Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the cardinality of the set of hyperreal numbers?

share|improve this question
2  
What do you think it is? Also, you could have checked on Wikipedia before asking here. –  t.b. Jul 27 '11 at 12:43
5  
Your question literally asks about the cardinality of hyperreal numbers themselves (presumably in their construction as equivalence classes of sequences of reals). You probably intended to ask about the cardinality of the set of hyperreal numbers instead? –  joriki Jul 27 '11 at 12:49
1  
@joriki: Either way all sets involved are of the same cardinality: $2^\aleph_0$. –  Asaf Karagila Jul 27 '11 at 13:02
    
Yes, I was asking about the cardinality of the set oh hyperreal numbers. Thank you. –  zar Jul 27 '11 at 16:50

1 Answer 1

up vote 13 down vote accepted

The Hyperreal numbers can be constructed as an ultrapower of the real numbers, over a countable index set. I will assume this construction in my answer.

The hyperreal field $^*\mathbb R$ is defined as $\displaystyle(\prod_{n\in\mathbb N}\mathbb R)/U$, where $U$ is a non-principal ultrafilter over $\mathbb N$.

Informally, we consider the set of all infinite sequences of real numbers, and we identify the sequences $\langle a_n\mid n\in\mathbb N\rangle$ and $\langle b_n\mid n\in\mathbb N\rangle$ whenever $\{n\in\mathbb N\mid a_n=b_n\}\in U$.

Since $U$ is an ultrafilter this is an equivalence relation (this is a good exercise to understand why).

Since the cardinality of $\mathbb R$ is $2^{\aleph_0}$, and clearly $|\mathbb R|\le|^*\mathbb R|$. On the other hand, $|^*\mathbb R|$ is at most the cardinality of the product of countably many copies of $\mathbb R$, therefore we have that $2^{\aleph_0}=|\mathbb R|\le|^*\mathbb R|\le(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\times\aleph_0}=2^{\aleph_0}$. Therefore the cardinality of the hyperreals is $2^{\aleph_0}$.

A usual approach is to choose a representative from each equivalence class, and let this collection be the actual field itself. However we can also view each hyperreal number is an equivalence class of the ultraproduct.

Suppose $[\langle a_n\rangle]$ is a hyperreal representing the sequence $\langle a_n\rangle$.

Since $U$ is non-principal we can change finitely many coordinates and remain within the same equivalence class. So for every $r\in\mathbb R$ consider $\langle a^r_n\rangle$ as the sequence:

$$a^r_n = \begin{cases}r &n=0\\a_n &n>0\end{cases}$$

We have only changed one coordinate. Therefore the equivalence to $\langle a_n\rangle$ remains, so every equivalence class (a hyperreal number) is also of cardinality continuum, i.e. $2^{\aleph_0}$ (as it is at least of that cardinality and is strictly contained in the product, which is also of size continuum as above).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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