I'm trying to come up with a bijection between $\mathbb{R}$ and a complete totally ordered field $F$. Bearing in mind that $F$ is arbitrary, is it still OK to use operators such as $+$, $-$, $\cdot$, $/$ and $\sum$ on elements of $F$, and have it be understood from context that these are supposed to be $F$'s operators?

  • $\begingroup$ This generally works. If you want to be extra careful, you can use $\oplus$ and $\otimes$ instead, and still use $(-a)$ for the additive inverse of $a$, and $a^{-1}$ for the multiplicative inverse of $a$. I wouldn't bother with this unless you have equations that mix $+$ and $\oplus$, or $\times$ and $\otimes$. Note that $\sum$ can only be a finite sum without additional assumptions. $\endgroup$
    – Mark
    Jan 16, 2018 at 10:39
  • $\begingroup$ If you really want to be unambiguous, you can use something like $+_{\Bbb R}$ and $+_F$, or $+$ and $\oplus$, but it's usually easy to tell from context whether you're adding elements in $\Bbb R$ or elements in $F$. $\endgroup$
    – Arthur
    Jan 16, 2018 at 10:40
  • $\begingroup$ Great, thanks guys :) $\endgroup$
    – Simplex
    Jan 16, 2018 at 10:41

1 Answer 1


Provided that it's clear from context whenever you write $x+y$ which field $x$ and $y$ come from, then it's clear which field $+$ belongs to, so it's fine. (And similar for other operators.)


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