Simple example for a set with cardinality of $\aleph_2$

Question

What is a simple or commonly known set that has a cardinality equal to $\aleph_2$ or greater?

Answer 1

The problem with alephs, other than $\aleph_0$, is that they don't really live in the 'commonly known' world. Think of the sets that are commonly known by mathematicians:

• Finite sets;
• Number sets $\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$, which have cardinality $\aleph_0$ or $2^{\aleph_0}$;
• Power sets, which have cardinalities of the form $2^{\kappa}$;
• Function sets, which have cardinalities of the form $\kappa^{\lambda}$;
• Sets of sequences (see function sets);
• Products and unions of sets, which have cardinalities of the form $\kappa\lambda$ or $\kappa + \lambda$;
• ...and so on.

The realm of the 'commonly known' consists of cardinals formed from finite cardinals and $\aleph_0$ by taking powers, sums and products, not cardinal successors.

I'd say the most commonly known set of cardinality equal to $\aleph_2$ is the (von Neumann) ordinal $\omega_2$, or indeed any ordinal $\alpha$ for which $\omega_2 \le \alpha < \omega_3$.

However, we can certainly construct cardinals which are greater than or equal to $\aleph_2$. Indeed, we know that $2^{\kappa} > \kappa$ for all cardinals $\kappa$, so $$2^{2^{\aleph_0}} > 2^{\aleph_0} > \aleph_0 \quad \Rightarrow \quad 2^{2^{\aleph_0}} \ge (2^{\aleph_0})^+ \ge \aleph_0^{++} = \aleph_2$$ and hence any set of cardinality $2^{2^{\aleph_0}}$ will do for you. For instance $\mathcal{P}(\mathbb{R})$ or $\mathbb{R}^{\mathbb{R}}$, both of which are sets appearing in (say) real analysis.