Lesson 10: Infinity times 2

Last lesson we used our abstract way of counting the number of elements in a set to extend the “normal” number system to include a number which we called “infinity type 1”.  In particular, analogously to how we define a set A to have N elements if we can define a bijective function from the set \{1, 2, 3, ..., N\} to A, we also define a set A to have “infinity type 1” elements if we can define a bijective function from the set \{1, 2, 3, ...\} to A.  It is implicit in my calling this new number “infinity type 1” that there is likely an “infinity type 2”.  This is indeed the case (and much more!), but before we can truly appreciate this fact we need to first get used to what this definition of “infinity type 1” really means.

We all already know that “infinity plus anything is infinity” and “infinity times anything (other than 0) is infinity”, and other sort of “obvious” statements like these.  However, if I claim that there are different kinds of infinity (which is what I’m claiming), then we need to be more careful when we make statements like “infinity plus anything is infinity”.  In particular, we need to know which kind of infinity plus anything is which kind of infinity.  Suppose for a second that there is an “infinity type 2” (there is, but since we haven’t proved it yet let’s just suppose it’s there).  Then we need to know whether “infinity type 1 plus anything is infinity type 1” or if “infinity type 1 plus anything is infinity type 2”.  Or maybe it’s the case that there are some things that we could add to “infinity type 1” to give us “infinity type 1” again, and other things which would give us “infinity type 2”.

While all of this might sound a bit strange or a bit “fake”, we’ll soon see that these ideas are in fact completely rigorous.  Indeed, the rigor and beauty of these ideas are such that we’ll be forced to somehow believe that there “really are” many kinds of infinities (scare quotes because I won’t discuss here what it means to “really be”).  So enough with the fluffy philosophy—let’s do some math.  What we’ll focus on in this lesson is giving precise meaning to the phrase “infinity times 2 is infinity”.  Actually, what we’ll show, is that “infinity type 1 times 2 is infinity type 1”.

Let us define a set that is in a very obvious way “infinity type 1 times 2”.  Namely, let’s define a set by taking every element in \{1, 2, 3, ...\} and “creating another copy of it”.  We can make this rigorous by adding into our set every single negative number.  Thus if A=\{1, 2, 3, ...\}, let’s let B=\{1, -1, 2, -2, 3, -3, 4, -4, ...\} and so on.  The set B is in some clear way “twice as big” as A, since we define B by essentially adding another copy of A to it (and just adding a trivial “minus” sign in front of some of the numbers).

Recall, however, that our idea of what “twice as big” means now has a very formal meaning.  Namely, if I have a set SET1 which can be put into bijection with \{1, 2, 3, ..., N\}, and I have another set SET2 which can be put into bijection with \{1, 2, 3, ..., M\}, then SET1 is “twice as big” as SET2 only if N=2\times M.

So let us not worry about our intuitive notion of what “twice as big means”, and rather just inquire about our rigorous definition of cardinality.  Namely, let us ask whether or not A=\{1, 2, 3, ...\} and B=\{1, -1, 2, -2, 3, -3, ...\} have the same cardinality.  I claim that they do in fact have the same cardinality, but before proving this let’s reflect on why this is a non-trivial statement.  Half of me wants to say that they have the same cardinality because \mathrm{infinity}=2\times \mathrm{infinity}, whereas the other half of me wants to say no, they don’t have the same cardinality because B clearly has “more” elements than A.  Thankfully we don’t need to rely on my intuition for this problem.

I’ll prove that A and B have the same cardinality by doing the only thing that I can do: define a bijective function from A to B.  This is, after all, the very definition of “having the same cardinality”.  Defining this function and seeing that it is indeed bijective is actually quite simple.

First, let us take the subset of A that contains only its odd positive whole numbers.  Namely, take the subset \{1, 3, 5, 7, ...\}, and also do the same for the even ones: \{2, 4, 6, 8 \mathrm{(who\ do\ we\ appreciate)},...\}.  Now, let’s define our function to map, in order, the odd elements in A to the positive elements in B, and the even elements in A to the negative elements in B.  In other words, if F is our function from A to B, then F(1)=1, F(3)=2, F(5)=3, F(7)=4, and so on, and F(2)=-1, F(4)=-2, F(6)=-3, F(8)=-4, and so on.  Clearly, no two elements in A get mapped to the same element in B (so F is injective), and every element in B is eventually “hit” by something in A (so F is surjective).  Thus, F is bijective, and therefore A and B have “the same number” of elements—namely, they both have “infinity type 1” elements!

This same logic lets us quickly prove that “infinity type 1 plus anything is infinity type 1”, where by “anything” I mean “any finite number of elements”.  We can see this as follows.  Let A=\{1, 2, 3, ...\}, as usual, and let APPLES be a set of apples which contains some finite amount of apples.  Since there are finitely many apples in APPLES, we know that there is some N such that \{1, 2, 3, ..., N\} can be put into bijection with APPLES.  Let us therefore just label the apples in APPLES as A_1, A_2, A_3, ..., A_N, so that APPLES=\{A_1, A_2, A_3, ..., A_N\}.  Now we can formalize the notion of “infinity type 1 plus anything” by considering the set with all of the elements in A=\{1, 2, 3, ...\} as well as all of the elements in APPLES.  This is clearly a set with “infinity type 1 plus N” elements in it.  Let’s denote this set by

NEW-SET=\{A_1, A_2, A_3, ..., A_N, 1, 2, 3, ...\}.

Now let’s define a bijective function from \{1, 2, 3, ...\} to NEW-SET, thus showing that the two sets actually have “the same number of elements” in them.  This is easy!  Let F be the function from A to NEW-SET defined by sending the first N numbers in A to the N apples in NEW-SET, and then “shifting” the rest of the numbers in A so that “N+1” in A gets sent to “1” in NEW-SET, and “N+2” in A gets sent to 2 in NEW-SET.  In symbols, this would read

F(1)=A_1, F(2)=A_2, F(3)=A_3, ..., F(N)=A_N,

F(N+1)=1, F(N+2)=2, F(N+3)=3,

and now we just continue that on forever! This is clearly bijective, and so we’ve shown that “infinity type 1 plus any finite number is infinity type 1”!

I’ll leave one more cool result for you to think about in the exercises, although I’ll spell most of it out (no need to pull out a pen and paper).  We need to get used to the ideas behind these maps because we’ll need to get fancier and fancier in the next two lessons to prove some even cooler things.  In the next lesson we’ll show that, in a certain sense, “infinity type 1 times infinity type 1 is infinity type 1”.  The proof of that statement is absolutely gorgeous (again, I didn’t come up with it), but not nearly as gorgeous as the proof that we’ll see in lesson 12 that there is in fact an infinity that is greater than infinity type 1.

So far all we’ve done is provide more evidence that “there is only one infinity”, but seeing the techniques used here and gaining respect for the logic that is at play is what we need to do before proving the existence of a whole ladder of new infinities!

Exercises

1) Prove that “infinity type 1 divided by two is still infinity type 1”.  Do this by realizing that the set of even numbers is a meaningful way of describing “infinity type 1 divided by two”, since we can split A=\{1, 2, 3, ...\} evenly into odd and even parts.  Then define a bijective function from A=\{1, 2, 3, ...\} to B=\{2, 4, 6, 8, ...\}, thus showing that A and B have the same cardinality!

Solutions to Exercises

A quick note before lesson 11

Previous Lesson

4 Responses to Lesson 10: Infinity times 2

  1. Anonymous says:

    What’s infinity•2

    • Gotta read the lesson to find out! In short, though, I simply use “infinity times 2” to refer to the fractions, since for each numerator there are infinitely many possible denominators (which can be viewed as “infinity times 2”).

      • Anonymous says:

        What you just said makes absolutely no sense

        • If you’d like you can think of the set of fractions as “infinity times infinity,” since for each numerator (of which there are infinitely many possibilities, one for each integer) there are infinitely many possible denominators (again one for each possible integer). The only important thing here is that the fractions appear to be MUCH more infinite than the integers (whole numbers). Have you read the lesson and the ones preceding it? If not I’m afraid there’s not much I can do to help. If this still doesn’t make sense perhaps you can give me a bit more constructive guidance as to what is confusing you.

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