Quick easy question that I don't understand

[cinder dome]

is this (Z −,≥) a well ordered set ?

[mellow totem]

Do you know what it means to be well-ordered set?

[cinder dome]

aproximately, like it needs to be a totally ordered

[mellow totem]

total order is a start, yes

[cinder dome]

and there needs to be a least element in the set

but least should depent on the comparaison element like here it's ≥

according to me here in any set we should find a maximum, and it works because any set that is on Z- has a maximum

[mellow totem]

correct

[mellow totem]

according to me here in any set we should find a maximum
what?

[cinder dome]

the thing I don't understand is that if the minimum is defined by the element composing the poset or not, like if the poset is defined by the division and suppose it's totally orderet when we're looking the for the smallest element we're really looking the the minimum ? min{2,5,7,3,5,6} = 2 or will it depend on the type of comparaisons we're making ?

that's the thing bothering me

if that's not true and it doesn't depends on the type of compaisons and it's really the minimum, then (Z −,≥) is not a well ordered set but (Z +,≥) is a well ordered set

[mellow totem]

Your grammar is making this hard to parse, but I think I get what you are saying. The minimum is based on your ordering relation. For any ordering $(X, \prec)$, we define precedence between two elements $x,y\in X$. We say $x$ preceeds $y$ if $x\prec y$. An oredered set $(X, \prec)$ has a minimum element $m$ if, for every other $x\in X$, $m\prec x$.

[quiet hatch]

SWR

[mellow totem]

In your problem, $(\bZ^-,\ge)$ has a minimum element $m$ if, for every other $z\in\bZ^-$, $m\ge z$.

[quiet hatch]

SWR

[cinder dome]

and for $(\bZ^-,\le)$ the miminum is $z\in\bZ^-$, $m\le z$ ?

[quiet hatch]

nusoa 🍉

[cinder dome]

Sorry, English isn't my first language, so sometimes I get mixed up when trying to understand things. It makes me mix up my English too, haha 😅

[mellow totem]

fyi the math you wrote here makes no sense. But by the context I know what you mean. Yes. For $(\bZ^-,\le)$, if $m\in\bZ^-$ is a minimum, then $m\le z$ for every other $z\in\bZ^-$

[quiet hatch]

SWR

[mellow totem]

I understand. The same thing happens to me.

[cinder dome]

You're killing me, haha! I can totally tell you've got a background in math—probably a master's or bachelor's!

[mellow totem]

I am simply meticulous.

[cinder dome]

though so like $(\bZ^-,\le)$ is a well ordered set but $(\bZ^-,\ge)$ isn't, ok I think I understood. Thank you !

[quiet hatch]

nusoa 🍉

[mellow totem]

You have this backwards

[cinder dome]

oh, ok yes I get it, my mind was a bit twisted around, you're right

[mellow totem]

So long as you understand the difference, then you're all set

[cinder dome]

yes, thank you ❤️

[cinder dome]

Also yes your right my think didn’t made sense bc I didn’t understand