set with cardinality $3^{\aleph_0}$

Question

I am having trouble understanding the cardinal number $3^{\aleph_0}$.
I understand basic cardinal numbers and properties but I am unable to construct any set with cardinality $3^{\aleph_0}$.
A little help would be highly praised.

Answer 1

$3^{\aleph_0}$ is not really a thing. The main reason why people write $2^{\aleph_0}$ is because the cardinality of the power set of $\{1,2,\ldots,n\}$ is equal to $2^n$ for all positive integer $n$, and hence $2^{\aleph_0}$ is used as a short form to indicate the cardinality of the power set of the natural numbers $\mathbb N$.

Answer 2

The cardinality of the set of functions from $\mathbb{N}$ to $\{-1,0,1\}$ is $3^{\aleph_0}$, which is the same thing as $2^{\aleph_0}$, since $2<3<4$ leads to $$ 2^{\aleph_0} \leq 3^{\aleph_0} \leq 4^{\aleph_0} = 2^{2\aleph_0} = 2^{\aleph_0}.$$

Answer 3

The most straightforward example of a set with cardinality $3^{\aleph_0}$ is the set of all functions from a domain of cardinality $\aleph_0$ (such as the set $\{0,1,2,3,\ldots\}$) into a set of cardinality $3$, such as the set $\{1,2,3\}.$

There are ways to show that $3^{\aleph_0}$ is the same as $2^{\aleph_0},$ including constructing an explicit one-to-one correspondence between the set mentioned in my first paragraph above and a similarly constucted set of cardinality $2^{\aleph_0}.$ But your question emphasizes the problem of exhibiting a set of cardinality $3^{\aleph_0}.$