Infinite sets and their Cardinality

Question

(I am a 13 year old so when you answer please don't use things that are TOO hard even though I actually can understand quite complex stuff)

I was studying Infinite sets and their cardinality (not in school, but just for fun) and I already know that the $|\mathbb N|$ is aleph naught $\aleph_0$ and $|\mathbb R|$ is aleph one $\aleph_1$. But I just have one question, does any set's cardinality ever get larger than aleph one? if so how?

Answer 1

You’re right when you say that $|\Bbb N|=\aleph_0$, but $|\Bbb R|$ is $2^{\aleph_0}$, often abbreviated to $\mathfrak{c}$, which may or may not be $\aleph_1$. The statement that $|\Bbb R|=\aleph_1$ is the so-called continuum hypothesis, often abbreviated to $\mathsf{CH}$; it’s known that the usual axioms of set theory do not imply that $\mathsf{CH}$ is true and also do not imply that it is false (assuming, as we do, that those axioms are themselves consistent). The great majority of mathematicians do not assume that $\mathsf{CH}$ is true.

Yes, there are larger cardinalities. Cantor’s theorem says that if $A$ is any set, $\wp(A)$, the set of all subsets of $A$, has larger cardinality than $A$. In particular, $\wp(\Bbb R)$, the set of all subsets of $\Bbb R$, has cardinality larger than $|\Bbb R|$; its cardinality is $2^{\mathfrak{c}}=2^{2^{\aleph_0}}$. The set of all sets of subsets of $\Bbb R$, written $\wp\big(\wp(\Bbb R)\big)$, is bigger yet: its cardinality is

$$\huge 2^{2^{\mathfrak{c}}}=2^{2^{2^{\aleph_0}}}\;.$$

Answer 2

You can always consider $P(X)$ of $X$, i.e., the set of subsets of $X$ and the cardinality of $X$ is smaller than that of $P(X)$.

Also, you can consider the space of functions $X \rightarrow X$.

You might be interested in the continuums hypothesis.

Answer 3

There is a famous theorem known as Cantor's theorem according to which for every set $S$, the cardinality of the power set $\mathcal P (S)$, that is the set of all subsets of $S$, is strictly larger than the cardinality of $S$. Thus, the set $\mathcal P (\mathbb R)$ of all subsets of real numbers has cardinality strictly larger than $\aleph_1$. You can find a proof of Cantor's theorem in basically any textbook as well as online. The proof is extremely elegant and short.

Answer 4

The cardinality of $\Bbb R$ is not $\aleph_1$. It is $2^{\aleph_0}$. Whether or not it is equal to $\aleph_1$ is known as The Continuum Hypothesis.

We cannot prove nor disprove that from the usual axioms of set theory (and mathematics). So it is possible that $|\Bbb R|=\aleph_1$, or that $|\Bbb R|=\aleph_2$; many many more values are possible.

There are canonical sets of size $\aleph_1$ and $\aleph_2$ and so on, whose construction is much too difficult to explain without having set a basic background on the topic. However Cantor's theorem tells us that $|X|<|\mathcal P(X)|$, where $\mathcal P(X)=\{A\mid A\subseteq X\}$ is the power set of $X$.

In the case $\mathcal P(\Bbb R)$ is a set whose cardinality is strictly larger than that of $\Bbb R$.