Countability of antichains

kindlychung

[这个贴子最后由kindlychung在 2010/09/25 11:13am 第 1 次编辑]

I think the set of antichains is a subset of P(P(B)). Intuition tells me it';s uncountable, yet I can';t provide a proof.

elimqiu

Sure there are uncountable antichains of P(N).
Let N1 be odd numbers of N, N2 = N \ N1
Let g: N1 → N2 be the natural 1-1 correspondence.
Define A = { E ∪ g(N1 \ E)  | E ∈ P(N1) }
You can see A is an antichain of P(N) that is uncountable.
Well this is a problem I never encounter before. Nice solutioin of mine!

kindlychung

Nice. Thanks.

elimqiu

Prove that A = { E ∪ g(N1 \ E)  | E ∈ P(N1) } is
(1) uncountable
(2) S, T ∈ A， S≠T → S△T is not empty   i.e. neither one is subset of the other.
Proof. Assume that E,F ∈ P(N1) and  E≠F, then
(1) (E ∪ g(N1 \ E))∩N1 = E ≠ F = (F ∪ g(N1 \ F))∩N1
    hence  (E ∪ g(N1 \ E))  ≠  (F ∪ g(N1 \ F))
    therefore |A| = |P(N1)| is uncountable
(2) If (E ∪ g(N1\E)) is a subset of  F = (F ∪ g(N1\F))
    then E = (E ∪ g(N1\E))∩N1 is a subset of (F ∪ g(N1\F))∩N1 = F
    So N1\E is not a subset of N1\F and
    g(N1\E) = (E ∪ g(N1 \ E))∩N2 is not a subset of g(N1\F)== (E ∪ g(N1\F))∩N2
    this implies that (E ∪ g(N1\E)) is not a subset of  F = (F ∪ g(N1\F))
    a contradiction. Therefore none pair of elements in A can be a chain in the sense of one is a subset of the other.
    and so

wangyangkee

为老外顶帖，为 elimqiu 顶帖，乐得赚积分，，，

moranhuishou

下面引用由wangyangkee在 2010/10/07 04:45pm 发表的内容：
为老外顶帖，为 elimqiu 顶帖，乐得赚积分，，，

是的，不为别的，外国朋友嘛。。

fleurly

眼神有问题
楼主多半是中国人
有可能是在外国留学的。猜测可能是美国