectively.

Please note that my knowledge of mathematical logic and set theory is very very limited. So I would like to have an answer to my question using only the concepts that Munkres has talked about in the first nine sections of his book.

The suggested questions are not formulated in this particular context.

Saaqib Mahmood
  • 26,762
  • Since there are bijections between ${0,1}^\omega$ and $\Bbb R$, and I have yet to see a well-ordering of the reals, I would assume the answer is no. – Arthur Dec 23 '15 at 11:13
  • @Arthur what have you edited in my original post? Moreover, I have not been able to see what you've stated in your comment. Can you please either elaborate it in an answer or else formulate an argument without using the concept of well-ordering? – Saaqib Mahmood Dec 23 '15 at 11:32
  • I edited the \emph{choice function} into _choice function_, which is how SE does it (click the "edited x mins ago" link, and you can see details of all edits done). Also, what I mean is that if you have a choice function on a set (i.e. a function $f$ that for every non-empty subset $B$ assign a value $f(B)\in B$), then you can use that to make a well-ordering. A well-ordering of ${0,1}^\omega$ would, by bijectivity, lead to a well-ordering of $\Bbb R$, which as far as I know, cannot be proven to exist without the axiom of choice. – Arthur Dec 23 '15 at 11:46
  • @Asaf Karagila please have a look at my question now. I have editted it to make it more specific. – Saaqib Mahmood Dec 23 '15 at 12:16
  • @Arthur please have a look at my question now. I'm sure you'll be answer it much better this time around. – Saaqib Mahmood Dec 23 '15 at 12:17
  • I'm sorry, not every "simple question" has an elementary answer. – Asaf Karagila Dec 23 '15 at 12:50
  • @AsafKaragila you're absolutely right. But can you please answer my question in the framework of the discussion in the first nine sections of Munkres' Topology, 2nd edition? That's where I've encountered this problem. – Saaqib Mahmood Dec 23 '15 at 13:12
  • @AsafKaragila can you please at least remove the "duplicate" qualifier from my question? You see, none of the answers to the questions you've pointed out serves my purpose. – Saaqib Mahmood Dec 23 '15 at 13:14
  • I am having trouble understanding something. What sort of answer will satisfy you? Giving a valid mathematical explanation is impossible because your knowledge in set theory is very limited; so it's either "Trust me, it's not possible without the axiom of choice" which frankly isn't an answer to begin with, or something else? What do you expect to have here as an answer? I am not entirely sure what you don't understand in the given answers to the duplicate (and its duplicate). Don't just say "Oh, I don't get it". Tell me what you don't get, and help me help you. – Asaf Karagila Dec 23 '15 at 13:47
  • 1
    (And while I am willing to go to very great lengths to help people understand set theory and the axiom of choice, I am not going to sit and read through the first nine chapters of Munkres just for that.) – Asaf Karagila Dec 23 '15 at 13:47
  • @SaaqibMahmuud I think you should read a proper book of set theory - namely Hrbacek, Jech, Introduction to set theory, or the less formal Halmos, Naive set theory - instead of asking others to review nine chapters of your topology book in search of all the tiny technical lemmas that you need. –  Dec 23 '15 at 14:04
  • @AsafKaragila I get your point. This question occurs in Section 9, which is in Chapter One, and not in Chapter 9 of Munkres. Yes, you can't be expected to go through Munkres just for answering my questions, but perhaps someone else,who might have happened to have gone through this very book, might read my question and answer it. Is this not likely either? – Saaqib Mahmood Dec 23 '15 at 16:21
  • @G.Sassatelli thanks for your valuable advice. I would certainly like to read Halmos sooner rather than later, but I'm so slow! And, I often find it rather difficult to concentrate. – Saaqib Mahmood Dec 23 '15 at 16:24
  • 1
    I didn't like Halmos' style. Try Enderton Elements of Set Theory... – Asaf Karagila Dec 23 '15 at 16:48
  • @AsafKaragila I'd certainly like to follow your advice. I'll also look into the answers already posted on SE and see what stumps me. Thank you very much. – Saaqib Mahmood Dec 24 '15 at 07:18

0 Answers0