w**********a
发帖数: 107 | 1 sigma algebra is closed under complement and countable union and Borel sigma
algebra is generated by open sets. So can we say that any Borel set is the
countable union of open sets and closed sets? |
j******w
发帖数: 690 | |
w**********a
发帖数: 107 | 3
why can you give me a counterexample?
thanks
【在 j******w 的大作中提到】
: No.
|
B****n
发帖数: 11290 | 4 You can consider the famous example of Cantor set.
Each time you remove 1/3 of the interval on [0,1]
sigma
the
【在 w**********a 的大作中提到】
: sigma algebra is closed under complement and countable union and Borel sigma
: algebra is generated by open sets. So can we say that any Borel set is the
: countable union of open sets and closed sets?
|
w**********a
发帖数: 107 | 5
but Cantor set is a closed set
【在 B****n 的大作中提到】
: You can consider the famous example of Cantor set.
: Each time you remove 1/3 of the interval on [0,1]
:
: sigma
: the
|
j******w
发帖数: 690 | 6 the set of irrationals.
proof: Otherwise, there is a countable sequence open sets U_n, n\in \omega so that
\cap_n U_n=Q (Q is the set of rationals). Since each U_n is comeager, Q is comeager
which is a contradiction.
Actually there is a hierarchy of Borel sets. Each level doesn't collapse.
【 在 wakakayikaka (哇咔咔一咔咔) 的大作中提到: 】 |
B****n
发帖数: 11290 | 7 你每次挖的是closed interval 那cantor set就一定不是closed set
【在 w**********a 的大作中提到】
:
: but Cantor set is a closed set
|
w**********a
发帖数: 107 | 8
so that
comeager
哦 大牛写的俺都没怎么看懂。。。你意思说set of irrationals is a Borel set?
【在 j******w 的大作中提到】
: the set of irrationals.
: proof: Otherwise, there is a countable sequence open sets U_n, n\in \omega so that
: \cap_n U_n=Q (Q is the set of rationals). Since each U_n is comeager, Q is comeager
: which is a contradiction.
: Actually there is a hierarchy of Borel sets. Each level doesn't collapse.
: 【 在 wakakayikaka (哇咔咔一咔咔) 的大作中提到: 】
|
w**********a
发帖数: 107 | 9
I am sure Cantor Set is closed since its complement is the countable union
of open intervals which is open.
【在 B****n 的大作中提到】
: 你每次挖的是closed interval 那cantor set就一定不是closed set
|
k******2
发帖数: 111 | 10 借宝地问下，borel set 到底是啥？
Union of Countable h-intervals? |
Q***5
发帖数: 994 | 11 It is trivial to prove that the set of irrationals is Borel.
I think what he means is the following:
Proof by contradition.
Suppose the set of irrationals (denoted by A) is a countable union of open
sets and closed sets. Since A does not contain any open segments, A must be
a countable union of closed sets, that means, its compensation set Q (the
set of all rationals) is the intersection of countable open sets. Each of
these open sets has the following property: the compensation of the open set
(which is a subset of Q) is of measure zero. As a result, Q also has this
property -- which we know is wrong.
【在 w**********a 的大作中提到】
:
: I am sure Cantor Set is closed since its complement is the countable union
: of open intervals which is open.
|
