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Find the cardinality of set $A$: $$A = \{x \subseteq \mathbb R : |x| < \aleph_o \}$$

I have thought about it for a while and I've come to the conclusion that the cardinality of this set will be at least continuum, because every real number singleton is in this set. However, the only upper-bound of this cardinaltiy that I can think of is $\mathbb R^{\mathbb R}$ whose cardinality is more than continuum.

Any suggestions and hints would be most appreciated.

Aemilius
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  • Do you know that $\mathbb R^2$ has the same size as $\mathbb R $? That easily gives you that the collection of subsets of size 2 has the same size. The general result (that the collection of all finite sets also has the same size) follows along the same lines. – Andrés E. Caicedo Feb 04 '18 at 14:41

2 Answers2

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This is the number of finite subsets of $\Bbb R$. For a given natural number $n$, the number of $n$-element subsets of $\Bbb R$ is bounded above by $|\Bbb R^n|=|\Bbb R|$. So $|A|\le\sum_{n=1}^\infty|\Bbb R|=\aleph_0|\Bbb R|=|\Bbb R|$.

Angina Seng
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Recall that the cardinality of $\mathbb{R}$ is $2^{\aleph_0}$.

We can in fact show that the set of all countable subsets of $\mathbb{R}$ is $\mathbb{R}$-sized, and then the result follows because you're just asking for a subset of those.

Note that $(2^{\aleph_0})^{\aleph_0}$ is equal to $2^{\aleph_0 \times \aleph_0} = 2^{\aleph_0}$, so the number of countable subsets of $\mathbb{R}$ is just the cardinality of $\mathbb{R}$ again.

Patrick Stevens
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  • The problem with this answer is that it hides the difficulties. While it is true without using the axiom of choice that $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0} $, you need the axiom of choice to conclude that there are $2^{\aleph_0}$ countable sets of reals: it is consistent that there are more. On the other hand, you can argue directly about the number of finite subsets. – Andrés E. Caicedo Feb 04 '18 at 15:48
  • @AndrésE.Caicedo Fair enough; in questions of cardinal arithmetic I'm happy with Choice, but you're right to point this out. – Patrick Stevens Feb 04 '18 at 15:49