- #26

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organic:

General:

We say that B > A if A not= B but there is an injection between A and subset of B .

furthermore, what i didn't mention in my last post was that since there is a bijection between E and N, it wouldn't make sense to say that E<N.phoenix:i'm having second thoughts about this definition. if we let N be the set of natural numbers and E be the set of even natural numbers, then, by this definition, E<N with the injection f(x)=x.

using the definition hurkyl quoted, A<=B if there is an injection from A to B. (technically, it's |A|<=|B|, where |A| is the cardinal number of A.) in this case, we can say that "A is dominated by B." furthermore, we can say that A and B are "equipollent" if there is a bijection from A to B and write without ambiguity that |A|=|B| and A=B with some ambiguity. the standard way of then exctracting the < from the <= is to say < means "<= but not =." in other words,

**A<B if there is an injection from A to B but there is no bijection from A to B**. i think this is equivalent to saying that there is a function from B onto A but no onto function from A to B.

using the definition in bold which reflects hurkyl's quote, one can proceed to prove that for any set X, X<P(X):

1. f(x)={x} is an injection from X to P(X). this proves that X<=P(X).

2. to prove X!=P(X), note that any function g from X to P(X) fails to be a bijection because it is not surjective. g is not surjective (onto) because the element of P(X) defined as {x in X such that x is not an element of g(x)} is not in the image of g.