M*********m
发帖数: 2024 | 1 【 以下文字转载自 Literature 讨论区 】
发信人: Microsystem (clam), 信区: Literature
标 题: [合集] 给会calculus的人出一道题吧
发信站: BBS 未名空间站 (Sat Nov 1 19:28:39 2008), 站内
☆─────────────────────────────────────☆
seiya (天馬流星拳 - 飛呀飛呀我的馬) 于 (Sat Nov 1 14:03:43 2008) 提到:
这个可能比较难，限本科学过calculus的吧
定义一个连续函数，把[0,1]上的有理数一对一地映到(0,1)上的有理数，
只要求在有理数上连续，无理数不管它
☆─────────────────────────────────────☆
lhr (麻辣呆牛筋) 于 (Sat Nov 1 14:05:17 2008) 提到:
我学过亚，但是现在只记得calculus这个词
还是来点幼儿园的题目吧，大星期六的，太打击人了
☆───────────────────────────────────── | M*********m
发帖数: 2024 | 2 能不能帮我们做道题？：）多谢。
【在 M*********m 的大作中提到】
: 【 以下文字转载自 Literature 讨论区 】
: 发信人: Microsystem (clam), 信区: Literature
: 标 题: [合集] 给会calculus的人出一道题吧
: 发信站: BBS 未名空间站 (Sat Nov 1 19:28:39 2008), 站内
: ☆─────────────────────────────────────☆
: seiya (天馬流星拳 - 飛呀飛呀我的馬) 于 (Sat Nov 1 14:03:43 2008) 提到:
: 这个可能比较难，限本科学过calculus的吧
: 定义一个连续函数，把[0,1]上的有理数一对一地映到(0,1)上的有理数，
: 只要求在有理数上连续，无理数不管它
: ☆─────────────────────────────────────☆
| G******i
发帖数: 163 | 3 f(x)= { 0.1+x*sqrt(2) }
{a}=the fractional part of a. | G******i
发帖数: 163 | 4 Then, construct a continuous g
from f(Q\cap[0,1]) to Q\cap (0,1).
【在 G******i 的大作中提到】
: f(x)= { 0.1+x*sqrt(2) }
: {a}=the fractional part of a.
| d*z
发帖数: 150 | 5 这样的函数显然不存在。
我们非常容易可以证明这样的函数在有理数集上单调。
假设不单调，不妨设有有理数0
f(a)
由于函数f连续，由介值定理，存在实数u in (a,b)使得f(u)=f(c)为有理数。
由于f在[0,1]上所有取值有理数的点的原像为有理数，得到u为有理数。
所以有理数u和c的取值相同，同一一映射矛盾。
而得到这个函数在有理数集上单调以后，很显然可以看出函数不存在了
【在 M*********m 的大作中提到】
: 能不能帮我们做道题？：）多谢。
| s******n
发帖数: 876 | 6 f(在实数集)上不一定连续啊
【在 d*z 的大作中提到】
: 这样的函数显然不存在。
: 我们非常容易可以证明这样的函数在有理数集上单调。
: 假设不单调，不妨设有有理数0: f(a): 由于函数f连续，由介值定理，存在实数u in (a,b)使得f(u)=f(c)为有理数。
: 由于f在[0,1]上所有取值有理数的点的原像为有理数，得到u为有理数。
: 所以有理数u和c的取值相同，同一一映射矛盾。
: 而得到这个函数在有理数集上单调以后，很显然可以看出函数不存在了
| s******n
发帖数: 876 | 7 数学系的说话太简练了吧? 具体点?
【在 G******i 的大作中提到】
: Then, construct a continuous g
: from f(Q\cap[0,1]) to Q\cap (0,1).
| K****Y
发帖数: 74 | 8 such function doesn't exist.
[a,b] in Q is compact, f is continuous, which means f([a,b]) is compact,
while (a,b) is not closed. | M*********m
发帖数: 2024 | 9 所以这个命题是否定的？顺道问问大家，我喜欢复分析，喜欢几何，不过都是大学水平
，如果要进一步学习，应该怎么走呢？
【在 K****Y 的大作中提到】
: such function doesn't exist.
: [a,b] in Q is compact, f is continuous, which means f([a,b]) is compact,
: while (a,b) is not closed.
| K****Y
发帖数: 74 | 10 For mathematical analysis, the book by Tom Apostol or Rudin is a good start.
No idea for Geometry.
【在 M*********m 的大作中提到】
: 所以这个命题是否定的？顺道问问大家，我喜欢复分析，喜欢几何，不过都是大学水平
: ，如果要进一步学习，应该怎么走呢？
| | | G******i
发帖数: 163 | 11 首先记住不要轻信别人的话，包括我的这句话
【在 M*********m 的大作中提到】
: 所以这个命题是否定的？顺道问问大家，我喜欢复分析，喜欢几何，不过都是大学水平
: ，如果要进一步学习，应该怎么走呢？
| G******i
发帖数: 163 | 12 你看了Apostol 和 Rudin，怎么还没搞懂compactness的概念?
start.
【在 K****Y 的大作中提到】
: For mathematical analysis, the book by Tom Apostol or Rudin is a good start.
: No idea for Geometry.
| K****Y
发帖数: 74 | 13 There is something wrong in the Proof? [a,b] in Q is compact, since for any
open cover, you can find finite
number of opens sets to cover this. But (a,b) is not compact since it's not
closed. I didn't see anything
wrong. But please correct me if I am wrong. Thanks a lot
【在 G******i 的大作中提到】
: 你看了Apostol 和 Rudin，怎么还没搞懂compactness的概念?
:
: start.
| v********e
发帖数: 1058 | 14 "[a, b] in Q" 是指[a, b] \cap Q? 那这集合是closed的么？
any
not
【在 K****Y 的大作中提到】
: There is something wrong in the Proof? [a,b] in Q is compact, since for any
: open cover, you can find finite
: number of opens sets to cover this. But (a,b) is not compact since it's not
: closed. I didn't see anything
: wrong. But please correct me if I am wrong. Thanks a lot
| K****Y
发帖数: 74 | 15 In metric space Q, [0,1] is closed since all it contains all its limit
points, right? | s*******n
发帖数: 740 | 16 (0,1)内有理数可列
1/2,1/3,2/3,1/4,3/4.......
So
映射
0->1/2
1->1/3
1/2->2/3
1/3->1/4
.
.
.
.
. | s*******n
发帖数: 740 | | K****Y
发帖数: 74 | 18 yes, the bounded set in Q is at most countable, but the function for this is
not continuous and 1-to-1, (hence monotonic)
【在 s*******n 的大作中提到】
: (0,1)内有理数可列
: 1/2,1/3,2/3,1/4,3/4.......
: So
: 映射
: 0->1/2
: 1->1/3
: 1/2->2/3
: 1/3->1/4
: .
: .
| K****Y
发帖数: 74 | 19 f(0)=0. but f([0,1])=(0,1) | K****Y
发帖数: 74 | 20 for this one, let i=0, j=n, then f(0)=1/n, which is not a function, since f(
0) is not unique. | | | l******o
发帖数: 1550 | 21 不连续
【在 s*******n 的大作中提到】
: (0,1)内有理数可列
: 1/2,1/3,2/3,1/4,3/4.......
: So
: 映射
: 0->1/2
: 1->1/3
: 1/2->2/3
: 1/3->1/4
: .
: .
| w********9
发帖数: 8613 | 22
f(
As a math major, you should have understood what I really meant, which was
why I demanded a 90+ for myself.
【在 K****Y 的大作中提到】
: for this one, let i=0, j=n, then f(0)=1/n, which is not a function, since f(
: 0) is not unique.
| w********9
发帖数: 8613 | 23 我没有数学专业背景,错了也要给九十分以上。:)
For a rational in its irreducible form i/j (0 being expressed as 0/1),
i/j -> (i+1)/(j+2)
where integers i and j satisfy：j >= i >= 0 and j > 0.
我要九十五分以上。:） | K****Y
发帖数: 74 | 24 HAHA.
Well, I am not a math major, I am in engineering, just interested in Math.
For this one, f(x)=1/3, f(y)=1/4, and you have x=y. So it's not a function
【在 w********9 的大作中提到】
: 我没有数学专业背景,错了也要给九十分以上。:)
: For a rational in its irreducible form i/j (0 being expressed as 0/1),
: i/j -> (i+1)/(j+2)
: where integers i and j satisfy：j >= i >= 0 and j > 0.
: 我要九十五分以上。:）
| w********9
发帖数: 8613 | 25
Notice that I had already allowed a rational to only have a UNIQUE
representation form. So it can't have two or more mapped values.
Why hadn't you decalred so earlier? ;)
Wrong.
【在 w********9 的大作中提到】
: 我没有数学专业背景,错了也要给九十分以上。:)
: For a rational in its irreducible form i/j (0 being expressed as 0/1),
: i/j -> (i+1)/(j+2)
: where integers i and j satisfy：j >= i >= 0 and j > 0.
: 我要九十五分以上。:）
| K****Y
发帖数: 74 | 26 In metric space Q, f is continuous, [a,b] is connected, which means f([a,b])
need to be connected, hence
range of f contains all the rational value (0,1).
Choose f(x)=1/3, f(y)=1/4, you will have x=0/2, y=0/3, hence x=y.
【在 w********9 的大作中提到】
:
: Notice that I had already allowed a rational to only have a UNIQUE
: representation form. So it can't have two or more mapped values.
: Why hadn't you decalred so earlier? ;)
: Wrong.
| w********9
发帖数: 8613 | 27
])
You can't have that, since I had said earlier: "(0 being (only) expressed as
0/1)".
If you do the mapping from x (with a unique i/j representation) to y, it has one and only one mapped value.
【在 K****Y 的大作中提到】
: In metric space Q, f is continuous, [a,b] is connected, which means f([a,b])
: need to be connected, hence
: range of f contains all the rational value (0,1).
: Choose f(x)=1/3, f(y)=1/4, you will have x=0/2, y=0/3, hence x=y.
| K****Y
发帖数: 74 | 28 But the function need an x such that f(x)=1/3, and a y such that f(y)=1/4,
right?
So what's the x and y?
as
has one and only one mapped
value.
【在 w********9 的大作中提到】
:
: ])
: You can't have that, since I had said earlier: "(0 being (only) expressed as
: 0/1)".
: If you do the mapping from x (with a unique i/j representation) to y, it has one and only one mapped value.
| w********9
发帖数: 8613 | 29
,
Solve for integers i and j satisfying those conditions in
(i+1)/(j+1) = r = k/l (k/l is r's irreducible form as defined earlier)
It turns into an elementary number theory problem, with i and j being
relatively prime to each other.
Solve for ax+by =c (a, b, and c are all integers) in integers x and y. Key
word: congruence. Solutions always exit. So don't worry about its non-
existence.
x=1/4 and 1/6 for y = 1/3 and 1/4, respectively.
【在 K****Y 的大作中提到】
: But the function need an x such that f(x)=1/3, and a y such that f(y)=1/4,
: right?
: So what's the x and y?
:
: as
: has one and only one mapped
: value.
| K****Y
发帖数: 74 | 30
Okay, I am lost. hehe
you said i/j -> (i+1)/(j+2), right? and i thought x=i/j, and f(x)=(i+1)/(j+2
).
x should be [0,1], which is right since you claim j>=i>=0 and j>0. Also i
and j have no common factor.
And You are right about f(1/4)=1/3. But then how do you define f(2/7)?
【在 w********9 的大作中提到】
:
: ,
: Solve for integers i and j satisfying those conditions in
: (i+1)/(j+1) = r = k/l (k/l is r's irreducible form as defined earlier)
: It turns into an elementary number theory problem, with i and j being
: relatively prime to each other.
: Solve for ax+by =c (a, b, and c are all integers) in integers x and y. Key
: word: congruence. Solutions always exit. So don't worry about its non-
: existence.
: x=1/4 and 1/6 for y = 1/3 and 1/4, respectively.
| | | w********9
发帖数: 8613 | 31
+2
The same value of 1/3, but it doesn't affect the fact that the mapping
correspondes to a function.
It is a function, but it doesn't map to an irrational value, so it is not
continuous. So I was wrong.
How about i/j -> lg(i+1)/lg(j+2)? It covers a lot of irrational values, but
it may still not be helping the whole situation. :)
【在 K****Y 的大作中提到】
:
: Okay, I am lost. hehe
: you said i/j -> (i+1)/(j+2), right? and i thought x=i/j, and f(x)=(i+1)/(j+2
: ).
: x should be [0,1], which is right since you claim j>=i>=0 and j>0. Also i
: and j have no common factor.
: And You are right about f(1/4)=1/3. But then how do you define f(2/7)?
| c*******g
发帖数: 509 | | w********9
发帖数: 8613 | 33 The problem is misleading. It really has no solution.
We can't find a continuous function, because there are more irrational
numbers than rational numbers. There are always irrational number no being
mapped to for any given mapping.
This should conclude it. | s*******n
发帖数: 740 | 34 哦，忽略了还有continuous这个限制
但是it is really a bijection
is
【在 K****Y 的大作中提到】
: yes, the bounded set in Q is at most countable, but the function for this is
: not continuous and 1-to-1, (hence monotonic)
| s*******n
发帖数: 740 | 35 lz说只要在Q上连续
【在 w********9 的大作中提到】
: The problem is misleading. It really has no solution.
: We can't find a continuous function, because there are more irrational
: numbers than rational numbers. There are always irrational number no being
: mapped to for any given mapping.
: This should conclude it.
| f******k
发帖数: 297 | 36 depends on the topology you choose on Q.
【在 l******o 的大作中提到】
: 不连续
| d*z
发帖数: 150 | 37 这个题目定义的有点不是很清楚,我觉得需要重新澄清一下.
比如这里连续指得就是普通微积分里面得连续概念(所以我们不需要重新定义有理数集上
的拓扑),而这里只需要在所有的有理点连续,而在无理点是否连续没有关系.
同样,另外一个问题式[0,1]上的有理点一一对应到(0,1)上的有理点.这里也有个不清楚
的地方.比如有一个[0,1]中的有理点a使得f(a)=1/2.那么如果还有一个无理点b使得f(b
)=1/2,这个函数是否算满足条件呢?那么这里我们假设这样的函数是算满足条件的.
也就是说这个函数会将[0,1]中的所有有理点映射到(0,1)中的有理点,而且不同的有理点
被映射到不同的值.另外(0,1)中每个有理点,肯定有唯一一个[0,1]中有理点被映射成这
个值(但是还可以允许存在无理点被映射过来).
那么在这样的约束之下,这样的函数是否存在呢?
【在 M*********m 的大作中提到】
: 【 以下文字转载自 Literature 讨论区 】
: 发信人: Microsystem (clam), 信区: Literature
: 标 题: [合集] 给会calculus的人出一道题吧
: 发信站: BBS 未名空间站 (Sat Nov 1 19:28:39 2008), 站内
: ☆─────────────────────────────────────☆
: seiya (天馬流星拳 - 飛呀飛呀我的馬) 于 (Sat Nov 1 14:03:43 2008) 提到:
: 这个可能比较难，限本科学过calculus的吧
: 定义一个连续函数，把[0,1]上的有理数一对一地映到(0,1)上的有理数，
: 只要求在有理数上连续，无理数不管它
: ☆─────────────────────────────────────☆
| m****e
发帖数: 130 | 38 作为R的子空间，[0,1]\cap Q 和 (0,1) \cap Q是同胚的
集上
(b
理点
【在 d*z 的大作中提到】
: 这个题目定义的有点不是很清楚,我觉得需要重新澄清一下.
: 比如这里连续指得就是普通微积分里面得连续概念(所以我们不需要重新定义有理数集上
: 的拓扑),而这里只需要在所有的有理点连续,而在无理点是否连续没有关系.
: 同样,另外一个问题式[0,1]上的有理点一一对应到(0,1)上的有理点.这里也有个不清楚
: 的地方.比如有一个[0,1]中的有理点a使得f(a)=1/2.那么如果还有一个无理点b使得f(b
: )=1/2,这个函数是否算满足条件呢?那么这里我们假设这样的函数是算满足条件的.
: 也就是说这个函数会将[0,1]中的所有有理点映射到(0,1)中的有理点,而且不同的有理点
: 被映射到不同的值.另外(0,1)中每个有理点,肯定有唯一一个[0,1]中有理点被映射成这
: 个值(但是还可以允许存在无理点被映射过来).
: 那么在这样的约束之下,这样的函数是否存在呢?
| j******w
发帖数: 690 | 39 I didnt' check the details. But the following construction should work.
I am waiting for the simpler proof.
Let p_n be an enumeration of rationals in (0,1)
Let u_n be an enumeration of rational open intervals in [0,1].
Define f on rationals as follows:
At stage 0. f(0)=p_1, f(1)=p_2. Moreover, let u_i be the least i so that
the following forced condition is consistent : f 1-1 maps rationals in u_i
onto the rationals in u_0
At stage n+1, if f(p_n) is undefined, let f(p_n) be p_i where i is the l | R*****1
发帖数: 173 | 40 Here is my opinion:
If we do not require the map to be onto. It is easy, just gives out a linear
map like x+(0.5-x)/2.
However, when we requires the map to be onto. The claim is false.
A={rational numbers in [0,1]} B={rational numbers in (0,1)}.
A is a close set. B is not. 1-1,onto , continuous function keeps topology
properties. That means A and B should be close or open at the same time.
I think many people confuse about this because of the concept "1-1". In
China, I used to understand it as | | | d*z
发帖数: 150 | 41 那么作为R的子空间,(0,1)中任意两个不同的稠密可列子集是否同胚呢?如果同胚,那么这
个题目的结果就好构造.
【在 m****e 的大作中提到】
: 作为R的子空间，[0,1]\cap Q 和 (0,1) \cap Q是同胚的
:
: 集上
: (b
: 理点
| w********9
发帖数: 8613 | 42 "函数“和”一对一“就是onto or surjective.
The original problem really has no solution, because there are infinite
number of other rational numbers between any two given rationals. If the two
boundaries of y HAD definite immediate neighbors, it would have a trivial
solution. But they don't really have. They each don't have a definite
immediate neighbor. |
|
