Edward357
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How do I prove that the set of all finite subsets of an infinite set has the same cardinality as that infinite set?
Edward357 said:How do I prove that the set of all finite subsets of an infinite set has the same cardinality as that infinite set?
Careful. The OP wants to prove that the set of all finite subsets of X has the same cardinality as X. This is certainly true whenever X is infinite.Focus said:You can't because its not true. Take the reals which have aleph-one cardinality, the subset {1} has cardinality 1 which is not aleph-one. Same goes for infinite subsets, naturals are a subset of reals but naturals have cardinality aleph-null.
Focus said:You can't because its not true. Take the reals which have aleph-one cardinality, the subset {1} has cardinality 1 which is not aleph-one. Same goes for infinite subsets, naturals are a subset of reals but naturals have cardinality aleph-null.
morphism said:What have you tried?
Your proof looks right, including this assertion here as well. But from the way you phrased it, I'm not entirely sure your justification for this assertion is correct. Would you elaborate?The union of all K^n is equivalent to lKl + lKl with the operation repeated countably many times. The sum must have cardinality lKl.
Focus said:Sorry I misread the post. You could try it this way (assuming CH): you know that if S is the set of all finite sets then a_i \in X \qquad \{\{a_1\},\{a_2\},...\} \subset S thus the cardinality of S is greater of equal to the cardinality of X. Now S is a subset of P(X) thus the cardinality of S is less than or equal to 2^{|X|}. I think you may be able to construct a proof to make that last inequality strict (perhaps by Cantors diagonal argument), then you have that |S|=|X|.
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