推 zako1113 :Q2: any intersection of closed sets is closed 11/04 19:18
→ zako1113 :bounded by x_1-1 and x_1+1 11/04 19:20
推 zako1113 :Q3: cantor's intersection theorem? 11/04 19:30
1.
Def:
A nonempty subset X of R is called perfect provided it is closed
and each neighborhood of any point in X contains infinitely many
points of X.
Q:By this kind of Def of perfect set.
Show that every perfect subset X of R is uncountable.
<pf>
X is perfect subset of R, by Def. X must be infinite.
=>Thus X must be countable or uncountable.
Suppose X is countable.
Let X={x1,x2,x3,....}
The interval U1 = (x1 - 1, x1 + 1) is a neighborhood of x1 ,and there
are infinitely many elements of X contained in U1.
Take one of those elements, say x2 and take a neighborhood U2 of x2 such that
closure( U2 ) is contained in U1 and x1 is not contained in closure( U2 ).
Again,the neighborhood U2 contains infinitely many elements of X.
Select an element, say x3, and take a neighborhood U3 of x3 such that
closure( U3 ) is contained in U2 but x1 and x2 are not contained in
closure( U3 ) .
/*why we need to let closure(Un),and U1 be open ? */<<Q1
and we have following three condition.
1. closure( Un+1 ) in Un
2. xj is not contained in Un (for all 0 < j < n
3. xn is contained in Un
Now consider the set
V = ∩( closure( Un ) ∩ S )
Then each set closure(Un)∩S is closed and bounded, hence compact.
/* why is closed? why is bounded? */<<Q2
and ( closure(Un+1)∩S ) in (closure(Un)∩S ).
Therefore, by the above result, V is not empty.
/* why we can know V is not empty? */<<Q3
But which element of S should be contained in V ? It can not be x1, because
x1 is not contained in closure( U2 ). It can not be x2 because x2 is not in
closure( U3 ), and so forth.
Hence, none of the elements { x1, x2, x3, ... } can be contained in V.
But V is non-empty, so that it must contain an element not in this list. That
means, however, that S is not countable.
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