5
$\begingroup$

From the four postulates of the Dedekind cuts, namely (for (a,b) denoted as the cut, a,b being subsets of the rationals):

  1. Every rational number lies in exactly one of the sets a,b,
  2. a,b are not empty,
  3. Every element of a is smaller than every element of b,
  4. a has no biggest element,

it seems that Conway keeps only 3. Since 4. guarantees, that a real number x cannot be given by different cuts, can a surreal number be then given by different generalized cuts? (I'm just beginning to get into the Surreal numbers, so I don't have much preknowledge!)

$\endgroup$

2 Answers 2

6
$\begingroup$

Two different surreal numbers $x,y$ are equal if $x\leq y$ and $y\leq x$, where that order is defined in terms of their recursive parts. So you can have $$2=\{1\mid\}=\{0,1\mid\}=\{1\mid4\}=\{-17,1.5\mid\pi\}$$

It's good to think of Dedekind cuts only as far as "Ah, we can use sets of relatively simple numbers to build more complex numbers", but not a lot further than that.

$\endgroup$
0
$\begingroup$

Matthew Daly's answer is correct. I want to clarify the effect of dropping conditions 1 and 4 when dealing with rationals in the left and right sets of surreals.

If $r$ is real, then (using intervals of dyadics or rationals or reals) $\{(-\infty,r)\mid(r,\infty)\}$ is the surreal version of $r$ (note that condition 1 is violated). But $\{(-\infty,r)\mid[r,\infty)\}$ is less than $r$ by properties of this construction of surreals. In fact, it's infinitesimally less than $r$. Similarly, $\{(-\infty,r]\mid(r,\infty)\}$ is infinitesimally more than $r$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .