

Different sizes of infinity - mgrouchy
http://timetobleed.com/different-sizes-of-infinity/

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ColinWright
What really tends to upset people is this:

* Between any two irrationals there is (at least one) rational

* Between any two rationals there is (at least one) irrational

* There are more irrationals than there are rationals.

~~~
ionfish
I think this becomes much more intuitive once one understands that the
cardinality of any nondegenerate closed interval is the same as the
cardinality of the continuum.

~~~
ColinWright
It becomes perhaps even _more_ obvious when you realize that any non-empty
open interval (a,b) with a<b is homeomorphic to the reals. Once that really
sinks in, much of this becomes easier.

------
shenberg
One point I never see discussed in these things is that you need to take a bit
of care with digit-based representations because 0.1111... = 1.000...,
0.0111.... = 0.1000... - an argument could be made that perhaps while you know
the number you're constructing via diagonalization doesn't appear in the list
of numbers you created, but it may be a different representation of a number
that does appear in the list.

The simplest way I know to work around that is by excluding numbers ending in
infinite strings of 0s or 1s from the list. Anyone know a nicer way?

~~~
alphaBetaGamma
The ambiguity in digit representations only comes from numbers ending in 1's
(in base 2, or ending in 9's in base 10). If you do the argument in a
different base than 2 you can avoid the problem: you can construct a number
whose base 10 representation consists only of 5's and 6's and which is
different from any number in the list.

------
jowiar
Steven Rudich's lecture slides on the topic are pretty great, and how I was
first introduced to the topic:
[http://www.cs.cmu.edu/afs/cs/academic/class/15251-s04/Site/M...](http://www.cs.cmu.edu/afs/cs/academic/class/15251-s04/Site/Materials/Lectures/Lecture25/lecture25.html)

Not sure if the lecture itself is available anywhere.

------
qntm
A much harder question is "How many infinities are there?"

[http://everything2.com/user/sam512/writeups/How+many+infinit...](http://everything2.com/user/sam512/writeups/How+many+infinities+are+there%253F)

~~~
ionfish
Why is that a harder question? It's a direct corollary of Cantor's Theorem
that there is no largest cardinal number (assuming the powerset axiom, of
course).

~~~
qntm
There's no largest integer, but the integers still form a set with an
(infinite) cardinality, namely aleph-zero.

Likewise, there's no largest cardinal, but you'd think that the cardinals also
formed a set with an (infinite) cardinality of its own.

Fun fact: they don't. _There are too many of them._

~~~
ionfish
Like I said in my response to pndmnm, in my view if someone "[would] think
that the cardinals also formed a set with an (infinite) cardinality of its
own" then they haven't really grasped the theorem yet. It's built into it that
if we're always allowed to form the powerset of a set then regardless of how
large a cardinal you can find, you can always take its powerset and obtain a
larger cardinal.

------
scovetta
I never liked Cantor's argument here. I'm sure I'm missing something, but the
'grid' that you set up has infinity rows but 2^infinity columns. The diagonal
that you make just cuts off the top-right corner of the grid, and that missing
element is one of the remaining rows.

The diagonalization argument just seems like handwaving.

~~~
ColinWright
There aren't 2^oo columns - the number of columns is the same as the number of
rows. And the argument can be made a lot less handwavey, it can be made very
precise in set theoretic terms.

Do you want references?

~~~
scovetta
That would rock, yes.

(But why aren't there 2^oo columns? In order to fill up all possible ...<snap>

I think I got it...

You're not trying to use every possible oo-length binary string, you're just
mapping each natural number to an arbitrary (unique) oo-length binary string,
so the grid in a square in the end.

~~~
ColinWright
Yes.

When working more formally it's actually easier just to show that for every
set A the set 2^A is strictly bigger, where 2^A is a common representation for
the power set of A, the collection of all subsets.

As an aside, if A and B are sets, the set A^B is

    
    
        {f : f:B->A},
    

the set of all functions from B to A. You can quickly show that if A and B are
finite, and |A|=n and |B|=m then |A^B|=n^m, thus matching our intuition. A
function in 2^A says for each element of A, choose whether it's in or out.
That gives a subset of A, so it makes sense to call the set of subsets 2^A,
the power set.

In what follows you need to "read like math." This is not a novel - work
through it, don't simply try to read. Draw pictures, use simple examples, try
to fund counter-examples. _Work on understanding._

So, given a set A, how do we show that |A| < |2^A|? Simple, we take any
function from A to 2^A and show that some element of 2^A doesn't get hit. This
shows that for every function from A->2^A it's not a matching. Hence no
matching exists between A and 2^A, so 2^A is not the same size.

I'll leave it to the interested reader to show that 2^A is never smaller than
A.

So how do we show that any function from A to 2^A leaves some element of 2^A
unhit?

Let g:A->2^A be any function from A to 2^A. We construct h in 2^A such that h
is not in the image of g.

h needs to be a function from A->{0,1}, so for every a in A, we need to say
whether h(a)=0 or h(a)=1.

Remembering that g(a) is in 2^A, and hence is a function A->{0,1} we define
h:A->{0,1} as:

    
    
        h(a) = 0 if (g(a))(a) = 1
        h(a) = 1 if (g(a))(a) = 0
    

Chase the definitions and you'll find that for every a in A, h != g(a).

------
mchahn
This is only tangentially relative, but I find the argument that real numbers
may not ever exist in nature to be fascinating. The argument is that all of
nature breaks down at the size of plank's constant so every real measurement
has a finite resolution.

------
tokenadult
If you like this you may also like "All about Infinity" (former title,
"Infinity Is Not a Number - It's a Free Man")

<http://nrich.maths.org/2756>

by Katherine Körner.

------
wkdown
I thought infinity is infinity; no varying degrees.

How is this different from saying counting to infinity with integers is
smaller than counting with floats?

~~~
ColinWright
This is not "infty" as per computer representations of numbers. This is
"infinity" as in "how many counting numbers are there" and "how many reals are
there".

We are not dealing with floats. Trivial, for any size being used, the number
of floats that can be represented is finite.

So what this is saying is this:

Consider the collection of counting numbers: 1, 2, 3, ... and so on. Call that
set N.

There is a set P which has the property that every time you try to match up
items of N with items of P, there are items of P left over.

If you have some cups and saucers, and every time you laid them out you had
some saucers left over, you'd say you had more saucers. Yes? Well every time
we try to match up things from N with things from P we can things from P left
over, so we should say we have more of them, right?

But there are infinitely many things in N, so P must be a bigger infinity.

------
learc83
I've always liked the explanation of different sizes of infinity that compares
the infinite set of whole numbers to the larger infinite set of fractions.

~~~
pndmnm
There are the same number of whole numbers as there are fractions/rational
numbers (my favorite quick proof -- every whole number is a rational number
(itself over 1). Map each n/m in lowest terms to 2^n*3^m for an injection of
rationals into the integers (by the fundamental theorem of arithmetic)).

~~~
JadeNB
> There are the same number of whole numbers as there are fractions/rational
> numbers (my favorite quick proof -- every whole number is a rational number
> (itself over 1). Map each n/m in lowest terms to 2^n*3^m for an injection of
> rationals into the integers (by the fundamental theorem of arithmetic)).

Notice that this shows only that there are no more whole numbers than rational
numbers, and no more rational numbers than whole numbers. To conclude that
there is the same number of each (more generally, that cardinal numbers are
totally ordered (EDIT: as ionfish
(<http://news.ycombinator.com/item?id=4251492>) points out, I should have said
that the inclusion relation on cardinals is antisymmetric)) is actually a bit
delicate; it's called the Schröder–Bernstein theorem (although Wikipedia
attaches more names, in a different order:
[https://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80...](https://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem)).

~~~
ionfish
Yes, it does rely implicitly on Cantor–Schröder–Bernstein. That might be a
downside, but I think when working informally (that is to say, when not
teaching a set theory course) one can simply assert that if there exist
injective functions from sets A and B into one another then they are
equinumerous.

That being said, although important in the theory of the order of the cardinal
numbers, Cantor–Schröder–Bernstein doesn't show that the cardinals are totally
ordered. That statement is actually equivalent to the Axiom of Choice, whereas
as far as I'm aware Cantor–Schröder–Bernstein holds in ZF.

~~~
pndmnm
That's absolutely correct -- trichotomy for arbitrary cardinals (any two
cardinals are cardinal-size-comparable) requires AC, but SB doesn't require
AC. Trichotomy for the cardinal numbers of well-ordered sets (e.g. ordinals)
doesn't require AC.

It's a little irrelevant to this thread... but as long as I'm quoting non-
proofs that require lots of extra machinery, I'll give my favorite appeal-to-
intuition equivalent of choice: the product of non-empty sets is non-empty
(any point in the product of a collection of non-empty sets is a choice
function on those sets).

~~~
ionfish
AC is equivalent to a lot of things. There's a collection of them on the
Wikipedia page.

<http://en.wikipedia.org/wiki/Axiom_of_choice#Equivalents>

Something I find pretty interesting is that some of these equivalences break
down in weak systems.

[http://www.math.uchicago.edu/~antonio/RM11/RM%20talks/mummer...](http://www.math.uchicago.edu/~antonio/RM11/RM%20talks/mummert.pdf)

~~~
pndmnm
Yup, I did a few projects on equivalents of AC back in the day. That's just my
favorite "appeal to intuition" one (my favorite "appeal to intuition" against
AC is: the identity function is the sum of two periodic functions (though this
is a consequence and not equivalent)).

Equivalence breakdown in alternate systems is a wonderful topic. I've been
trying for a couple years now to figure out how to get back into set theory
now that I'm out of academia. Maybe later this summer...

