I***e
发帖数: 1136 | 1 1.
1) If A’s limit point set B is countable. We know B has to be a closed set.
Thus, B- is an open set with countable open intervals. One of these intervals
contain uncountable elements of A. Let’s assume it is (0, 1). Then you can
argue that there is a k such that (1/k, 1-1/k) contains infinite number of A’
s elements thus have an limit in (0, 1)…
2) It suffices to prove that A has one limit point in itself. Assume A and B
are disjoint. Again B is a closed set so B- is a countable union of open | I***e
发帖数: 1136 | 2 2.
1) Prove by contradiction.
A={x, f-(x)=f+(x) != f(x)}
B={x, f-(x)!=f+(x)}.
If A is uncountable, then from 1., A has a sequence x1, x2, … such that f(xi)
differs from lim{x->xi} f(x) at least by epsilon >0, and that lim xi->X for
some X in A. However, lim {x->X} f(x) does not exist. Thus the contradiction.
If B is uncountable, then also from 1., B has a sequence x1, x2,…, such that
f-(xi) and f+(xi) differ by at least epsilon >0 for all xi’s. And that lim
xi -> X in B. But now you can see tha | F***t
发帖数: 412 | 3 1。1) 我先证明如果A是不可数的，那么在闭区域
...[-2,-1],[-1,0],[0,1],[1,2]...中必有一个含有
不可数的A中的点。则在该区域中必有A的聚点。但不能
只有一个，因为如果只有一个的话，我们用一个小的开
区间把它盖住，剩下的点只能是有限个，因为无限个又
要有聚点。如果我们不断地缩小这个开区间，r=1/2,1/4....
这个小区间每次都只能排除有限个点，这个极限过程只能
排除可数个点，这证明A可数，矛盾。所以聚点不能是一个，
同理也不能是有限个，也不能是可数个，只能是不可数个。
这个证明没你的好。
intervals
A’
sets.
countable…
are
值
3。试构造一单调递增函数，该函数在某闭区间[a,b]上的断点构成的集合在[a,b]上dense
【在 I***e 的大作中提到】
: 1.
: 1) If A’s limit point set B is countable. We know B has to be a closed set.
: Thus, B- is an open set with countable open intervals. One of these intervals
: contain uncountable elements of A. Let’s assume it is (0, 1). Then you can
: argue that there is a k such that (1/k, 1-1/k) contains infinite number of A’
: s elements thus have an limit in (0, 1)…
: 2) It suffices to prove that A has one limit point in itself. Assume A and B
: are disjoint. Again B is a closed set so B- is a countable union of open
| F***t
发帖数: 412 | 4 1。2)
我假设A中没有它本身的聚点，那么每一个点都是孤立点，
都可以用一个小的开区间把它盖住，并把所有其余的排除
在外。如果有任意两个小的开区间交叉的话，让它们的半径
减半，这样它们就不会交叉了。这样所有的点都被disjoint
的开区间一一覆盖了。下面这招是在书上看的，看了之后
觉得非常高明：每个开区间中都存在一个有理数，这些
有理数可以和这些开区间一一对应，而有理数是可数的，
所以那些开区间也是可数的，从而A集合是可数的。矛盾。
这招是Walter Rudin用来证明单调函数有可数个断点的，
我看了后觉得非常深刻。看来你也是用类似的招法:)
接下来证明如果A中只有有限个A的聚点也矛盾，进而
如果只有可数个聚点也矛盾，所以A中一定有不可数个
A本身的聚点。
intervals
A’
sets.
countable…
are
值
3。试构造一单调递增函数，该函数在某闭区间[a,b]上的断点构成的集合在[a,b]上dense
【在 I***e 的大作中提到】
: 1.
: 1) If A’s limit point set B is countable. We know B has to be a closed set.
: Thus, B- is an open set with countable open intervals. One of these intervals
: contain uncountable elements of A. Let’s assume it is (0, 1). Then you can
: argue that there is a k such that (1/k, 1-1/k) contains infinite number of A’
: s elements thus have an limit in (0, 1)…
: 2) It suffices to prove that A has one limit point in itself. Assume A and B
: are disjoint. Again B is a closed set so B- is a countable union of open
| F***t
发帖数: 412 | 5 1。3)
double limit point是我发明的术语，不知道书上有没有这个说法。
这个证明和前一个类似，关键也在于找到能够和点一一对应的开区间。
如果不是double limit point，这个开区间就一定存在，一定能推出
矛盾。所以double limit point 也有不可数个。
1。4)
我开始还以为如果一个集合有不可数无穷多的点，那么它一定在某个
小区间内dense。因为这样的话，证明单调函数有至多可数断点就容易了。
但是这个是不对的。cantor集合就是反例。实际上要证明单调函数的
断点定理，只要double limit point就够了。思考过程上1.3在1.4之后。
intervals
A’
sets.
countable…
are
值
3。试构造一单调递增函数，该函数在某闭区间[a,b]上的断点构成的集合在[a,b]上dense
【在 I***e 的大作中提到】
: 1.
: 1) If A’s limit point set B is countable. We know B has to be a closed set.
: Thus, B- is an open set with countable open intervals. One of these intervals
: contain uncountable elements of A. Let’s assume it is (0, 1). Then you can
: argue that there is a k such that (1/k, 1-1/k) contains infinite number of A’
: s elements thus have an limit in (0, 1)…
: 2) It suffices to prove that A has one limit point in itself. Assume A and B
: are disjoint. Again B is a closed set so B- is a countable union of open
| F***t
发帖数: 412 | 6 2。
我的证明好象和你的也类似。
首先第一类断点必须是左极限右极限和函数值本身这三个
不全相等的。假设大的和小的相差为r，则可以构造一个
新的函数r(x)，该函数在原函数第一类断点的地方的值
为r，在非断点的地方值为0，在第二类断点的地方无定义。
假设原函数有不可数无穷多第一类断点，则有不可数无穷
都的x使r(x)>0。接下来证明必存在一epsilon>0，
存在不可数无穷多的x使，r(x)>epsilon。这个用反证法。
它的否定是，任意的epsilon，r(x)>epsilon的x都只有
可数个。那么总的r(x)>0的数目必然可数。矛盾。在大于
epsilon的r(x)的x的集合中(不可数)，根据前一个题，必然
有聚点，必可以找到一个序列是r(x)在该点的极限>epsilon，
因为所有的点都>epsilon。
然后还要说明要使该点成为原函数的第一类断点，必须r(x)
在该点的极限为0。因为第一类断点必须有左极限，即从左边
的任意序列趋近于它极限相等。因此r(x)在该点左极限为0，
同理右极限也为0，所以r(x)在该点的极限为0(但本身不为0)。
这和r(x)的极限>epsilon
【在 I***e 的大作中提到】
: 2.
: 1) Prove by contradiction.
: A={x, f-(x)=f+(x) != f(x)}
: B={x, f-(x)!=f+(x)}.
: If A is uncountable, then from 1., A has a sequence x1, x2, … such that f(xi)
: differs from lim{x->xi} f(x) at least by epsilon >0, and that lim xi->X for
: some X in A. However, lim {x->X} f(x) does not exist. Thus the contradiction.
: If B is uncountable, then also from 1., B has a sequence x1, x2,…, such that
: f-(xi) and f+(xi) differ by at least epsilon >0 for all xi’s. And that lim
: xi -> X in B. But now you can see tha
| F***t
发帖数: 412 | 7 3。
我是按照cantor集合构造的。
把[0,1]区间从中间剖开，左半部份赋值为0，右半部份赋值2/3。
左右两半在从中剖开，左左部份赋值为0，左右部份赋值为2/9，
右左部份赋值2/3=6/9，右右部份赋值为8/9。....
这样做下去就是一个单调递增函数，在每一个取半的地方都是
断点，断点在[0,1]上dense。
用小数表示更简单。
用二进制表示[0,1)上的实数，每一个都是一序列
0.100100101001010111...
cantor集合用三进制表示，也是一序列，但由0，2组成：
0.200202020202020202222220...
现在我把[0,1)上的实数的二进制表示中的1变为2。
这个映射把实数映射到cantor集合，结果和前面的一样。
f(xi)
contradiction.
that
…
值
【在 I***e 的大作中提到】
: 2.
: 1) Prove by contradiction.
: A={x, f-(x)=f+(x) != f(x)}
: B={x, f-(x)!=f+(x)}.
: If A is uncountable, then from 1., A has a sequence x1, x2, … such that f(xi)
: differs from lim{x->xi} f(x) at least by epsilon >0, and that lim xi->X for
: some X in A. However, lim {x->X} f(x) does not exist. Thus the contradiction.
: If B is uncountable, then also from 1., B has a sequence x1, x2,…, such that
: f-(xi) and f+(xi) differ by at least epsilon >0 for all xi’s. And that lim
: xi -> X in B. But now you can see tha
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