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标题: "归纳法"(2) [打印本页]

作者: HF: 时间: 2009-1-23 17:16
标题: "归纳法"(2)

Can we extend the induction method from natural numbers to real numbers? i.e.

Suppose there is some property depends on t, where t >=0 can be any real number. Suppose we know:

(1)The property holds true when t = 0 and

(2) If the property is true for any s in [0 t), then the property also holds at t.

Can we then conclude that the property holds for all non negative real number? If not, give a counter example or give some stronger condition to `make' it true.

作者: idiot94 时间: 2009-1-23 20:11
标题: wow ... :)

The first part of your question is relatively easy, that is the counter-example, I believe it will be worked out fairly soon.

The second part of your question is quite difficult(at least to me..), actually honestly speaking, I have to admit that even now I am still very confused by the so-called "transfinite induction", which is somehow related to the Axiom of Choice. :(

作者: HF: 时间: 2009-1-23 20:39
标题: 回复：wow ... :)

I'm so ignorant. Would you give a Ke(1)Pu(3) introduction of "transfinite induction"?

作者: 学生 时间: 2009-1-23 21:46
标题: 超限归纳法

反例：设 f 是一个定义在 0 到无穷的左闭半直线上的实值连续函数且f(0)=0。用您所说的“归纳法”就可“证明”函数 f 必处处为 0。

有关数学知识可在网上查阅超限归纳法 。

作者: 只需填入未经注册笔名 时间: 2009-1-23 21:49
标题: First one

Consider function f(x) = -x and f is continuous

f(0) >= 0,

For any positive real number t, if f(x) >= 0 for all 0<= x < t, then by continuiuty, f(t) >= 0

By false "归纳法", f(x) >= 0 for all x >= 0

But we know the statement f(x) >= 0 for all x >= 0 is not true.

作者: salmonfish 时间: 2009-1-23 21:53
标题: 回复："归纳法"(2)

Is this contradictory?!

"If the property is true for any s in [0 t), then the property also holds at t."

Albeit the "distance" from right edge of [0 t) to t is infinitesimal, one side could be "Heaven" and other side could be "Hell" .

本贴由[salmonfish]最后编辑于：2009-1-23 14:40:50

本贴由[salmonfish]最后编辑于：2009-1-23 15:5:42

作者: 只需填入未经注册笔名 时间: 2009-1-23 21:57
标题: Second part

The set theory / math logic is kind of messing...Many talent mathematician in late 19 and early 20 century spent their lifetime in studying the math logic, attempted and failed to build a solid foundation for mathematics... A genius Kurt Gödel proved the famous incompleteness theory.

作者: 只需填入未经注册笔名 时间: 2009-1-23 22:02
标题: Another comment

It is well known that math has no solid foundations, the current finance system is built with math (mathematical finance). Maybe that is the reason for the credit crunch...

Just a joke. Please do not take it serious... No offense to any one...

作者: 只需填入未经注册笔名 时间: 2009-1-23 22:10
标题: 想这些东西

很容易走火入魔的。。。

CANTOR， Kurt Gödel 你们都知道。。。

不过，也有可能，你是GOD送人类的礼物，帮助人类理解这些东西的。。。

作者: 只需填入未经注册笔名 时间: 2009-1-23 22:23
标题: 估计老大你想得太高深了。。。

我们可以想想怎么证明这个"归纳法"

反证法，如果不对，有一个反例

我们知道初始时是对的，初始是排得最小。。。

自然想到最小的反例。。在一定的条件下（关于ordering)，我们一定能找到.

用
这样一来就得到矛盾了。。。

最小的存在跟 Axiom of Choice很是有关的。。。

Maybe this is not what you are thinking, sorry!

作者: idiot94 时间: 2009-1-23 22:42
标题: 没有没有，偶是真的不懂。HF兄弟，关于超限归纳法，全序集这些个玩意儿，

上次我师傅写选择公理的时候提到过，我就问过他，我小时候看过几本书，可是对于其中的核心想法一直没有弄明白，现在也没有明白，不敢乱讲。刚才在网上随便查了一下，有一些介绍（google可以出来一堆），可是看了看，还是不明白。可是主要是对基数/序数这对概念没有弄懂吧，所以真的不敢乱讲话，请专家们来补充吧。

作者: salmonfish 时间: 2009-1-23 22:47
标题: 俺学习了。

It seems to me that "transfinite induction" applys only to a property P defined on an well ordered set. Is a non-empty real number set an ordered set?

作者: 只需填入未经注册笔名 时间: 2009-1-23 23:15
标题: 1点很重要

every non-empty subset of S has a least element in this ordering.

real numbers with traditional ordering (<) is not a well defined set.

作者: 只需填入未经注册笔名 时间: 2009-1-24 01:02
标题: 老大，你太谦虚了。。。

老大，你太谦虚了。。。

作者: 冷眼看戏的Lili 时间: 2009-1-24 01:31
标题: 回复：俺学习了。

Yes. Any nonempty subset of the real line (i.e., the set of all real numbers) is totally (or say, fully) ordered by the common "=<" (the relation of "less than or equal to").

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