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. :( |
I'm so ignorant. Would you give a Ke(1)Pu(3) introduction of "transfinite induction"? |
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 。 |
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. |
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 |
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. |
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... |
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。。。 |
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! |
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。 |
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? |
every non-empty subset of S has a least element in this ordering. real numbers with traditional ordering (<) is not a well defined set. |
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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