11
$\begingroup$

(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?

$\endgroup$
9
$\begingroup$

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

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

$\endgroup$
11
$\begingroup$

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 $\Bbb R$.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.