http://at.yorku.ca/cgibin/bbqaforum=ask_an_analyst;task=show_msg;msg=0168.0001

in particular, "Now, Cantor does not prove that P(N) = aleph_1. Indeed he famously doesn't do this, as this is precisely the continuum hypothesis!" so it appears as though our claim that the cardinality of the power set of the naturals (let's refer to this as P(N), where P(N) is the power set of the naturals) is not equal to cardinality of the reals ( or simply R). What got me searching about this topic was the following thought on the matter:

suppose i take any irrational, say pi = 3.14159........... the question that came to mind, and seems to have an obvious answer, is "Is there a rational which has the same digits but just in a different order?" Well, by construction, i think of course that's the case... how about just trading the first 1 and 4... so we have 3.41159............ that will do it. So, given what i quoted above, how can it be that a non-trivial subset (not the null set and not the whole set) of P(N) is sufficient for being the same size as R. I think the answer lies in the fact that we can break that infinitely long string of digits in infinitely many ways... here's what i mean. consider the following 2 elements of P(N) (remember elements of this set are themselves sets... they are subsets of N):

Note: the order we list elements in a subset doesn't matter... any set only cares about it's elements, and not how we list them inside... the fact that we can order them is an independent consideration

first element to consider:

--------------------------

{3, 14, 15, 9, ...} - we see from this element we can get 3.14159...

second element to consider:

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{3, 141, 59, ...} - we see from this element, we can ALSO get 3.14159...

So, at least under that construction where we use the pattern above to identify an element of P(N) with an element of R... we will end up with many representations of a given irrational (in fact, infinitely many... which size of infinity

*that is,*is another good question). of course, this is just one construction, and although it may be an obvious one, perhaps another will prove this one to be not so dramatic. but for now, i can at least "see" how it's possible that P(N) > R. I mean, in the same way that at first we may think that the naturals are bigger than the even naturals... of course we know they have the same cardinality because the map f(n) = 2n and it's inverse g(e) = (1/2)*e prove that a bijection exists between them (any function f is a bijection if and only if it has an inverse function).

Ok, that's all i'll say for now, but there's more...

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