數學中的集合的閉集

第一條命題只有 T4 (normal) space 先正確. 所以我假設你問緊o既係 Euclidean Space R^n
(Definition of normal space === the statement you want to prove!!)
如果唔Normal, 就有emperorkeith2006 給的那個counter example.

My chinese is not good in math language, write in english.
1) F1 and F2 are closed (閉集), then distance D from F1 to F2 >0. (distance = infimum (最大下限) length from a in F1 to b in F2) because if distance =0 there is a sequence (a_i,b_i) in F1 and F2 getting closer and closer, and finally will converge to the same point, so F1 intersect F2 is nonempty.

Now let e=D/3>0. and let B(x,e) be an open ball (開集) of radius e around a point x. Then U B(x,e) where x in F1, will be open set (開集) G1, and U B(y,e) where y in F2, will be open set G2, and they will not intersect each other.

2) Let me clear up some language:
若有界 (bounded) 閉集(closed set) 族 (family){F(i)}中任何有限個元的交非空 (finite intersection property),求證這個閉集族的無限個元的交都是非空的.

bounded + closed = compact (緊集) in R^n
compact set has property: every open cover has finite subcover (任何開覆蓋都有有限子覆蓋)

Let F(i),i in A, be an infinite family.
Fix F(1) in your family.
If x in F(1) does not belong to F(i) for all i in A
then x belongs to F(i)* for some i in A (* = complement (補集))
so U F(i)* will be open cover for F(1)
but F(1) is bounded and closed => compact, it has finite open cover F(i_1)*, ..., F(i_n)*
so F(1) < F(i_1)* U ... U F(i_n)*
taking complement again, this means F(i_1) ∩ .... ∩ F(i_n) ∩ F(1) = empty set,
contradict finite intersection property.

So x is in F(i) for all i in A
so ∩ F(i), i in A, is nonempty.