Let $X$ be an infinite set of cardinality $|X|$, and let $S$ be the set of all finite subests of $X$. How can we show that Card($S$)$=|X|$? Can anyone help, please?


The cardinality is at least $|X|$, since $S$ contains all singletons.

Let $S_n$ be the subset of $S$ consisting of all subsets of cardinality exactly $n$. Then $S$ is the disjoint union of the $S_n$.

Now, for any positive integer $n$, the number of subsets of $X$ of cardinality $n$ is at most $|X|^n = |X|$ (equality since $|X|$ is infinite); because an $n$-tuple of elements of $X$ determines a subset of $X$ of cardinality at most $n$; and every subset with $n$ elements determines only finitely many distinct $n$-tuples of elements of $X$ (namely, $n!$). So $|S_n|\leq n!|X|^n = |X|$ for all $n$.

Therefore: \begin{align*} |X| &\leq |S| = \left|\bigcup_{n=0}^{\infty} S_n\right| = \sum_{n=0}^{\infty}|S_n|\\ &= \sum_{n=1}^{\infty}|S_n| \leq \sum_{n=1}^{\infty}|X|\\ &= \aleph_0|X| = |X|, \end{align*} with the last equality since $|X|\geq\aleph_0$.

Thus, $|X|\leq |S|\leq |X|$, so $|S|=|X|$.

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    $\begingroup$ Do I have it right that this requires some choice, as in the absence of AC, the reals could be a countable union of countable sets? $\endgroup$ – Ross Millikan Mar 15 '11 at 4:02
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    $\begingroup$ @Ross: I don't think so. In the absence of AC, you do not know that a countable union of countable sets is countable (the fact that $\mathbb{R}$ may be a countable union of countable sets show shows such unions need not be countable without AC). Here, we are assuming $X$ has a cardinal, and cardinal arithmetic does not usually require AC. Though I worked with $X$, you could simply work with the corresponding cardinals, which are already well-ordered, so that should allow you to avoid AC. But if we don't assume $X$ has a cardinality, I expect $S$ does not either. $\endgroup$ – Arturo Magidin Mar 15 '11 at 4:05
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    $\begingroup$ @Ross: The statement that for every two sets $X$ and $Y$ there is either an injection from $X$ to $Y$ or an injection from $Y$ to $X$ (sometimes called "comparability of cardinals") is equivalent to AC. However, one defines ordinals without having to have AC, and then one defines cardinals to be ordinals that cannot be bijected with any smaller ordinals. These definitions make sense even in the absence of AC, and all of cardinal arithmetic goes through without needing AC. It is the statement that every set is bijectable with a cardinal (in this sense) that is equivalent to AC. (cont...) $\endgroup$ – Arturo Magidin Mar 15 '11 at 4:38
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    $\begingroup$ @Ross: (cont...) It's the difference between "cardinality" and "cardinals." Cardinals are specific sets, well-defined in ZF. Saying that every set has a cardinality (i.e., is bijectable with a cardinal) is equivalent to AC, as is saying that for every $X$ and $Y$, either $X\preceq Y$ or $Y\preceq X$ ($\preceq=$there is an injection) which is probably what Shecter's saying. (Bernstein's Theorem, aka comparability of cardinals). See also Bergman's handout "The Axiom of Choice, Zorn's Lemma, and all that" at math.berkeley.edu/~gbergman/grad.hndts $\endgroup$ – Arturo Magidin Mar 15 '11 at 4:41
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    $\begingroup$ @Arturo, as far as I've seen the term cardinality means the equivalence class under the relation "there exists a bijection", while assuming AC it reduces to $\aleph$ numbers, and without AC it becomes a different beast altogether. $\endgroup$ – Asaf Karagila Mar 15 '11 at 13:55

This is an old post, but because Arturo's otherwise good answer is a bit cavalier on choice usage and the comments don't make the exact level of choice needed entirely clear, I thought I'd explain my own approach to the question in a ZF framework.

We can establish the following with no well-orderability assumptions:

Lemma: If $X$ is a nonempty set and $X\times X\approx X$ (meaning there is a bijection from $X\times X$ to $X$), then $\bigcup_{n\in\omega}X^n\approx\omega\times X$.

Proof: Fix a bijection $f:X\times X\to X$ and $x_0\in X$. Then we can define

$$g_0:x\in X^0\mapsto x_0$$ $$g_{n+1}:x\in X^{n+1}\mapsto f(g_n(x\restriction n), x_n)$$

Then by induction we have that $g_n$ is an injection from $X^n$ to $X$, and then $$h:x\in\bigcup_{n\in\omega}X^n\mapsto \langle \operatorname{dom}x,g_{\operatorname{dom}x}(x)\rangle$$

is an injection from $\bigcup_{n\in\omega}X^n$ to $\omega\times X$. (We use that $X^n$ for different $n$ are disjoint because $x\in X^n$ implies $x:n\to X$ so that $\operatorname{dom}x=n$.) For the reverse inequality, define

$$j:n\in\omega,x\in X\mapsto(z\in n+1\mapsto x).$$

Then if $j(n,x)=j(m,y)$ we have $(z\in n+1\mapsto x)=(z\in m+1\mapsto y)$ so the domains of the functions are equal, i.e. $n+1=m+1\Rightarrow n=m$, and also $x=j(n,x)(0)=j(m,y)(0)=y$.

Adding an assumption of well-ordering allows us to simplify the statement to what we are after:

If $X$ is an infinite well-orderable set, then the set $[X]^{<\omega}$ of finite subsets of $X$ is equipollent to $X$.

Proof: Since singletons are in $[X]^{<\omega}$ and naturally bijective with $X$, $X\preceq[X]^{<\omega}$. For the converse, we have $X\times X\approx X$ because $X$ is infinite well-orderable, so by the lemma $$\bigcup_{n\in\omega}X^n\approx\omega\times X\preceq X\times X\approx X;$$ thus $\bigcup_{n\in\omega}X^n$ is also well-orderable, so we can reverse the surjection $f:\bigcup_{n\in\omega}X^n\to[X]^{<\omega}$ which maps each function to its range to get an injection. Thus, $X\preceq [X]^{<\omega}\preceq\bigcup_{n\in\omega}X^n\preceq X$.


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