Announce:
  "The battle for accountable voting systems"
  David Dill, Computer Science, Stanford University
  2004.Feb.25 (Wed.) 16:00, Duncan Auditorium





Finish functions:
- Terms: "1-1"=injection; "onto" = surjection; "1-1,onto"=invertable=bijection
- Go through proof of 1-1,onto?  [Will do that today]
  Reading: Rosen 3.1, working through examples.  [A follow-up to 1.5.]





Cardinality
("Some infinities are bigger than others")
Reading: Rosen 3.2[cardinality only]


Hotel infinity (Due to Martin Gardner):
It has rooms numbered 1, 2, 3, ... .
When full up, how does a clever concierge add one more guest?  10 more?
How to double the #guests (say, each platonic guest labelled by N)?
A cute way to add a countable number of countably sets and still
have rooms left over:
When Buses 1,2,3, ... show up (each with people 1,2,3,...),
we can cleverly fit everybody into rooms by using powers-of-primes:
the existing guest in room i is now assigned to room 2^i.
The i'th person from the first bus is in room 3^i,
and the i'th person from the next bus is in room 5^i.
In general, the i'th person from the j'th bus goes
into room number (the j'th odd prime)^i.
This leaves many rooms empty (e.g. room 6=2*3, 15=3*5, etc.).
In fact, there are infinitely many empty rooms!
 [Note that the current guests can be viewed as just one additional busload,
  or equivalently the other buses as other hotel infinity franchises.]


Okay, let's address these questions by being more formal than a Hotel.
Def'n: Sets A and B have the same cardinality if there is a bijection
       between them.
My personal visualization: a set of domino tiles, with members of A
written on one side, and members of B written on the others.
The cardinality of N is sometimes written aleph-null.

What is a bijection between N and 2N?  Between N and Z?

Def'n: a set is countable or countably infinite or enumerable, if there
       is a bijection with N.
This is because an enumeration of the set (counting its elements)
is tantamount to a bijection with N.
   [Note: sometimes countable is used to include finite sets;
    always ask if it's unclear from context.]

Aside: sometimes we say countably-infinite sets are "isomorphic to N".
isomorphic means "the same up to re-labeling".
This use is appropriate if we're just considering the set,
w/o considering relations on the set.
If there are relations (like "<") that we want to preserve,
then "isomorphic" is stronger than just "same cardinality".



Hmmm, meta-clue: this term suggests there are infinite sets that 
aren't countable!  (Is there an aleph-one?)

Are the following sets countable (enumerable)?
- set of strings over {0,1}?
- set of strings over {a,..z}?
- set of infinitely-long strings?
- set of strings union infinite-strings?
- Q
- R
- P(N)


Diagonalization proof that N !~  [0,1].
  We'll show that any function f:N->[0,1] can *not* be onto.
    We construct (the decimal representation of) a number x,
    such that forall i, f(i) != x.
    Thus, there isn't an i such that f(i)=x, so f isn't onto.
    Let x be made so that its i'th decimal place is
      the i'th decimal place of f(i) plus 1.  (Wrap-around 9+1=10 to 0).
      

  Note one glitch: 
   We said, "any f(i) differs from x in the i'th
   decimal, so they're different".
  This seems clear, until we realize the implicit reasoning:
   If two numbers have different decimal representations, they're different.
  This isn't true: Consider 0.24999999..., and 0.250000....
  How to fix the proof?
  It turns out, the *only* non-unique representations stem from the 
     999.../000... different, so if the differing decimal-places
     don't involve 9 and 0 then indeed the numbers are different.
     [Math students can worry about proving this.] 
  So we can patch: instead of making our i'th decimal place
     by the f(i)[i]+1, we can make it f(i)[i]+2 (wrapping around -- "mod 10").

This is a "diagonalization proof".

Lemma: |[0,1)| = |[0,infty)|.
  We'll construct an exlicit f, and show it's 1-1,onto.
    Thinking...try f(x)=(x/1-x).  Yep, that'll do it.
  - f(x)=x/(1-x) is one-to-on and onto [0,infty).
    Onto: For any positive number z, z/(z+1) is in [0,1) and f(z/(z+1))=...=z.
    1-1: if f(y)=f(z), then y/(1-y)=z/(1-z), and y-yz=z-yz, so y=z.  Check.


Lemma: |[a,b)| = |[c,d)|.
  Proof: left to reader.

Lemma: |[0,infty)| = |(-infty,+infty)|.
  Proof: Left to reader.
    Hint: partition [0,infty) into intervals,
          then have each interval map to "an interval twice as wide",
          alternatingly positive and negative.
    One glitch: make sure 0 isn't mapping to the same thing twice!
 

Aleph-one: |P(N)| = |R|.



Th'm [Cantor]:  For all sets A, there is no bijection A --> P(A). 
    Proof left to the reader.
    The proof can't quite be the same as our proof for N vs R:
    Why not?  Consider showing there's no bijection in R-->P(R).
    We can't talk about the i'th row of the table (f(i)),
    because we've already shown R uncountable.

    Hint: try something like the Russel paradox "all sets not
       containing themselves", but somehow in terms of A and P(A).


Handy th'ms:
  Prove: If A is not countable, then AuB is not:
  Prove: If A and B are both countable, then AxB is:





Continuum hypothesis: Is R the smallest uncountable set?
  That is, is there an uncountable set A such that |N| < |A| < |R|?
This is one of Hilbert's 26 problems of 1900.
Resolution: take it or leave it; you get a consistent (sound)
  system either way.  (At least, as sound as the current system :-)







Looking back:
We use the term "cardinality" of sets, and have formally defined it.
Some of the results are strange.
For instance, the reals in [0,1] have the same cardinality of
the reals in [0,2] and even all R.  But in other senses we
don't want to throw away the fact that somehow there do seem to
be twice as many elements of [0,2] than of [0,1]; "Measure Theory"
is the branch of math dealing with trying to recapture this notion.
That doesn't invalidate the work here, since we can compare sets over
entirely different domains (e.g. real numbers, and infinite-length-strings.)

Also, beware doing arithmetic on infinite sums, etc.  Consider:
    1 - 1 + 1 - 1 + 1 - 1 + ...
Since addition is finite-associative, we might expect it to be
infinite-associative.  But if it were, then compare:
    (1-1)+(1-1)+(1-1)+... = 0+0+0+... = 0
with
    1+(-1+1)+(-1+1)+... = 1+0+0+0+... = 1
with
    1+1+1+1+(1-1)+(1-1)+(1-1)+... = 4+0+0+... = 4
    (where we have just shifted each -1's each eight places down --
     a common enough re-indexing technique).
     [Exercise: write the original and this version as 
      sums indexed to infinity.]

By re-associating and re-indexing, we can easily contrive to get
any integer we want as the ostensible sum of this series.
The upshot of course, that this sum is not defined, even though
it initially feels like it should be.
The bigger picture is that finite-associate doesn't imply
infinite-associate.