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Let Q denote the set of rational numbers and S = {x | x belongs N ; N; x>=10}

Consider the Following POSETs

I. (Q ∩ [0, 1], ≤)

II. (S, ≤)

Which of the above POSETs are well ordered?

### 1 comment

Both, right ?

Only option B should be True not First Why ???

A well ordered set is a total ordered set in which every finite subset has a least element.

Now (Q ∩ [0, 1], ≤) will eventually give ([0,1],<=) wouldn't it ?

Now [0,1] basically contains a subset (0,1) which doesn't have least element , because here if you choose any number then i can always give you a smaller number than that.

by

@Shaik Masthan brother please Verify it ?

got it
thank you so much :)
that is closed interval, right ?

Finite sets which are Totally ordered sets are well ordered

@Na462

still you didn't reply to my comment, then what is the use of pinging me ?

@Shaik Masthan

That is closed interval I agree

but by definition : Finite sets which are Totally ordered sets are well ordered

now Q denotes the set of rational numbers

now Q $\cap [0,1]$ <=

there's infinite numbers between 0 to 1 isn't ??

therefore it's not a finite set

i:e 3.22 <= 3.23

3.23<= 3.24......goes on !!

between intersection operator, then 0 must be least number, right ?
Yeah there should exist a least number

And it's a   total order finite  set

If those Two conditions where satisfied

Then it's well - ordered
It's somewhat complex

That's why I'm leaving for now :p

Waiting for someone who properly explain
edited by

Sorry for late reply brother i was out.

In total ordered set we look for the whole set as well as all the subsets so cant i say (0,1) is a subset of [0,1] for which there is no least element ?

Its important to see for whole set and every finite subset of it :

https://en.wikipedia.org/wiki/Well-order

ok