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wc596520206

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[求助] 大家看看给点思路也行，只要1和3.第一个看起来不难。

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1楼 2014-11-25 11:00:05

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wc596520206

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应该都是关于数论的题目

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2楼2014-11-25 11:25:18

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★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ...
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wc596520206(feixiaolin代发): 金币+10 2014-11-25 20:18:22
wc596520206: 金币+50, ★★★★★最佳答案 2014-11-28 17:02:40

1. Clearly $\sum_{i=1}^{N}\frac{1}{i}\leq \prod_{p\leq N}\sum_{j=0}^{\left \lceil\log_p{N}\right \rceil}\frac{1}{p^j}$ , which is $\leq \prod_{p\leq N}\sum_{j=0}^{K}\frac{1}{p^j}=\prod_{p\leq N}\frac{1-\frac{1}{p^{K+1}}}{1-\frac{1}{p}}/latex]. <br /> <br /> Now as [latex]\lim_{x\rightarrow 0+}\frac{x}{-\ln(1-x)}=1$ , thus two positive series $\sum_{p }\frac{1}{p}$ and $\ln(\prod_{p}\frac{1}{1-\frac{1}{p}})$ are either both divergent or both convergent. But the latter is divergent according to the above inequality, so both series are divergent.

3.Step (i) Let us show that if a metric space (X, d) has at least two distinct cluster points (limit point) p and q, then it has a subset which is neither open nor closed.

Take a sequence {x_n} approaching to (in a metric space, “approaching” means $d(x_n,a) \rightarrow 0$ ) p, and a sequence {y_n} approaching to q. Without loss of generality, assume $\{x_n, p, q, y_n\}_{n=1}^{\infty}$ are all distinct points of X. Then let the set $A=\{x_{2n}, p, y_{2n}\}_{n=1}^{\infty}$ .

If A is open, then the point p has an open neibourhood in A, which is impossible as the sequence x_{2n+1} is not in A but approaches to p. If A is closed, then A^c (A complement) is an open set. This is still impossible as y_{2n} is not in A^c but approaches to q in A^c.

Step (ii). If the space only has at most one limit point, we discuss by cases.

Case 1. No limit point. Then every point has an open neibourhood which contains only itself. Thus every single point is an open set. Because any union of open set is still open, hence arbitary subset of X is open (therefore is closed too).

Case 2. The point x is the only limit point. Then every point other than x is an open set. This means that any set which excludes x is open. If a set A contains the point x, then since A^c is open so A is closed.

Combining steps (i) and (ii), we prove the claim.

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3楼2014-11-25 14:13:46

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3楼: Originally posted by hank612 at 2014-11-25 14:13:46
1. Clearly \sum_{i=1}^{N}\frac{1}{i}\leq \prod_{p\leq N}\sum_{j=0}^{\left \lceil\log_p{N}\right \rceil}\frac{1}{p^j}, which is \leq \prod_{p\leq N}\sum_{j=0}^{K}\frac{1}{p^j}=\prod_{p\leq N}\frac{1- ...

the first part misses a [, then everything looks weird. I re-post it.

1.Clearly $\sum_{i=1}^{N}\frac{1}{i}\leq \prod_{p\leq N}\sum_{j=0}^{\left \lceil\log_p{N}\right \rceil}\frac{1}{p^j}$ , which is $\leq \prod_{p\leq N}\sum_{j=0}^{K}\frac{1}{p^j}=\prod_{p\leq N}\frac{1-\frac{1}{p^{K+1}}}{1-\frac{1}{p}}$ .

Now as $\lim_{x\rightarrow 0+}\frac{x}{-\ln(1-x)}=1$ , thus two positive series $\sum_{p}\frac{1}{p}$ and $\ln(\prod_{p}\frac{1}{1-\frac{1}{p}})$ are either both divergent or both convergent. But the latter is divergent according to the above inequality, so both series are divergent.

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4楼2014-11-25 14:19:00

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4楼: Originally posted by hank612 at 2014-11-25 14:19:00
the first part misses a [, then everything looks weird. I re-post it.

1.Clearly \sum_{i=1}^{N}\frac{1}{i}\leq \prod_{p\leq N}\sum_{j=0}^{\left \lceil\log_p{N}\right \rceil}\frac{1}{p^j}, which i ...

第一问能写点具体步骤吗写到纸上拍照片就行了

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5楼2014-11-25 14:28:33

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哥们还能看一下第二题吗

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6楼2014-11-26 08:04:37

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wc596520206

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请大家再做一下第二题

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7楼2014-11-26 08:07:23

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楼主写的一手好字

2.1 For any point x in the open set U, there exists an open interval (A,B) satisfying $x\in (A,B)  \subseteq  U$ . Shrink  this open interval (A,B) a little bit as $(a_x,b_x)$ (i.e.  A<=a<x<b<=B), such that both a_x and b_x are rational numbers.  Now do this operation for every point in U, then $U=\bigcup_{x\in U}(a_x,b_x)$

2.2. In the expression of $U=\bigcup_{x}(a_x,b_x)$ , we may assume that there is no duplication of the rational open intervals. Therefore the set of open interval has an injective map into $2^{\mathbb{Q}^2}$ ,  U is mapped to {(a_x,b_x)}_x.  Therefore the cardinality of open sets is less than or equal to $2^{\aleph_0}=\aleph_1$ . But clearly the cardinality of open sets is at least $\aleph_1$ (for instance, every x is mapped to (x-1,x+1)), so there exists a bijection between these two sets.

2.3. For any isolated point x of D (namely x is in D but is not a limit point of D), there is an open interval (A_x, B_x) which contains x but contains no other points of D.  If $x\neq y$ are two isolated points of D, then $(A_x,B_x)\cap (A_y, B_y)=\emptyset$ . Disjoint open intervals are at most countable many, because in each open interval (A_x, B_x), you can find a ration number q_x. The map
$(A_x,B_x) \mapsto q_x$ is injective into $\mathbb{Q}$ , hence is countable.

Now it is clear: D has uncountable many number of points, among them at most countable many points are isolated points. The remaining uncountable points are in $D\cap {D'}$ .

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8楼2014-11-27 07:33:50

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[求助] 大家看看 给点思路也行，只要1和3.第一个看起来不难。

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[求助] 大家看看给点思路也行，只要1和3.第一个看起来不难。