Mathematics & Theology

[PolyHedral]

Yes, but we're talking topology of 1-dimension for R, and P(N) basically creates a set like R out of N, because for any element n in N, there are an infinite number of elements within P(N) I can associate it with, such that if I did this for every n is an element of N, I would always (to borrow reptillian's use of the pigeonhole principle) be trying to cram an infinite number of pidgeons into every hole, exactly as I would with the reals.

But the pigeonhole principle is only concerned with cardinality. I'm not arguing that |P(N)| = |R|, but that you can map the two to each other.

Take away set notation (or, better yet, think of both sets from P(N) and elements of R as "points") and you basically have the reals. As you point out, you can pick a "between" for the rationals, and yet there is no one-to-one for this set with the reals. That is, any map from Q to R will involve cramming pidgeons. This is not true of P(N) or P(Q) and R.

However, it is still impossible to pick a "between" sets of integers, even once you assign an ordering. (which you haven't actually done, so considering them as a space doesn't make sense.) P(N) is discrete; R is continuous, and that means they behave very differently despite being the same size.

(Incidentally, if you can actually provide a one-to-one map between P(N) and R, then go ahead.)

 

[LegionOnomaMoi]

But the pigeonhole principle is only concerned with cardinality. I'm not arguing that |P(N)| = |R|, but that you can map the two to each other.

But that's what having the same cardinality means: "Two sets A and B have the same cardinality if there exists an invertable mapping A -> B." Hubbard and Hubbard, p. 23.
"The cardinality of N is defined to be d or ℵ0 (aleph null). Any set equivalent to N is said to have the same cardinality...We shall see that a simple diagonal argument will prove that the set R of real numbers is not denumberable. The cardinal number of R is defined to be c or ℵ 1 (aleph one). Any set equivalent to R also has the cardinality ℵ1. Thus the set I of irrational numbers has cardinality ℵ 1"
Chatterjee's Topology: General and Algebraic (p. 23).

(Incidentally, if you can actually provide a one-to-one map between P(N) and R, then go ahead.)

Sure. I'll use the one from Mathematics of Infinity (pp. 132-133), as it is basically the same as in other texts, but stated far less formally (I apologize for the use of picture, but scanning turned out to be easier than typing).

"Theorem 4.4.1 card(R) = card( P(N) )

Proof: We produce a one-to-one function
f: [0,1]----> P(N)

thus proving that card [0,1]--->card ( P(N) ). Then by [another theorem showing that this interval of R has the same cardinality of R] we will have shown that card(R)=card[0,1] is less than or equal to card( P(N) )....

In this age of computers, it should come as no suprise that each real number can be written as a binary decimal. That is, to each real number x is an element of [0,1] there is a string
x = .b1b2b3...
of bits b1, b2, b3,... is an element of {0,1}. Specifically, we can write down an equation x = .b1b2b3... exactly when

where the coefficients of the remaining fractions 1/2 to the n are 0. Thus we will write

 

[LegionOnomaMoi]

(Incidentally, if you can actually provide a one-to-one map between P(N) and R, then go ahead.)

Was that sufficient? I didn't include a proof which shows that a mapping of P(N) to [0,1] also proves that we can construct a one-to-one map from P(N) to R, as that is assumed. I don't know if you want a proof of this too (that is, a proof that the interval [0,1] has the same cardinality as the set R).

 

[A-ManESL]

I prefer the proof above with a small change: Think of a subset of the naturals as a sequence of zero's and one's. (A one indicates a particular element is in the subset, a zero indicates it isn't). Between every digit, put a 3 and in the begining put a decimal point. (Eg 11001... becomes 0.131303031...; this is done to overcome the problem of ambiguity in representations) Now every subset has been put in a one one correspondence with a real number. So |P(N)|<= |(0,1)|. Conversely every real number between 0 and 1, has a binary expansion albeit sometimes redundantly; and removing the point gives us a 0-1 sequence, i.e. a subset of the naturals; and hence that particular set of reals has cardinality <= |P(N)|. By Schroder Bernstein theorem we have |P(N)|=|(0,1)|. It is now trivial to conclude that |P(N)|=|R| (Hint: Think of the tan function.) As an added bonus by Cantor's theorem (|A|<|P(A)|) we have also established that the reals are uncountable.

 

[LegionOnomaMoi]

I prefer the proof above with a small change

The proof can be far shorter (one method is that which you refer to). However, while I suspect PolyHedral would have no trouble with a shorter, more succinct proof, I used the one I did for the sake of others. The several "examples" given are not necessary, but are helpful for those whose experience with analysis is not at all close to that of a mathematician. The proof is not designed as a "proof" (as the result is well established) but for educational reasons, and hence is longer and more "informal" than is necessary. PolyHedral has a background in mathematics, but many on this forum do not, so I figured I'd use a "proof" from a non-technical text.

 

[A-ManESL]

PolyHedral has a background in mathematics, but many on this forum do not, so I figured I'd use a "proof" from a non-technical text.

Agreed.