"归纳法"(2)

HF: · 发表于 2009-1-23 17:16:16

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.

冷眼看戏的Lili · 发表于 2009-1-24 01:31:34

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").

只需填入未经注册笔名 · 发表于 2009-1-24 01:02:43

老大，你太谦虚了。。。

只需填入未经注册笔名 · 发表于 2009-1-23 23:15:07

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

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

salmonfish · 发表于 2009-1-23 22:47:58

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?

idiot94 · 发表于 2009-1-23 22:42:36

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。

只需填入未经注册笔名 · 发表于 2009-1-23 22:23:30

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!

只需填入未经注册笔名 · 发表于 2009-1-23 22:10:12

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。。。

只需填入未经注册笔名 · 发表于 2009-1-23 22:02:04

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...

只需填入未经注册笔名 · 发表于 2009-1-23 21:57:49

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.

salmonfish · 发表于 2009-1-23 21:53:15

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

只需填入未经注册笔名 · 发表于 2009-1-23 21:49:38

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.

学生 · 发表于 2009-1-23 21:46:15

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 。

HF: · 发表于 2009-1-23 20:39:50

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

idiot94 · 发表于 2009-1-23 20:11:07

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. :(

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"归纳法"(2)

回复：俺学习了。

老大，你太谦虚了。。。

1点很重要

俺学习了。

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，

估计老大你想得太高深了。。。

想这些东西

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Second part

回复："归纳法"(2)

First one

超限归纳法

回复：wow ... :)

wow ... :)