5
$\begingroup$

It seems that I suffer the "too-much-logic-too-pedantic-too-confused"-disease. (You know? This very disease which lets you doubt everything and lets you yell for formalized proof. It's annoying, especially in real life.) The last days I've been trying to understand a certain argument which uses "iteration" (it is out of Hatcher's "Algebraic Topology", the proof for proposition 2B.1. a), page 169 -- but it doesn't matter, I would say).

The generalized problem

Generalized, what I want to proof are the following two claims:

1) For an intervall $I$, assuming $A(I)$, one can construct an intervall $J$, such that

  • $J \subsetneqq I$,
  • $length(I) = 2\cdot length(J)$ and that
  • $A(J)$ holds.

2) Then, by iteration, one has intervals

$$I_0 \supsetneqq I_1 \supsetneqq \ldots$$

with $A(I_m)$ for every $m\in\mathbb{N}$.

Where I stumble

While 1) just goes through by instantiating an arbitrary interval $I$ and constructing a $J$, part 2) seems quite hard to conclude, but just in a detail. In fact, 1) is the inductive step with the exception, that I have nothing which is a successor of something, as $J$ was constructed within the induction.

If I would have a sequence of intervals $\{I_m\}_{m\in\mathbb{N}}$, then of course, I could just run induction on it, using 1). But $I$ and $J$ are not related in this sense.

Ideas

The very first idea is: I'm just confused. This can happen from time to time, especially the more (mathematical) logic I consume. And I did.

The second idea: The sequence of intervals I would need to construct unravels (don't know if this is the right word here...). Take $I_0 = [0,1]$ for example. There are two possibilities (in Hatcher's proof) $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$. If it was the first, the sequence continues with either $[0,\frac{1}{4}]$ or $[\frac{1}{4},\frac{1}{2}]$. As I know the borders explicitly, maybe I could modify 1) into something where

$$J = [\frac{b}{2},b]\text{ or }[a,\frac{b}{2}]$$ and use this? Not sure.

The third idea: putting the borders as a "depending sequence", $a_{n+1} = f(a_n)$, but I have no clue what that should be.

Thanks for any help.

PS: Of course, this is just about formalizing as I cannot see how to give a more pedantic proof. I see the argument, but saying "then run the argument again", which has to happen infinite times, is not a proof at all, as it would need infinite steps.

$\endgroup$
  • 1
    $\begingroup$ @MauroALLEGRANZA Note the post says $I_{n+1}\subsetneq I_n$. $\endgroup$ – AlexR Jan 20 '15 at 11:29
3
$\begingroup$

It follows from the following.

Dedekind's Recursion Theorem: Given a set $X$, $x\in X$ and a map $f\colon X\to X$, there exists (one and only one) function $h\colon \mathbb N\to \mathbb X$ such that $h(0)=x$ and $\forall m\in \mathbb N(h(m+1)=f(h(m)))$.

(For a proof see, for instance, Thomas Jech's Introduction to Set Theory, (chapter 3, section 3). I finally found it on wikipedia too).

In the notation of the theorem, let $X$ be the set of all intervals, let $f$ be one of the maps given by the process described in 1 and let $x=I$. The function $h$ is your desired sequence of intervals.

$\endgroup$
  • $\begingroup$ I don't think this is a disease at all. Some people like to hurl the term pedantic at other's people feet as if it is something dirty, but I will pick up that label and wear it as a badge of honor. Your question is very pertinent and I wish I had thought of that before having been presented with the theorem above. Only after seeing it, did I realise its importance and how much humans take for granted. $\endgroup$ – Git Gud Jan 20 '15 at 12:06
  • $\begingroup$ Thank you very much for your answer (and also for your nice comment). In the middle of "working" mathematicians, my kind of interests seem to always miss the point – so it's good to see another perspective; and a recognition of my very interests: rather logic then topology. ; ) $\endgroup$ – aphorisme Jan 20 '15 at 12:34
2
$\begingroup$

You are actually using induction in step 2) as well, start with $I_0: A(I_0)$ (I assume $A$ is a predicate?). The step 1) gives you a new interval $I_1$ s.t. $A(I_1), I_1 \subsetneq I_0, l(I_1) = \frac12 l(I_0)$.
Now you've shown (in step 1)) that given $I_n, A(I_n)$ you can construct $I_{n+1}\subsetneq I_n, A(I_{n+1}), l(I_{n+1}) = \frac12 l(I_n)$. By induction (in step 2)) you obtain a sequence of intervals $$(I_k)_{k=0}^\infty \text{ such that } A(I_k), l(I_k) = 2^{-k} l(I_0) \text{ and } I_j \subsetneq I_k\qquad\forall k,j\in\mathbb N, k<j$$ What you want to do with this sequence (build it's limit?) is up to you now.


The predicate used for induction will be $$\phi(n): \exists! (I_k)_{k=0}^n, \underbrace{l(I_k) = 2^{-k} l(I_0), I_j \subsetneq I_k, A(I_k), R(I_k) = I_{k+1} \ \forall k,j\in\{1,\ldots, n\}, k<j}_{P((I_k))}$$ Where $R$ is the construction in step 1).
Then $\phi(n)$ gives you a sequence $(I_k)_{k=0}^n$ and you extend it to a sequence $(I_k)_{k=0}^{n+1}$ in the inductive step. This new sequence is a "witness" for $\phi(n+1)$ by construction, so you show $\phi(n) \Rightarrow \phi(n+1)$

The last step now is a short proof of the claim $$\exists! (I_k)_{k=0}^\infty : P((I_k)_{k=0}^N)\ \forall N\in\mathbb N$$ Existence follows by defining $I_k := R^k(I_0)$ the $k$-times application of the reduction.
Uniqueness is given by contradiction: Assume $(I_k)$ and $(J_k)$ satisfy the claim but $I_N \ne J_N$. Then since $P((I_k)_{k=0}^N)$ and $P((J_k)_{k=0}^N)$, $\phi(N)$ implies $(I_k)_{k=0}^N = (J_k)_{k=0}^N$ contradicting the assumption.

$\endgroup$
  • $\begingroup$ Thanks for your answer! I thought about this, but the problem is, induction only gives you an universal quantifier in front of a fixed formula. Something like: if $\phi(0)$ and $\forall n. \phi(n)\Rightarrow\phi(n+1)$ then $\forall n. \phi(n)$. But you cannot obtain a "sequence of intervals" or anything "by induction." You would need to know all the $I_m$ before, to define such a $\phi(n)$. So what I was asking for (now I realized) was a $\phi(n)$ which fits my needs, but it seems that this is not possible in this very case. (At least as long as I want to go with Hatcher's argument.) $\endgroup$ – aphorisme Jan 20 '15 at 12:28
  • $\begingroup$ @aphorisme Indeed the predicate will be $$\phi(n): \exists (I_k)_{k=0}^n, l(I_k) = 2^{-k} l(I_0), I_j \subsetneq I_k, A(I_k) \ \forall k,j\in\{1,\ldots, n\}, k<j$$ $\endgroup$ – AlexR Jan 20 '15 at 12:31
  • $\begingroup$ ! I'll definitely have a look into this idea – as I want to understand it. But give me some time. $\endgroup$ – aphorisme Jan 20 '15 at 12:38
  • $\begingroup$ I might be mistaken, but: for every $n$ we chose, we might get a different sequence. Hence we cannot deduce $I_0 \supsetneqq I_1 \ldots$. Or? And if so, this is exactly the problem I was struggling with. $\endgroup$ – aphorisme Jan 20 '15 at 12:44
  • $\begingroup$ @aphorisme The key is that $\phi(n)$ incorporates the complete sequence $(I_k)$ with $I_0 \supsetneq I_1 \ldots \supsetneq I_n$, not only the last element, so we can construct a longer sequence reusing all of the intervals given to us by $\phi(n)$. $\endgroup$ – AlexR Jan 20 '15 at 12:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.