～～集合的基本問題～～

若 X 和 Y 為集合，且 X 的所有元素都是 Y 的元素，則有：

X 是 Y 的子集（或稱包含於 Y ）；
X ⊆ Y;
Y 是 X 的父集（或稱包含 X ）；
Y ⊇ X.
(1)
可以﹐因為Z的每一個完素都屬於Z
所以set Z 是它自己的子集
2)
證明：給定任意集合 A ，要證明
圖片參考：http://upload.wikimedia.org/math/d/0/9/d096fc15d57854ec89d746709b02e52e.png
沒有元素。
對有經驗的數學家們來說，推論 "
圖片參考：http://upload.wikimedia.org/math/d/0/9/d096fc15d57854ec89d746709b02e52e.png
沒有任何元素，如何使"這些元素"成為別的集合的元素？ 換一種思維將有所幫助。
為了證明
圖片參考：http://upload.wikimedia.org/math/d/0/9/d096fc15d57854ec89d746709b02e52e.png
一定是 A 的子集。
Proof: Given any set A, we wish to prove that ø is a subset of A. This involves showing that all elements of ø are elements of A. But there are no elements of ø.
For the experienced mathematician, the inference " ø has no elements, so all elements of ø are elements of A" is immediate, but it may be more troublesome for the beginner. Since ø has no members at all, how can "they" be members of anything else? It may help to think of it the other way around. In order to prove that ø was not a subset of A, we would have to find an element of ø which was not also an element of A. Since there are no elements of ø, this is impossible and hence ø is indeed a subset of A.
3)
一定﹐因為對一個non-empty set 來說﹐空集和它自己都是其subset
4)
可以﹐考慮實數集R
則任一區間(a,b)都是其subset﹐好明顯有無限多subset

I love the part showing that the empty set is a subset of any set.
Particular impressed that the author understands the difficulty of understanding the argument by beginers.

4) 一個 set 可否有無窮多個 subset?
可以，除非這個集只有 有限個元，因為任何有限元集(設有n個元)只有 有限個子集(2^n 個)。