F***t
发帖数: 412 | 1 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 |
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F***t
发帖数: 412 | 2 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 |
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F***t
发帖数: 412 | 3 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 |
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g*******e
发帖数: 14 | 4 map A*A*... to R, a non-countable set. |
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B****n
发帖数: 11290 | 5 I believe we can assume it's a differential curve that divides a sphere into
two equal areas. But maybe someone can give a more rigous proof for this
approximation.
In the following, I denote the curve as (t,f(t)) t in [0,1] and assume it's a
unit sphere with center at 1/2
1. We can only consider bounded variation curve otherwise its length can not
be minimal.
2. For bounded variation curve, its root is countable. Here I define root as a
real number a such that f(a)=0 and there exists a epison s |
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C********n
发帖数: 6682 | 6 dont think so
x^m m =[0,1] |
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l*****e
发帖数: 238 | 7 surely uncountable
define
f_a = x, for x \in [0,a]
a, for x > a
each of the above is continuous, and the collection {f_a} is uncountable,
due to the natural 1-1 correspondence with {a} for a \in [0,1] |
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j******w
发帖数: 690 | 8 has size 2^{\aleph_0}.
Just notice that every continuous function is decided by the
function restricted to rational numbers. |
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B********e
发帖数: 10014 | 9 如果你的表达足够精确，同时数列按我的理解是countable的话，我觉得
这个集合除了必须可数，没有别的要求 |
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b****t
发帖数: 114 | 10 Hi all,
suppose a sequence of functions {f_n} defined on a countable set of vectors
X
(e.g. integer vectors Z^n), these functions pointwise converge to function f
defined on set X. Is it true that f_n converge to f uniformly?
Thanks,
Beet |
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c****n
发帖数: 2031 | 11 Given a smooth closed 2-manifold S in R^3. Define the function d(x), x\in R^
3,
to be the unsigned distance function to the surface S. （For example in 1D,
if S is a point x0, then d(x)=|x-x0|.）The function d(x) is only continuous
but not smooth, e.g. it's not differentiable on S. Now the question is: is
there a sequence (countable or uncountable) d_\epsilon(x)->d(x) uniformly (i
.e. under supremum norm), where d_\epsilon are smooth and {x:d_\epsilon(x)=0
}=S, for all \epsilon(i.e. preserve zero |
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x*****d
发帖数: 427 | 12 no. S split to a countable union U {f(x)>1/k}
x |
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j******w
发帖数: 690 | 13 come on.
Lots of Number theory problems (GH, Fermat theorem...) look stupid but they
require really much, right?
"The image of an open set under a continuous 1-1 map is a Borel set" Even
this is not trivial if you consider Baire Space (For R^n, the nontrivial case shows up at G_{\delta} level).
Kechris' proof require much fewer but it's weaker.
Actually the previous proof shows more:
If A is a Borel set and f is a borel function with countable to 1, then the
image of A under f is a Borel set.
I |
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K****Y
发帖数: 74 | 14 yes, the bounded set in Q is at most countable, but the function for this is
not continuous and 1-to-1, (hence monotonic) |
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Q***5
发帖数: 994 | 15 I think you can prove (1) by the following arguments:
1.1 It is true when A is any open segment -- This is just by defination of
cdf.
1.2 It is true for any open subset of [0 1]. An open subset is just a
uninon of countably many opens segments.
1.3 It is true for any Borel measurable subset A: m_n(A) = inf_{A\subset D,
D open}m_n(D)<= B\inf_{A\subset D, D open}m(D) = B m(A)
measure |
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b**g
发帖数: 335 | 16
which
If you know some algebra, then you should have heard of field extension, i.e.
given a field F, insert something, say, x, to create a new field F[x]
Similarly, begin with a model of axiomatic set theory (e.g. ZFC), say M,
insert
something, say G (G means "generic" set), to it. This G is specially
chosen such that M[G] is still a model of ZFC but violates continuum
hypothesis. (Note: this model M is also special, it has to be countable &
transitive)
This whole process (called "forcing") is |
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r********e
发帖数: 103 | 17 separable (or second countable)? |
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m*******s
发帖数: 3142 | 18 For subsets A and B of R, define A+B={a+b|a in A, b in B}
suppose that B is a Borel set, Prove that A+B is a Borel set if A is open
我的思路 大概是 這樣的. 利用 Lindelof定理 ,
A+B= \bigcup_{b\in B}(A+b)=\bigcup_{i=1}^{\infty}(A+{b}_{i}) where {b}_{i}\
in B
再利用Borel set countable union的封閉性,似乎已經完成了證明,里頭根本沒有用到B
is a Borel set. 請問我的 證明是否正確? |
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m*******s
发帖数: 3142 | 19 Let I be a nonempty set and {x_l}l∈I an indexed collection of nonegative
real numbers, that is, x_l≥0 for each l∈I ,Define
\sum_{l\in I}x_l=sup \left \{\sum_{l\in F}x_l: F finite, F\subset I \right
\}
where each sum in the set on the right is the ordinary sum of a finite
collection of real numbers
Show that if \sum_{l\in I}x_l<\infty, then {l∈I:x_l>0} is countable.
这题看起来好像很显然，但是我说不清理由，请大家说说证明关键的地方。 |
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F**S
发帖数: 13 | 20 "the image of an F_\sigma set in R^n under a continuous transformation from
R^n to itself is also an F_\sigma set"
I was trying to use the property that the image of an arbitrary (countable
when applied to the F_\sigma set) union of sets is the union of the images
of those sets. but a continuous transformation need not be closed, so I'm
puzzled...
any hints? thanks! |
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s**e
发帖数: 15 | 21 你看这个是不是：let X be the set of all real numbers. Y={A|A is a subset of
X, A or X\A is countable}. (X,Y) is not a topological space. Suppose it is a
topological space, then any one point subset is open. It follows that
interval [0,1] is open, which is not in Y. \sigma algebra 要求可数并， 拓扑
空间要求任意并，这大概是二者的本质差别。 |
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n******t
发帖数: 4406 | 25 怎么觉得uncountable 是愚不可及呢？ |
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B****n
发帖数: 11290 | 26 大哥 我很想解釋給你聽 但是赫然發現 解釋一個笑話 如果還是不怎麼好笑的笑話 可
能才真是愚不可及 所以還是算了 |
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a***s
发帖数: 616 | 27 过奖了，别这样讲。
sigma algebra满足algebra的性质，但是algebra不一定有sigma algebra的性质。
比如通常讲的Borel sigma algebra当然是algebra: closed under complement, finit
e union, finite intersection。
但是[0,1)上由[a,b)型区间的finite union构成的collection只是一个algebra而不是s
igma algebra: not closed under countable union/intersection. e.g. intersecti
on over all n from 1 to infinity [0.5,0.5+1/n) gives {0.5} which is not in t
his collection. |
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p*****n
发帖数: 758 | 28 suppose X is a banach space and X* be its dual. suppose X* is separable and
f_n be a countable subset of it. by definition we can find unit vector x_n
in X such that |fn(x_n)|>1/2|f_n|.
show that span{x_n} is dense in X. |
|
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w**********a
发帖数: 107 | 30
why can you give me a counterexample?
thanks |
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B****n
发帖数: 11290 | 31 You can consider the famous example of Cantor set.
Each time you remove 1/3 of the interval on [0,1]
sigma
the |
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w**********a
发帖数: 107 | 32
but Cantor set is a closed set |
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B****n
发帖数: 11290 | 33 你每次挖的是closed interval 那cantor set就一定不是closed set |
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w**********a
发帖数: 107 | 34
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? |
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z****e
发帖数: 702 | 35 对于概率空间(Omega,F,P),
若Omega为countable,
则Omega上的函数列{Xn}的almost surely收敛和in probability收敛等价。
我想由a.s.-->i.p.为显然，但是i.p.-->a.s.不知道怎么证？请达人指教。 |
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z****e
发帖数: 702 | 36 设B为(0,1]上的Borel sigma代数，
如何找出一组可数的{Fi},使得B为{Fi}生成的sigma代数？
thanks！ |
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Q***5
发帖数: 994 | 37 Let A be the set of points who have non-zero P measure. Let B = (0,1]-A
E(X|G)(t) = X(t) on A
E(X|G)(t) = E(X 1_B)/P(B) on B |
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z****e
发帖数: 702 | 38 牛人啊。谢谢！
还是有一点不懂：你这里给出的，是P可能有质点/P对于lebesgue测度不绝对连续的情
况，那么如果P是对于lebesgue测度绝对连续的测度，那么是不是就是E(X|G)=E(X)? |
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z****e
发帖数: 702 | 39 牛人啊。谢谢！
还是有一点不懂：你这里给出的，是P可能有质点/P对于lebesgue测度不绝对连续的情
况，那么如果P是对于lebesgue测度绝对连续的测度，那么是不是就是E(X|G)=E(X)? |
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Q***5
发帖数: 994 | 40 Lebesgue measure is a special case (A is empty), and you are right, E(X|G)
is constant E(X) in that case. |
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s*****e
发帖数: 115 | 41
This statement is false!
It's easy to construct a continuous function f : R-->R such that X={t|f(t)=0
} is exactly the Cantor set, which is NOT the union of countably many
disjoint closed intervals (or points). |
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Q***5
发帖数: 994 | 42 You can construct a Cantor-like set with measure >0 and that should fulfill
the requirements.
The idea is simple: instead of digging a hole of lengh 1/3 of the internal
each time, you can make that ratio progressively smaller: for example, in
the first round, the length of the hole is 1/3 of the interval; the second
round, make it 1/3^2 of the intervals; the third round, 1/3^3, and so on...
Now, let F be the set of all the end points of the intervals during the
construction, then F is countabl... 阅读全帖 |
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D*****r
发帖数: 6791 | 43 http://golem.ph.utexas.edu/category/2013/05/bounded_gaps_betwee
Guest post by Emily Riehl
Whether we grow up to become category theorists or applied mathematicians,
one thing that I suspect unites us all is that we were once enchanted by
prime numbers. It comes as no surprise then that a seminar given yesterday
afternoon at Harvard by Yitang Zhang of the University of New Hampshire
reporting on his new paper “Bounded gaps between primes” attracted a
diverse audience. I don’t believe the paper is... 阅读全帖 |
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B****n
发帖数: 11290 | 44 條件不完整巴
diagonal sequence最多也只可能證明到countable個點的pointwise convergence
如果要整個函數sequence pointwise convergence至少需要函數的其他條件 |
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p******e
发帖数: 1151 | 45 I think I have made myself very clear but apparently you did not agree.
I will try another time and it is my last post for this topic. If I am
understood well, what I said implies that the answer to your question is "NO
".
A measure theory (whatever you call, probability etc) needs to satisfy
obvious axiom such as
m(A+B)=m(A)+m(B) etc. (A, B has no intersection). (A nice generalization is
countable additivity)---that is what we mean it is the generalization of
measuring length and volume.
If one... 阅读全帖 |
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m****m
发帖数: 2211 | 46 那个时代是不是还没有countable和measure的概念? |
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m****m
发帖数: 2211 | 47 那个时代是不是还没有countable和measure的概念? |
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p***c
发帖数: 2403 | 48 你说的那样子选，建立了从自然数集到实数集的单射，但它不是满的 |
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s****b
发帖数: 2039 | 49 请问＂但它不是满的＂是什么意思？
难道上面那种数实数的方法不是一一对应？ |
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p***c
发帖数: 2403 | 50 把你的数法列出来，就是x1,x2,x3,把它们全部写成小数，如果是有限小数，用999999
代替，比如2.1写成2.09999999
取一个小数，它的第n位是1-9里面的一个数，并且它的第n位不等于xn的第n位，这个数
将不等于你取的任何一个xi |
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