1

Forgive me for my lack of formal notation, I haven't taken any classes on set theory, or any advanced math topics for that matter.

From my understanding based on the wikipedia entries, a well-ordered set must have a least element. I have seen posts where people will explain contradictions with this statement, such as the least element of R is a contradiction, as you can divide by two to get a smaller real number. How does this not contradict AoC?

Here's a quote from another post, where somebody was explaining how well-ordering makes sense.

"...don't confuse "ordered" and "countable". You demand to know ..."

Which explained (I think, anyways) why you can't find the nth element of the real numbers, because the cardinality of [1,2,3,...n] must be less than the cardinality of real numbers.

Maybe I misunderstood the premise, but doesn't this logic also apply to countable infinities? The smallest rational number, for example, can be divided by 2 to get a smaller number. This set can be 1:1 mapped the integers, so there must be a set of positive rational numbers [q1, q2, q3...] that maps 1:1 to the natural numbers [1,2,3,...], which does have a clear least element. Both share a cardinality of aleph0, so they have to map 1:1, right? Why does AoC not contradict itself?

Edit: So this has been marked as a duplicate, but I haven't found any questions that answer this for me. Allow me to rephrase? These two sets share cardinality. How can a least element exist in Q if we can definitively prove is does not exist? Does this mean my intuitive definition of least elements only apply to some sets? If so, how do we redefine the definition to talk about the least element of positive rational numbers, as it has to exist under the theorem. Is there anything we can say about this number? If not, doesn't this contradict the idea of the theorem?

Andrés E. Caicedo
  • 81,687
Tyler
  • 11
  • The well-ordering does not need to agree with any "natural ordering". You can explicitly define two different linear orderings on $\Bbb N$ itself, even more. If that doesn't break your brain, this shouldn't either. – Asaf Karagila Mar 08 '19 at 18:56
  • 2
    The usual correspondences between rational numbers and integers do not preserve order – J. W. Tanner Mar 08 '19 at 18:56
  • Is there anything that we can say about the order of rational numbers, for example? I suppose it makes sense that there can be different orderings of natural numbers, but the fact that they're of the same cardinality is what doesn't make sense to me intuitively. It sounds like you can have different cardinalities of the same magnitude, which I also believe contradicts AoC – Tyler Mar 08 '19 at 19:01
  • 1
    @MichaelBurr: The Axiom of Choice is equivalent to the statement that every set can be well-ordered. I don't know how much closer you can get. – Asaf Karagila Mar 08 '19 at 19:02
  • "In mathematics, the well-ordering theorem states that every set can be well-ordered. A set X is well-ordered by a strict total order if every non-empty subset of X has a least element under the ordering. This is also known as Zermelo's theorem and is equivalent to the axiom of choice" - Wikipedia Isn't this saying the two are identical? – Tyler Mar 08 '19 at 19:02
  • @Tyler There are two senses of "least" here, which may be confusing. There's "least" in the sense of smallest, and "least" in the sense of first in an ordering. The set $(0,1)$, for example, has no least element in the first sense because indeed for any $x$ we can find a smaller element, e.g., $x/2$. But we might have an ordering on that set (let's say that $1/2$ is the "first" element), so it could have a least element in the second sense. – Théophile Mar 08 '19 at 19:11
  • Well, here's what I make of that. In any finite set, the two "senses of a least element" are the same. When things get infinite, the two somehow differ. But this seems to defy the logic of the axiom for me. Zemelo called AoC an "unobjectionable logical principle", but this seems rather objectionable to me, which implies it's outside the scope of ideas Zemelo had when the theory was created in the first place, right? (slightly unrelated) Is there anything at all we can say about the smallest rational number? Is it the same as the smallest real number, or do we not know? – Tyler Mar 08 '19 at 19:18
  • 1
    @Tyler No, the ordering could be arbitrary even for a finite set. For example: given ${1,2,3,4,5}$, let me choose the ordering $3\prec2\prec4\prec5\prec1$. You can see that any subset has a least element with respect to the ordering (e.g., the least element of ${1,5}$ is $5$), but that this is different from how big or small the number itself is. – Théophile Mar 08 '19 at 19:21
  • Sorry, I meant for any well-ordered finite set! – Tyler Mar 08 '19 at 19:24
  • @Tyler Of course, any subset of the positive integers has a natural ordering that happens to correspond to the size of the numbers: $a \prec b$ if and only if $a < b$. So there are cases where the two senses of "least" coincide. But consider an ordering on $\Bbb Z$, for example. You would have to do something like $0 \prec 1 \prec -1 \prec 2 \prec -2 \prec \ldots$, whereas $1 > -1$ arithmetically. – Théophile Mar 08 '19 at 19:24
  • @Tyler Every finite set is well-ordered: you can just write them in any order you like. The same goes for countably infinite sets. The problem is with uncountably infinite sets. – Théophile Mar 08 '19 at 19:26
  • 1
    Well, thank you for your help, I hadn't considered that. I feel I have a better understanding of the subject, although not perfect. I'll try to investigate a bit more on properties and differences between countably and uncountably infinite sets! – Tyler Mar 08 '19 at 19:30
  • @Tyler Glad to help. Good luck! – Théophile Mar 08 '19 at 19:33

0 Answers0