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I'm currently reading about surreal numbers from here.

At multiple points in this paper, the author has stated that if $x\not\leq y$, then $x\geq y$.

Shoudn't the relation be "if $x\not\leq y$, then $x>y$"?

Hasn't the possibility of $x=y$ already been negated when we said $x\not\leq y$?

Thanks in advance.

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    It's not wrong to conclude $x\geq y$, since $x>y$ implies $x\geq y$. – Thomas Andrews Jun 25 '13 at 15:15
  • @ThomasAndrews- true. but when we say $x\not\leq y$, we are saying $x$ is not less than $y$, and neither can it be equal to $y$. When we say $x\geq y$, we are saying $x$ is either greater than, or equal to $y$. When the possiblity of equality has already been negated, is there much to be gained by also including equality amongst the list of possible relations? –  Jun 25 '13 at 15:18
  • Although I understand that it is mathematically sound, I don't quite see the point. –  Jun 25 '13 at 15:19
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    It's true that there is a stronger conclusion, I'm just saying he might not need it. – Thomas Andrews Jun 25 '13 at 15:19
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    Although there is nothing to be gained by including possible equality, there is also nothing lost, and that, I believe, is Thomas' point. If $x$ is indeed greater than or equal to $y$, then a priori $x \geq y$. – Chris Leary Jun 25 '13 at 15:22
  • I'd have to see a specific example to argue for sure. I'm not inclined to browse through a PDF to see why he might do this in any particular case, but sometimes all a mathematician needs is $x\geq y$ :) – Thomas Andrews Jun 25 '13 at 15:22
  • @ChrisLeary I haven't said nothing is lost. But I guess it depends on what you mean by "lost." – Thomas Andrews Jun 25 '13 at 15:24
  • @Thomas Andrews - Perhaps my quick response was phrased loosely. I was just trying to emphasize that the stronger strict inequality implies the weak inequality. We surely "lose" the fact that $x$ is, in fact, strictly greater than $y$. – Chris Leary Jun 25 '13 at 15:29
  • I note that there is also some odd usage of the symbol $\not\geq$ right up front in that article - it is used to compare surreal numbers $x$ and what looks like left and right "cuts" of numbers - so sets of numbers. In that sense, there might be something odd in this particular article about how it defines surreal number inequality. – Thomas Andrews Jun 25 '13 at 15:33

4 Answers4

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The paper defines $\leq$ and it is not necessarily the "standard" $\leq$ that you and I are used to. It is often better to substitute some other symbol like $x\preceq y$, and realize we have to prove stuff about this symbol before we can use it.

In particular, in that paper, it is proved in Theorem 1.4 that $x\not\leq y$ implies $y\leq x$. It does not prove that $y<x$.

So trying to reason about $\leq$ as it if means "less than or equal" is a mistake. There is no definition of "less than" to use here. There is only $\leq$.

There are examples where $x\leq y$ and $y\leq x$. This means that if we define $x<y$ in terms of $x\leq y$ and $x\neq y$ then we get $x<y$ and $y<x$ and therefore we find that $<$ defined this way is non-transitive.

I think the name "surreal" should give a strong hint at the non-standard nature of these "numbers."

Thomas Andrews
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  • I did not find $x-{X_L|X_R}$ anywhere but did find $x={X_L|X_R}$ (for instance between Axioms 1 and 2). Are you having difficulty viewing equals signs in the pdf? (This is not in joke, it happened to me once.) – Marc van Leeuwen Jun 25 '13 at 16:42
  • @MarcvanLeeuwen Really? That might just be my fonts or a sizing issue, I guess. Yep, my fonts. Zooming in, all those "-" characters were "=" signs :) Didn't even look like a thick hyphen, just a "-" sign. – Thomas Andrews Jun 25 '13 at 16:44
  • I also see that some of the times when I was seeing $x<y$ that it was $x\leq y$, depending on magnification. Arrrgh. In any event, I've corrected my answer. – Thomas Andrews Jun 25 '13 at 17:39
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In the paper linked to, there appears to be only a single primitive relation, written $x\leq y$, of which $y\geq x$ is just a notational variant. The equals part of the symbol is present because $x\leq x$ holds for all $x$ (theorem$~1.2$) but this is just a mnemonic device without formal implications. The negation of the relation $x\leq y$ is $x\not\leq y$, and it is written like that to remind the reader that it only affirms the absence of the relation $x\leq y$, not the presence of the relation $x\geq y$. In fact $x\not\leq y$ does imply $x\geq y$, but that is only shown in theorem$~1.4$.

Equality is not an explicitly mentioned relation (at least it seems so scanning rapidly), and so there is no reason to introduce $x>y$ as as shorthand for $x\geq y\land x\neq y$; for this reason theorem$~1.4$ is formulated the ways it is. But of course the hypothesis $y\not\leq x$ does exclude the possibility that $y$ actually is $x$; indeed the notation $x>y$ is introduced just after that theorem. However I do not agree with the given motivation "Theorem 1.4 means that the the surreal numbers are totally ordered", because they are not even partially ordered, merely preordered. In view of theorem 1.2 it is what I would call a total preorder (not sure the term is in use), a relation in which every pair is comparable at least one way, possibly both ways.

This is the next point; the paper states that $x\leq y\land x\geq y$ is not only possible when $x=y$. It introduces the relation $x\equiv y$ for $x\leq y\land x\geq y$, and given transitivity, this is an equivalence relation. What is really totally ordered is the set of equivalence classes for this equivalence relation.

Marc van Leeuwen
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You are correct, if we were speaking of $\leq/\geq$ relations we know and love, as standard ordering relations on the reals: The negation of $x \leq y$ is exactly $x > y$, and that would be the correct assertion if we were talking about a "trichotomous" ordering, where we take that for any two real numbers, one and only one of the following hold. $x\lt y, \lor x = y, \lor x>y$.

But your text is not wrong that $x \nleq y \implies x\geq y$, (that is, the right hand side is implied by the left hand side, and this would be a valid implication in even in the standard real numbers). And it seems your text is using strictly $\leq$ and $\geq$ so that for any two numbers $x, y$, we have one and only one of the following relations to consider: $x \geq y$ or $x \leq y$, andthese relations do not necessarily hold the same properties we know and love with respect to their standard meanings on the reals.

amWhy
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    Everybody here is assuming that this $\leq$ is some standardly defined $\leq$, but it is defined quite explicitly in the paper, and it is quite possible that is difficult or impossible to prove that $x\leq y$ and $y\leq x$ implies $x=y$. In particular, there are lots of cases where the author writes $x\not\leq Y$ where $x$ is a surreal number and $Y$ is a set of surreal numbers. – Thomas Andrews Jun 25 '13 at 15:41
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Here's another way of looking at it.

$$(x\le{y})\leftrightarrow (x<y \lor x=y)$$

$$(x\not\le{y})\leftrightarrow(x>y\land{x\ne{y}})$$

$$x\not\le{y}\rightarrow{x\ne{y}}$$

Therefore: $$x\not\le{y}\not\rightarrow{{(x\ge{y})}}$$

rurouniwallace
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  • Your last implication is problematic: $x\geq y \equiv (x \gt y) \lor (x = y)$, so the fact that $x \neq y$ doesn't imply the negation of $(x \geq y)$, because if $x \gt y$ is true, even if $x \neq y$, $x \gt y \lor x = y$ is nonetheless true, simply by virtue of the fact that $x > y$ is true. – amWhy Jun 25 '13 at 15:38
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    Everybody here is assuming that this $\leq$ is some standardly defined $\leq$, but it is defined quite explicitly in the paper, and it is quite possible that is difficult or impossible to prove that $x\leq y$ and $y\leq x$ implies $x=y$. In particular, there are lots of cases where the author writes $x\not\leq Y$ where $x$ is a surreal number and $Y$ is a set of surreal numbers. – Thomas Andrews Jun 25 '13 at 15:41
  • @amWhy problematic implication fixed. – rurouniwallace Jun 25 '13 at 15:43
  • @ZettaSuro: Problematic implication replaced by probelmatic (i.e., incorrect) non-implication – Marc van Leeuwen Jun 25 '13 at 15:48
  • @MarcvanLeeuwen Would you care to point out what this supposed "problematic non-implication" is? I'm all ears. – rurouniwallace Jun 25 '13 at 15:51
  • One has $x\not\le{y}\rightarrow{{(x\ge{y})}}$, which is the implication that your last expression (incorrectly) denies the truth of. Indeed $x\not\le{y}\leftrightarrow x>y\rightarrow{{(x\ge{y})}}$. (By the way in your second line "$x>y\land x\neq y$" the second part is a useless addition.) – Marc van Leeuwen Jun 25 '13 at 15:57
  • If it is not true that $x\not\leq y \rightarrow x\geq y$, then you can come up with a counter example - a case where $x\not\leq y$ but it is not true that $x\geq y$. – Thomas Andrews Jun 25 '13 at 16:20