Talk:Countable set/Archive 1

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1 Whether countable implies infinite (part 1)
2 Countable ordinals
3 Proof of the countability of Q
4 Whether countable implies infinite (part 2)
5 Whether countable implies infinite (part 3)
6 Query
7 Axiom of choice
8 Numbers or not numbers
9 Question
10 1-on-1 correspondences

[edit] Whether countable implies infinite (part 1)

This page defined countable to mean "countably infinite". I've changed this, because in my experience it usually means "countably infinite or finite" (i.e., the opposite of uncountable). Note that we already have an article on countably infinite, so there is some duplication here.
Zundark 2001-08-12

[edit] Countable ordinals

This page address cardinals, but what does it mean to say an ordinal is countable or uncountable?

A countable ordinal is just an ordinal which is countable as a set, i.e. its cardinal is aleph_0. An uncountable ordinal is an ordinal which is not finite and not countable (as a set). --AV

[edit] Proof of the countability of Q

Why would we prove the countability of Q by taking it as a countable union of countable sets? The usual way is just to present an ordering, and is much simpler, IMO.

0, 1, -1, 2, -2, 1/2, -1/2, 3, -3, 1/3, -1/3, 4, -4, 5, -5, 1/5, -1/5, 6, -6, 3/2, -3/2, 2/3, -2/3, 1/6, -1/6, ...

What system are you following here? When will 1/4 and 5/2 show up? --AxelBoldt

[edit] Whether countable implies infinite (part 2)

I checked about a dozen random textbooks I have, and all but 2 defined "denumerable" as "countably infinite", although the 2 that defined it as "countable" are quite well-known, so I'll put a comment indicating it's not completely universal agreement. Revolver 19:12, 9 Jul 2004 (UTC)

[edit] Whether countable implies infinite (part 3)

By definition of some books (i.e. Walter Rudin's Principles of Mathematical Analysis) a countable set must be infinite, and a finite set is not countable. However, he uses the terminology (not mentioned in Wikipedia): 'at most countable'. Both countable (infinite) and finite sets are 'at most countable'. An infinite sum of finite sets, for example, is at most countable. To prove that this sum is strictly 'countable' one must prove that it is infinite.

Nerses Ohanyan

It's a bit strange that the use of countable to mean countably infinite wasn't mentioned in the article. I've added it now. --Zundark 07:15, 17 Sep 2004 (UTC)

???? But it's in the previous sentence? "A countable set is a set which is either finite or countably infinite." I've reverted. [[User:Noisy|Noisy | Talk]] 09:58, 17 Sep 2004 (UTC)

No, that's a different definition. By that definition every finite set is countable, whereas by the definition Nerses Ohanyan was talking about, no finite set is countable. So I'm going to revert back, perhaps with some extra clarification. --Zundark 10:09, 17 Sep 2004 (UTC)

OK - I think I understood on the third or fourth reading. Boy, I'm glad I'm not a mathematician. ;-) [[User:Noisy|Noisy | Talk]] 10:31, 17 Sep 2004 (UTC)

[edit] Query

Does countable mean that an element in set B can be indexed? Or am I missing something In other words set A and set B are bijective if I can write array[n]=Bk, where n is a natural number, is this sufficient for B to be countable? 59.145.141.130 10:49, 19 January 2006 (UTC)

Um, yes. If I understand you correctly, that's the definition of countability. —Keenan Pepper 17:46, 19 January 2006 (UTC)

[edit] Axiom of choice

"THEOREM: (Assuming the axiom of choice) The union of countably many countable sets is countable."

I think we don't need the axiom of choice here, as we can choose by order of counting. The Infidel 09:45, 22 January 2006 (UTC)

You do need at least a weak form of the Axiom of Choice. The result can't be proved in ZF. --Zundark 12:48, 22 January 2006 (UTC)

But where? There's scetch of proof in the article. I can't see anything that can not be done by the enumeration functions of these countable sets (and set of sets). 84.160.220.72 14:49, 22 January 2006 (UTC)

But first you must choose those infinitely many enumeration functions. --Zundark 16:04, 22 January 2006 (UTC)

I'm not sure about that. If a set is countable, there exists an enumeration. Any one will do.And I don't need a choice function that gives an enumeration for every subset of enumerations but only one enumeration out of the full set of all enumerations. The Infidel 19:06, 22 January 2006 (UTC)

Let the original sets be A_n, for n in N. You need an enumeration for each A_n. For each n, let E_n be the set of all enumerations for A_n. Each E_n is non-empty (as you point out). You need to choose an element from each of the infinitely many non-empty sets E_n. It's the Axiom of Choice that says you can do this. --Zundark 20:28, 22 January 2006 (UTC)

Look at it this way: let $E = \times E_n$ be the cartesian product of all the sets of enumeration on the A_n. Now for any element ${\Phi} = (\varphi_1, \varphi_2, \ldots ) \in E$ there is an enumeration $\varphi_1(1), \varphi_1(2), \varphi_2(1), \varphi_1(3), \varphi_2(2), \ldots$ . If we have the element $Φ$ , we have the enumeration. No need to choose. If we don't have the element, we at least have the set of all the elements, which is not empty. So we know that an enumeration exists, even if we should not be able to name it. Thus the union of our contably many countable sets is countable. The Infidel 19:04, 23 January 2006 (UTC)

You're implicitly assuming the Axiom of Choice in deciding that it's possible to choose an element out of the Cartesian product of infinitely many sets (since that's equivalent to choosing one element from each of the infinitely many sets). Ruakh 20:55, 23 January 2006 (UTC)

It's not necessary to choose an element, it suffices to know it's in there. No need to choose an enumeration, as long as we can show there is one. The Infidel 21:47, 23 January 2006 (UTC)

The point is that you don't know there's an element in the Cartesian product if you don't have the Axiom of Choice. Without the Axiom of Choice, a Cartesian product of non-empty sets can be empty. --Zundark 22:44, 23 January 2006 (UTC)

That was realy a point. I'm pretty sure I could construct a cartesian product as a restriction of some power set, but I never checked my "tools" if they depend on the axiom of choice or not. Can you provide any source, evidence or hint for your statement? The Infidel 18:26, 24 January 2006 (UTC)

If the Axiom of Choice were false, then there would be a set of non-empty sets without a choice function. The Cartesian product of these non-empty sets would be empty, because any element of the Cartesian product gives a choice function. --Zundark 19:26, 24 January 2006 (UTC)

Ok, for the moment, I will not dispute that the axiom of choice would be necessary to proof the theorem, as I am not able to state such a proof. On the other hand, I did not see a proof that it is necessary.

I feel your proof, if correct, is strong evidence that the axiom of choice holds. Otherwise, the product of some non-zero cardinal numbers is zero or undefined (and as a relation rather than a function, includes zero).

Thinking ever more strictly about sets and proofs, doubting more and more of the usual ways of deduction, how exactly is it that an element of a Cartesian product as you described will give a choice function? You will need a projection to the components, but how do you do that? I would not be astonished to find you will need the axiom of choice here.

PS: is this talk page still the appropriate place for this discussion? Shouldn't we switch to our own talk pages?

The Infidel 20:33, 24 January 2006 (UTC)

PS PS: Don't take too literally what I said about the product of cardinal numbers, I reckon we are talking about classes here rather than sets. The Infidel 20:46, 24 January 2006 (UTC)

Remember that the Axiom of Choice is known to be true for any finite collection of sets; the difficulty comes in when we're talking about an infinite collection of sets. So, 1*2 is well-defined and nonzero, but 1*2*3*... might not be. Now, obviously 1*2*3*... is going to be greater than zero, because obviously the Axiom of Choice is true, but that doesn't change the fact that we're relying on the Axiom of Choice (or some equally non-ZF assumption). Ruakh 20:56, 24 January 2006 (UTC)

I see that for infinite cardinals this is not (provably) incorrect, but still very counterintuitive. After all, as we can't proof axioms, they are not the least choosen because they are intuitive. The Infidel 21:10, 24 January 2006 (UTC)

To answer some of The Infidel's points:

Set theorists have constructed models of ZF in which the union of countably many countable sets is sometimes uncountable. This shows that the Axiom of Choice (or a weak form of it) is necessary for this theorem.

As far as I know, there is no reasonable way to define products of infinitely many cardinal numbers without the Axiom of Choice.

An element of a Cartesian product is a function from the index set I into the union of the base sets A_i such that each i maps into A_i. So an element easily gives you a choice function if the A_i are all distinct. Projection maps can be defined without the Axiom of Choice, but they aren't needed here.

--Zundark 22:16, 24 January 2006 (UTC)

That's cool. I didn't know that. This clarifies a lot. Thank you. The Infidel 18:10, 26 January 2006 (UTC)

[edit] Numbers or not numbers

Someone recently added:

Note that the number of elements of the set of natural numbers is not itself a number in the sense that you can do the usual kind of addition and subtraction.

I couldn't tell what this meant exactly, so I removed it. You can add cardinalities of sets, simply by taking two disjoint sets with those cardinalities and taking the cardinality of their union. —Keenan Pepper 17:17, 22 January 2006 (UTC)

After your reverting, the article states "In mathematics the term countable is used to describe the size of a set, i.e. the number of elements it contains.". For infinite sets, this "number" is not a number in the "usual" sense: you cannot add up, subtract or divide it in any way we are used to do with natural numbers or the elements of a field or ring. For the "number" of elements of a countably infinte set, the equation $\aleph_0 + 1 =\aleph_0$ holds. If you remove anything in wikipedia that you don't understand exactly and without asking, there'll probably be nothing left. The Infidel 18:56, 22 January 2006 (UTC)

The new version by Ruakh is better all around. —Keenan Pepper 21:58, 22 January 2006 (UTC)

Thanks! I try. :-) Ruakh 22:51, 22 January 2006 (UTC)

It looks better now. My main point was that someone without knowledge about cardinal numbers would use everyday maths on equations like $\aleph_0 + \aleph_0 =\aleph_0$ , subtracting aleph on both sides and ask on the help page if zero equals infinity. The Infidel 18:38, 23 January 2006 (UTC)

[edit] Question

Normally with infinite sets you work in the direction of infinity (the big end).
Let's go the other way.

Suppose you have an infinite set S.
Now you remove one element.
and another
and another
...

As I see it, you will never come back into the finite world again in this way.
Is this logically valid? Can anybody explain this to me?
Landheha 21:25, 6 March 2006 (UTC)

It seems perfectly valid. What do you need explained? —Keenan Pepper 22:39, 6 March 2006 (UTC)

Well, when you keep removing elements from a set (infinite or not) you should be able to come back to the empty set at one time or another, shouldn't you?
I don't see this happening for infinite sets. Do you? Landheha 22:53, 6 March 2006 (UTC)

If you start with the empty set and add one element at a time at a fixed, finite rate, you'll never have an infinite set. Conversely, if you start with an infinite set and remove one element at a time at a fixed, finite rate, you'll never have an empty set. It makes sense to me; what do you find odd about it? Ruakh 22:56, 6 March 2006 (UTC)

Sorry, but I haven't done any college studies on this topic. OK, so adding or removing an element to an infinite set is invalid?Landheha 23:01, 6 March 2006 (UTC)

No, it's completely valid — but adding or removing one element won't change the cardinality. You'd need to add or remove infinitely many elements, and even then it depends on exactly which elements. (Infinities aren't always very intuitive, I'll admit. |N| = |2N|, but |N \ N| = 0, while |N \ 2N| = |N|. Adding infinities is slightly better, though: if A and B are infinite sets, then |A U B| = |A| and/or |A U B| = |B|.) Ruakh 23:08, 6 March 2006 (UTC)

I AM removing infinitely many elements, one at the time. I think you are again focusing on the big end in your examples by adding, multiplying or dividing. I am subtracting, removing elements infinitely. You say "it depends on exactly which elements you remove". I don't understand this. How can you point to certain elements in an infinite set? Landheha 23:48, 6 March 2006 (UTC)

Re: "I AM removing infinitely many elements, one at the time.": No, you're not. If you remove elements, one at a time, at a fixed, finite rate, then you'll never have removed an infinite number of elements. Ruakh 00:10, 7 March 2006 (UTC)

Re: "I think you are again focusing on the big end in your examples by adding, multiplying or dividing. I am subtracting, removing elements infinitely.": Dividing? Who divided anything? Note that \ is the symbol for relative complement — i.e., the "subtraction" of one set from another. Ruakh 00:10, 7 March 2006 (UTC)

Re: "How can you point to certain elements in an infinite set?": That depends on the set. With the set of natural numbers (which is what I used in my examples), there are plenty of easy-to-point-to elements, such as 1, 2, or 3. Ruakh 00:10, 7 March 2006 (UTC)

I thought there were generic production rules for infinite sets which start from the empty set. It seemed logical to me that you should be able to come back to the empty set again. I know not all things can be reversed, but here I could not see a reason why not. Landheha 23:48, 6 March 2006 (UTC)

Re: "I thought there were generic production rules for infinite sets which start from the empty set.": Yes, but not by adding one element at a time. Ruakh 00:10, 7 March 2006 (UTC)

OK, thank you very much for your patient answers, I think I got it now. Have done some extra reading and thinking and I see where I went wrong. Thinking this way indeed takes some more getting used to than I thought :-)

this "Question"-part has played it's part, at least for me and I don't think it is very instructive for others. Is it OK if I remove it altogether?

[edit] 1-on-1 correspondences

There might be another possible way to look at these counter-intuitive 1-on-1 correspondences between different infinite sets. To do this you can make an analogy with fractals.

Picture before you a fractal like the famous Mandelbrot set on a computer screen. You can zoom in to any smaller part of the screen and you will get to see the small part filling the entire screen with the same detailed resolution. You can keep zooming in forever without losing detail.

Suppose now mathematics is a tool with a certain precision.
It can be pointed at an object of any size.

Say you look at the set N.
Math will let you look at it with it's inherent precision.
Then you zoom in on the subset of the even numbers.
Math will let you look at it with it's inherent precision.
Yes, you can make an 1-on-1 correspondence.

Say you look at the set Q.
Math will let you look at it with it's precision.
Then you zoom in on N.
Math will let you look at it with that same precision.
Yes, you can make an 1-on-1 correspondence.

You can also simulate this recursive zooming in on the number of points on any piece of line. Math allows you to keep zooming in, so you arrive (or better: you will not arrive :-) at R points. Each step of zooming in will give you again the precision math has.

Being a non-mathematician, I think this is a fun idea.
I can't find fault with it and it might make life much easier.

Anybody???? Landheha 21:25, 6 March 2006 (UTC)

I don't think the fractal analogy is really helpful, simply because I don't think the fact that any part of a fractal is equivalent to the whole is no more intuitive than the fact that the interval (0,1) is equivalent (in some ways) to R. Also, because I think most people would have a hard time mentally zooming in from the natural numbers to the even numbers, seeing as the even numbers are not really a region of the natural numbers. (Actually, I'm more inclined to zoom out to see the even numbers — when you're zoomed in, your scale is smaller, but as you move out, your scale increases.) Ruakh 23:01, 6 March 2006 (UTC)

Zooming in or out doesn't matter for the resolution of the tool. This is just a matter of starting point. You are still able to make 1-to-1 correspondences. I thought it clearer to paint the picture starting with "zooming in on a subset, being a part of the whole".

Excuse me for making a non-correct statement where you say the even numbers are not really a region of the natural numbers. I am just refering to the many cases where just this example is used to show the counter-intuitive aspects of infinity in that the cardinality of a subset is the same as its superset. If this is not mathematically correct, please don't blame me. I am just echoing someone else's examples here Landheha 00:07, 7 March 2006 (UTC)

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