- #1

MathematicalPhysicist

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well i thought about it, and perhaps bacuase X is infinite it has a subset which is infinite countable, let denote it by A={x1,x2,x3,...}, let f:X->X be any function, and let's look on f:A->X, f(A) is alos infinite countable, by definition: f(A)={f(x)|x in A}, we can't have A to be a proper subset of f(A), cause if it does then: |f(A)|>|A| but then we have aleph-null is bigger then itself which is a contradiction.

now then, we have for every f:X->X there exists a nonempty proper subset A of X, such that f(A) is a subset of A.

if X is finite then i need to show that there's a bijection between X and a proper subset of it (which is a contradiction), we have f(X) a subset of X, and f(A) a subset of A, but i dont know how to procceed from here, any advice?