Cardinal logarithmetic

By Luke on

I wanted to write this down before I went to bed, so I could think about it in the morning: 2^aleph-0 is aleph-1, so clearly the “logarithm” of aleph-1 is aleph-0. Does aleph-0 have a logarithm? A more precise way of phrasing that is: does there exist a set such that its power set is isomorphic to the natural numbers?

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8 Responses

  1. Trieu

    Huo, you alway come up with something cause me unable to sleep.
    I think logarithm 2 of aleph-1 is aleph-0. Now, we ask ourself that does aleph-1 really have
    another logarithm, ie. log3aleph-1 = ?
    So, it is seemingly more precise when we ask does aleph-0 have a log2 ?
    Well, the existing of Beth-1 is just to compare the size of 2 infinite sets, and one of them is called
    the smallest infinite set, denoted as aleph-0.
    Let A be a set that the number of its subset is equal to a size of a smallest infinite set.
    Suppose aleph-luqui is the cardinality of A.
    Then, A is a set such that: 2^aleph-luqui = aleph-null. —(*)
    If A is a finite set, it’s cardinality could be written down as a certain number, hence,
    2^aleph-luqui is very different from aleph-null.
    If A is an infinite set, then contrast to definition, aleph-null is no longer a
    cardinality of a smallest infinite set in this case.
    Thus, A is not exist.

  2. Luke

    Nice. Your proof doesn’t quite work, only because you neglected to mention that theorem that says that there is never a 1-1 map from P(A) into A. With that, it is quite a nice proof by contradiction.

  3. Trieu

    It’s pretty nice to mention that, we are sure that there is no bijection between P(A) and A,
    since, 2^n-n elements in P(A) will left over, except in one case where A={} for instance, may be right, but that’s not an infinite set. :-)

  4. Thomas

    Interesting idea. A while ago I actually worked out (I think…) how exponentiation of cardinal numbers works. In the following I write “f” for a finite number (more precisely an integer >= 2; I leave the special cases 0 and 1 for the reader ;). The transfinite numbers I simply write without \aleph, i.e. just 0, 1, 2, …
    Then we have:
    f^f = f
    i^f = i (where i=0,1,2,… is a transfinite number)
    f^i = i+1

    i^j = i (if j < i) (both i and j transfinite numbers)
    i^j = j+1 (if j >= i)

    This gives the following logarithms:
    log_f(f) = f
    log_f(0) = does not exist (Luke’s initial question)
    log_f(1) = 0
    log_f(2) = 1
    log_f(3) = 2
    …
    log_0(f) = does not exist
    log_0(0) = f
    log_0(1) = 0
    log_0(2) = 1
    …
    log_1(f) = does not exist
    log_1(0) = does not exist
    log_1(1) = {f, 0}
    log_1(2) = 1
    log_1(3) = 2
    …
    log_2(f) = does not exist
    log_2(0) = does not exist
    log_2(1) = does not exist
    log_2(2) = {f, 0, 1}
    log_2(3) = 2
    log_2(4) = 3
    …

  5. Thomas

    oops, I guess the ‘less than’ sign caused trouble. So let’s restate:
    i^j = i (if j ‘less than i)
    i^j = j+1 (otherwise)

  6. Thomas

    Alright, I really got to go to bed soon, but one more thing: I finally figured out a satisfactory (for me at least) answer why 0^0 should be 1 (instead of say 0). Can you guess why? (hint: it’s related to cardinal numbers and functions).
    Feel free to take guesses or give your own ideas on what 0^0 should be and why.

  7. Luke

    Well, gee, neglecting the piles of mathematical evidence and just limiting my scope to cardinals, I’m going to say that there’s one function that maps the empty set to itself. But maybe you had one of the other aleph-omega reasons in mind? :-)

  8. Thomas

    Yep, that’s basically it. The more general statement is that the number of functions that map from a set A to a set B is card(B)^card(A). Amazingly this holds for the empty set as well. To see that, keep in mind that a function is a relation R (R subset A x B) such that for_all a in A there is exist exactly one b in B with (a,b) in R. If A is the empty set , we have exaclty one relation (R={}) satisfying this definition of a function. Thus n^0 = 1. If A is nonempty but B is empty, there is no relation R that satisfies the definition. Thus 0^n = 0 for n at least 1.

    Also, I know that it is mathematical ‘common sense’ that 0^0 should be 1. However, when I was asking my undergrad calculus professor he gave an explanation with the continuity of the function x^0 (which is definitely 1 for all x!=0 so it should be 1 for x=0 as well). But for me this argument is not that strong since one can give counterexamples based on continuity:
    0^(x^2) (seems to be a nice function, since both basis and exponents are in C^\infty). And it would suggest 0^0 = 0
    +/- (exp(-c/(x^2))^(x^2)) could suggest an arbitrary value for 0^0.
    Anyway, I think this discussion is more a matter of taste about is beautiful / elegant in mathematics, it’s just something I was occasionally wondering about…

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