# Is ordering done component wise in the hyperreals?

In the ultrafilter construction of the hypperreals, is ordering done component wise? I.e. if $a,b$ are hyperreal numbers then $a = [x_n]$ and $b = [y_n]$ where $[x_n]$ and $[y_n]$ are equivalence classes of the set $\mathbb{R}^\mathbb{N}$ under the equivalence realtion: $(r_n) \sim (s_n)$ if $\{n: r_n = s_n\}\in \mathcal{F}$ where $\mathcal{F}$ is a non-principle ultrafilter on $\mathbb{N}$.

Then do we say $[x_n]<[y_n]$ if $\{n: x_n<y_n\}\in \mathcal{F}$?

• Exactly! All properties of the order are easy verifications and the linearity follows from the ultrafilter property. – Bib-lost May 6 '17 at 17:16
• Thanks, and if we say $|[x_n]|$ do we mean that we take the modulus of the components? i.e. if $[x_n] = \{-1,-2,-3,...\}$ then is $|[x_n]| = \{1,2,3,...\}$? – fosho May 6 '17 at 17:16
• @Dman Yup, everything is done componentwise. (Although you should write e.g. "$\vert [x_n]\vert=[\{1, 2, 3, ...\}]$", etc.) – Noah Schweber May 6 '17 at 17:38