# The cardinality of the set of all finite subsets of an infinite set

by Edward357
Tags: cardinality, finite, infinite, subsets
 P: 6 How do I prove that the set of all finite subsets of an infinite set has the same cardinality as that infinite set?
 HW Helper Sci Advisor P: 2,020 What have you tried?
P: 284
 Quote by Edward357 How do I prove that the set of all finite subsets of an infinite set has the same cardinality as that infinite set?
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.

HW Helper
P: 2,020

## The cardinality of the set of all finite subsets of an infinite set

 Quote by Focus 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.
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.

Also, the cardinality of the reals is 2^aleph-0, and if the continuum hypothesis isn't assumed, this is generally not aleph-1.
P: 6
 Quote by Focus 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.
I know that it is not true that every finite subset of an infinite set has the same cardinality as the infinite set. No finite subset of an infinite set has the same cardinality as the infinite set. The question was about the set of ALL finite subsets of an infinite set.
P: 6
 Quote by morphism What have you tried?
O.k.

If K is an infinite set, then form U K^n for natural numbers n. We know that if card K = lKl, then K^2 has card lKl. We know that for any finite natural number, K^n has cardinality lKl. The union of all K^n is equivalent to lKl + lKl with the operation repeated countably many times. The sum must have cardinality lKl. It is clear that for every subset of elements {b1,...., bm}, there is injection that maps that set to an ordered set <b1,....,bm>. There are "more" ordered sets actually. It has to be true that the set of finite subsets is < or = the set of ordered n-tuples, which has cardinality lKl.
 P: 284 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|. Hope this is more helpful
PF Patron
 Quote by Focus 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|. Hope this is more helpful