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What is the cardinality of the set of hyperreal numbers?

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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).

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