antisymmetry (Zorich)

[gefest_md]

I will write in english for a while. Answer in russian if you wish.
Is it true that (in Zorich) the axiom of antisymmetry is redundant (depends on the axiom of continuity)? I suspect it is. But I just know that some authors of analysis or other write redundant axioms.

[Oleg Zubelevich]

gefest_md в сообщении #1048478 писал(а):

axiom of antisymmetry

whats this

[gefest_md]

Here is how I think. I prove that LUB is unique by using the characteristic property of LUB and no antisymmetry. Then suppose $x\leqslant y$ and $y\leqslant x$. Let them be positive. Consider the segment $[0,y].$ The number $y$ is LUB. Then I show that $x$ is also LUB. Then by uniqeness $x=y$.

[Oleg Zubelevich]

gefest_md в сообщении #1048478 писал(а):

But I just know that some authors of analysis or other write redundant axioms.

even if you are right who will invent a new relation for reals in the standard course of analysis just to get rid from redundant axioms

[gefest_md]

Some history. I have written to the authors of a school handbook a remark on how the uniqueness of LUB have to be proved. They prove it by using the theorem of characteristic property of LUB but they put that theorem in the handbook after the place they used it. I proposed at first to use antisymmetry to prove uniqueness of LUB. Now I think the two theorem should be switched.

[arseniiv]

Удобнее всё-таки иметь $\leqslant$ отношением нестрого порядка сразу по определению, т. е. иметь три соответствующие аксиомы — рефлексивность, антисимметричность и транзитивность (0, 1, 2 у Зорича). Ну и заодно линейного (4 у Зорича). А до супремума ещё скакать надо!

[gefest_md]

I am wrong, because I can not prove trichotomy without antisymmetry. Or in order to prove $x<y\to\neg (y\leqslant x)$ I need antisymmetry.