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Is there an example of an ordered, uncountable set in $\Bbb R$?

My Calculus professor, who likes to keep things simple, defined a sequence in $\Bbb R$ as an "ordered and infinite list of real numbers". "Ordered" here means there is a definitive first, second, ..., $n$ th term in the list.

My initial reaction was the definition was ambiguous. But could not come up with an example. Is the definition in fact strong? Does the definition of "ordered" in this context imply a denumeration?

Ishfaaq
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    If you are sticking with the typical order on the reals there are no uncountable ordered subsets of the reals. If there are no restrictions on ordering then every subset can be ordered in ZFC. –  Mar 09 '14 at 03:28
  • With you definition every subset of $\Bbb R$ is ordered. – sas Mar 09 '14 at 03:29
  • @Paul Plummer: So do you mean to say that it is fine to define a sequence this way? – Ishfaaq Mar 09 '14 at 03:34
  • A sequence can be seen as an ordered list and is typically considered countable but there are generalizations to include uncountable lists. –  Mar 09 '14 at 03:40
  • If you are familiar with functions a definition I like for a sequence of reals is a function $f: \mathbb{N} \to \mathbb{R}$, and the $n^{th}$ term in the sequence would be $f(n)$. If taken as the definition implies all sequences are countable in length. (Sorry if my first comment was a tad confusing, my eyes skimmed over the word sequence in the post) –  Mar 09 '14 at 03:44
  • @Paul Plummer: I like this definition too. It's the one you find in most Analysis books. Simple and unambiguous. I have no issue with it. My issue is with the definition given by my Prof. Is it equivalent? – Ishfaaq Mar 09 '14 at 03:51
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    I would bet that is what the professor means, but one can have "sequences" that are longer than the natural numbers, or even uncountable (I put quotes since sequences are typically assumed to be countable). I guess technically your prof's definition includes these longer sequences and as others have pointed out, within ZFC, you can order the whole real line, or any subset, so there is always a next thing in the list. –  Mar 09 '14 at 04:11
  • In the context of real analysis, particularly studying continuity of the the real numbers, I don't think these longer sequences add anything due to some nice topological properties of the reals. In more general topological situations you do need a more general concenpt than sequences though. You may want to look into nets and filters. –  Mar 09 '14 at 04:16
  • Thanks everyone for the answers. Really appreciate it. But I think my doubts were largely cleared by the comments from @Paul Plummer . I'm going to find my teacher tomorrow and tell him that I find "ordered" too ambiguous a term and to see if he agrees to replace it with "eumerated" so a sequence in $\Bbb R$ is an "enumerated and infinite list of real numbers" which I think reflects the formal definition – Ishfaaq Mar 09 '14 at 05:22

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From what I understand, your question comes down to why a set being ordered does not imply it being countable.

The slightly subtle notion here is the difference between cardinal and ordinal numbers. In general, there are two standard ways to compare the sizes of two sets. You can construct a bijection (as you've probably seen), or you can construct an order-preserving bijection.

For instance, you've probably seen a bijection between $\mathbb{N}$ and $\mathbb{Q}$, but there is no order preserving bijection, so we say both sets have cardinal number $\aleph_0$ but ordinal numbers $\omega$ and $\omega^2$ respectively.

More visual perhaps is the difference between $A=\{1,2,3,...\}$ and $B=\{1,3,...,2,4,...\}$. In one case all the natural numbers are listed sequentially. In the second, all odd numbers come before the first even. We have clearly that $|A|_{card}=|B|_{card}$, but we say that $|B|_{ord}=\omega * 2$.

Now, the notion you want here is well ordering. A set is well ordered if every subset has a least element. In that sense, if we use an assumption called the axiom of choice, we can construct a well-ordered set which is uncountable, but we will never be able to completely describe it (and that's one reason why the axiom of choice has been somewhat controversial).

Given these assumptions, a well ordering of a uncountable set will contain lots and lots of ...'s. So for instance, we can write the open interval $(0,1)=\{0.1, 0.11, 0.111,...0.2, 0.22, 0.222, ..., 0.21, 0.211,...\}$, but its clear we will never be able to list all the terms in that set in a way we can assign a natural number to every term. Nonetheless, it can be shown that there existis some well ordering of every uncountable set, but this ordering will certainly not be the same as the regular ordering of $\mathbb{R}$

Felipe Jacob
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  • There's an inaccuracy here: the set of all rational numbers is not well-ordered by the usual order relation, so its order type is not ${\omega}^2$ or any other ordinal number. (In fact an exact opposite is true: the set of all rational numbers is dense under the usual order.) – Mike Rosoft Jan 02 '21 at 08:33
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The interval $(0,1)$ is an ordered, uncountable set in $\mathbb{R}$.

It is perhaps slightly tighter to restate his definition as a map from the natural numbers $\{1,2,3,\dots\}$ to the set in question, $\mathbb{R}$ in this case. So while there are uncountable ordered sets in $\mathbb{R}$, they are not sequences since there are not enough natural numbers to completely cover the uncountable set with their images under the function.

I introduce the term "well-ordered", meaning "every subset has a least element". If a set is well-ordered, we can put it into correspondence with the natural numbers by listing, first, the least element of the whole set. Then we list the least element of the whole set without its least element. Then we continue, removing one element at a time, adding it to our list, and considering the least element of the remainder of the set. Zermelo (of Zermelo-Fraenkel set theory) proved that, assuming the Axiom of Choice, every set can be well-ordered. See Well-ordering theorem at the English Wikipedia.

The Well-ordering theorem is surprising. It's sufficiently surprising that it causes some philosophical disagreements among some mathematicians about the validity of the Axiom of Choice. This is also one of the important differences between cardinals and ordinals. You and I have both made arguments comparing the cardinalities of the natural numbers versus uncountable sets. The only thing we've said about ordinals is that the Axiom of Choice makes it possible to provide a well-ordering to any set, including uncountable sets.

Knuth's "Surreal Numbers" has a very readable introduction to the ordinals, as well as explaining how there can be enough of them to well-order any set. (In short, there are a lot of ordinals per cardinal.)

Eric Towers
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  • Yes I know the formal definition of a sequence. All that is fine. ordered here is in the sense that I have mentioned in the question. What then is the $n$th element in $(0, 1)$?? – Ishfaaq Mar 09 '14 at 03:24
  • @Ishfaaq: Every set is ordered by your definition (in ZFC). [ http://en.wikipedia.org/wiki/Well-ordering_theorem ] – Eric Towers Mar 09 '14 at 03:27
  • Thanks for that. I'll keep reading. But my knowledge on Set Theory is pitiful. So pardon me. But this still does not answer my question. There is no definitive $n$th term in $(0, 1)$ is there. Suppose there is. Then there are first, second,... $(n - 1)$th terms using which I can construct an incredible many more. – Ishfaaq Mar 09 '14 at 03:32
  • @Ishfaaq: don't confuse "ordered" and "countable". You demand to know $n$th element, which in fact the same as defining it "countable". Definition of "ordered" is existing relation ">" for every two elements of set. Now, assuming confusing of two terms, re-word your question as "An example of countable, uncountable set?" and you'll get the answer yourself. – Van Jone Mar 09 '14 at 03:49
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A sequence can be seen as an ordered list and is typically considered countable, especially in real analysis/calculus, but there are uncountable lists.

I would bet that you professor means countable list (ordered by the natural numbers), but one can have "sequences" that are longer than the natural numbers, or even uncountable (I put quotes since sequences are typically assumed to be countable). I guess technically your prof's definition includes these longer lists and as others have pointed out, within ZFC, you can order the whole real line, or any set so there is an $n^{th}$ element (where $n$ is an ordinal).