Math Is Fun Forum

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Set , Help

To question II , I did this .   

Check it please.

Last edited by Stanley_Marsh (2007-04-23 20:06:02)

---

Numbers are the essence of the Universe

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Set , Help

A partial ordering has to be (i) reflexive (ii) antisymmetric and (iii) transitive.

Therefore ~ is a partial order on A.

Also, since a ≤ b or b ≤ a for any negative real numbers a,b, we have (a,a)~(b,b) or (b,b)~(a,a) for any (a,a),(b,b) ∈ S, i.e. ~ is linear (total) on S.

Last edited by JaneFairfax (2007-04-23 22:38:12)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Set , Help

Can’t think of another one, really.

Last edited by JaneFairfax (2007-04-24 00:11:16)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Set , Help

Okay, I found another one. Try

Then ~ is also linear on U (easily checked).

To prove it’s a maximal linearly ordered subset, consider any element (x,y) not in U.

Hence adding another element to U will make U no longer linearly ordered, i.e. U is a maximal linearly ordered set.

Last edited by JaneFairfax (2007-04-24 02:04:23)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Set , Help

Note that

also works.

In fact, it turns out that there are infinitely many maximal linearly ordered subsets containing S! Given any real number c ≥ 0, the set

is also a maximal linearly ordered subset containing S.

Last edited by JaneFairfax (2007-04-24 02:57:37)

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Set , Help

I don't really understand what is maxiaml linearly ordered subsets   , and the way to find it.
It says I need to use Hausdorff Maximality Principle , I have no idea.

Last edited by Stanley_Marsh (2007-04-24 09:22:39)

---

Numbers are the essence of the Universe

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Set , Help

The Hausdorff Maximality Principle only tells you that a maximal linearly ordered subset must exist. It doesn’t tell you how to find such a subset. You must find it yourself.

A maximal linearly ordered set is a linearly ordered set that is not a proper subset of any other linearly ordered set. In our example, S is linearly ordered set, but it is not maximal because it is a proper subset of T. On the other hand, T is maximal. For, suppose T is a proper subset of V. Then V contains an element not in T, i.e. an element (a,b) such that a ≠ b. If you set c = (a+b)⁄2, then (c,c) and (a,b) would be elements in V that are not comparable – that is to say, V is not linearly ordered. Hence T is maximal, since it is not a proper subset of any linearly ordered set.

For the other maximal sets I’ve found, it will help a great deal if you can visualize them in your head. That was what I did – most of the time I rely heavily on intuitive visualizing when I’m proving something.

Last edited by JaneFairfax (2007-04-24 10:00:50)

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Set , Help

About question 2 , if it asks what are the equivalent classes .
Then , it's  A={m-n=0(mod 3)} , B={s-t=1 (mod 3) }, C={a-b = 2 (mod 3) }, Since the union of A,B,C = Z , right???

Last edited by Stanley_Marsh (2007-04-25 02:50:18)

---

Numbers are the essence of the Universe

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Set , Help

Yes, that’s correct. The equivalence classes are

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Set , Help

Thx ,Jane

Last edited by Stanley_Marsh (2007-04-25 11:37:31)

---

Numbers are the essence of the Universe