What is the basis of the hyperreal numbers? Let us consider the Hyperreal numbers as a vector space over the real numbers. This vector space is quite interesting.
Here are some interesting subspaces:


*

*The finite numbers

*The infinitesimal numbers


My question is, what is a basis of the Hyperreal numbers?
 A: Well, if one is allowed to consider infinite sums in $*\Bbb{R}$ (that is, a sum indexed in $*\Bbb{N}$), I believe I found one generating set that is almost a base.
First, considering binary representations, for each positive real number $x$, there is a sequence $(..., a_1, a_0, a_{-1}, ...)$, $a_n \in \{0, 1\}$ of bits such that $$x = ... a_2 4 + a_1 2 + a_0 + a_1 2^{-1} + a_2 2^{-2}...$$ There exists an $n_0$ such that $a_{n_0}=1$ and $a_n=0$ for $n > n_0$.
Applying the transfer principle, we conclude that the set $$B = \{2^n : n \in *\Bbb{Z}\}$$ is a generating set of vectors. But we can filter it a little more.
Consider two nonzero hyperreals $u$ and $v$ and say that they are equivalent if $u/v$, $v/u$ are both finite. This equivalence tells, roughly, that $u$ and $v$ have comparable magnitude, and in particular they are either both infinite, both real, or both infinitesimal.
Now, let $\beta$ be a set with one unique element of each equivalence class in $B$.
For any hiperreal $x$, assume w.l.g. that it is positive and write it as above. There is a $2^{m_0}$ in $\beta$ which is equivalent to $2^{n_0}$. Since $x - 2^{n_0} \leq 2^{n_0}$, $2^{m_0}$ is also equivalent to $x$. Therefore, $x 2^{-m_0} = b_0 + \delta$ where $b_0$ is real and $\delta$ is an infinitesimal. Then, $x = b_0 2^{m_0} + x_1$, where $x_1$ is infinitesimal with respect to $x$.
Now we apply transfinite induction or Zorn's lemma (by ordering representations of $x$) and write
$$x = b_0 2^{m_0} + b_1 2^{m_1} + b_2 2^{m_2} + \dots$$
