Why is the $<$ relation not antisymmetric?

Question

I'm a bit confused. Consider the relation $<$ on $\mathbb N$. Then $<$ is per definition antisymmetric if

$$a<b \land b<a \implies a=b.$$

The problem which I have is that the premise of the implication can never be true. It is not possible for a natural number to be less and greater than another number at the same time. But when the premise is false (which is the case), the implication is always true. So it follows that $<$ is an antisymmetric relation - which is obviously wrong. Where is the fallacy?

Why is it obviously wrong to say it's antisymmetric? Words are defined by their definitions, not by what it seems to you they sort of really mean... — David C. Ullrich, May 05 '20 at 17:41
Because I know that $<$ is not antisymmetric but only $\leq$ - for the reason of missing antisymmetry. — ATW, May 05 '20 at 17:42
HOW do you "know" it's npt antisymmetric? "for the reason of missing antisymetry" doesn't make much sense to me... — David C. Ullrich, May 05 '20 at 17:42
@Thomas perhaps you are mixing up concepts. $<$ is antisymmetric, but it fails to be reflexive which means that it does not define a partial order — Ben Grossmann, May 05 '20 at 17:47
Sorry @DavidC.Ullrich, my mind was elsewhere. I meant $<$ does not define a partial order but $\leq$ for the reason of antisymmetry. But Omnomnomnom pointed out the problem. That was the fallacy. Thank you both. — ATW, May 05 '20 at 17:52

egreg · Accepted Answer · 2020-05-05T17:47:41.123

2

It is antisymmetric. The statement

for all $x,y$, if $x<y$ and $y<x$, then $x=y$

is true. It is true because, for every $x$ and $y$, the premise is false, but where's the problem?

Look at it from a different direction. Can you prove it is not antisymmetric? You should find $x$ and $y$ such that $x<y$, $y<x$ and $x\ne y$.

Are you perhaps surprised in learning that equality is antisymmetric? I hope not. Indeed, equality is a partial order relation, as well as an equivalence relation. Just apply the definitions.

Can you? Of course not. So it is antisymmetric.

edited May 05 '20 at 17:47

answered May 05 '20 at 17:42

egreg

238,574

Okay, I always thought I would know. But yeah, it does make sense. So there is no fallacy. Would the relation e.g. "$A$ is an anchestor of $B$" on the set of all humans (squared) be antisymmetric as well? Would be the same case, wouldn't it? – ATW May 05 '20 at 17:47
1

@Thomas Yes, it's antisymmetric for the same reason (excluding time travel, that is ;-). Of course this adds nothing to our knowledge about the relation, because it'a a “vacuously true” statement. – egreg May 05 '20 at 17:48
If $x\lt y$ and $y\lt x$ then isn't it also true that $x\ne y$? – John Douma May 05 '20 at 18:00
1

@JohnDouma Yes, of course. But this doesn't affect the truth of “if $x<y$ and $y<x$, then $x=y$”. – egreg May 05 '20 at 20:05

Why is the $<$ relation not antisymmetric?

1 Answers1