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For $D \subset\mathbb{R}^n$ with property $S$ we define $s(D) \in [0,1]$ which is a unique value based on some geometric properties of $D$. I need to find a function $f$ such that $f(D)$ has also property $S$ and $s(f(D)) = 0$.

I already showed that for every $D \subset \mathbb{R^n}$ with property $S$ and $s(D) > 0$ there exists a function $\varphi$ such that $\varphi(D)$ has also property $S$ and $s(D) > s(\varphi(D))$.

We can now define a sequence $(\varphi_n)$ in $D$ such that every $\varphi_n(D)$ has property $S$ and $$s(D) > s(\varphi_1(D) > \ldots > s(\varphi_n(D)) > s(\varphi_{n+1}(D)) > \ldots$$ for every $n \in \mathbb{N}$. I also showed that $(\varphi_n)$ has a subsequence which converges to a function $\Phi$ such that $\Phi(D)$ has also property $S$ and $s(\Phi(D)) < s(\varphi_n(D))$ for all $n \in \mathbb{N}$. However this does not imply $s(\Phi(D)) = 0$.

Im asking myself: Does this mean that a function $f$ exists where $f(D)$ has also property $S$ and $s(f(D)) = 0$? Since $\Phi(D)$ has also the property $S$ there is another function such that $s$ gets shrinked. So we can continue to make $s$ smaller and it doesnt have to stop anywhere right? I need to find a valid argumentation for this. Or do I miss something?

I guess we could argue like this: Assume that $s=0$ is not impossible. Then there is a smallest $s = s_0$. But this is a contradiction since we can shrink $s$ further because theres always another function..

Arji
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1 Answers1

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Edit:

As clarified in the comments to this answer, the construction of $\Phi$ can be applied to any sequence of functions $\phi_n$ with strictly decreasing values of $s(\phi_n(D))$. In that case $0$ can indeed be reached, which can be proved as follows:

The set of all values $s(f(D))$ for all functions $f$ has an infimum $I$. We can construct a sequence that converges to this infimum (see here). (This doesn’t require the axiom of choice.) Applying your construction of $\Phi$ to this sequence yields $g$ with $s(g(D))=I$ If $I\gt0$, we could now find $h$ with $s(h(D))\lt I$, contradicting the fact that $I$ is the infimum.

joriki
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  • I dont agree. I understand that you may not reach a certain value at all with finite steps. But I showed that you can choose a sequence which converges to a specific value. From there you can continue to apply those functions. In other words: The family of those shrinking functions is compact. So after infinitely many shrinking functions you can start over and shrink again. You could even apply another converging shrinking sequence and continue to apply one after that. This is not the same as your set does not contain the value $0$. – Arji Apr 23 '20 at 10:29
  • @Arjihad: I don't understand how this intermediate step through $\Phi$ is supposed to change anything. Say you have a sequence of functions, each of which shrinks the value of $s$ closer to $\frac12+\frac14$, and you find a function $\Phi$ that actually brings the value to $\frac12+\frac14$. Now you start over and find functions that bring the value of $s$ closer to $\frac12+\frac18$, and again find a function that actually attains $\frac12+\frac18$; and so on. You'll get closer and closer to $\frac12$, but never beyond, and never close to $0$. Or am I missing something here? – joriki Apr 23 '20 at 10:45
  • @Arjihad: If you still disagree, it would help if you could say where exactly in your argument $0$ cannot be replaced by $\frac12$, and why. – joriki Apr 23 '20 at 10:58
  • So we will get as close as we want to $\frac{1}{2}$ but we never reach it. Okay. But that means for every $\varepsilon > 0$ there is a $n \in \mathbb{N}$ such that $\frac{1}{2}+\frac{1}{n} < \varepsilon + \frac{1}{2}$. This means we have convergence to $\frac{1}{2}$. Since I showed that we can get to the limit in one step we know that we can get to $\frac{1}{2}$ and continue. Since we are now at $\frac{1}{2}$ there is a function such that we can get to a smaller value $<\frac{1}{2}$. Doesnt this mean we can get as small as we want? – Arji Apr 23 '20 at 12:26
  • @Arjihad: It wasn't clear to me from the post that you can get to any limit in one step. In the post you only say that you can get a $\Phi$ for some particular sequence $\phi_n$. If you can do this for any limit, then indeed there must be a smallest element and it must be $0$. A rigorous argument for this would be: The set of all values $s(f(D))$ for all functions $f$ has an infimum $I$. We can construct a sequence that converges to this infimum (see here). – joriki Apr 23 '20 at 15:11
  • Applying your construction of $\Phi$ to this sequence yields $g$ with $s(g(D))=I$. If $I\gt0$, we could now find $h$ with $s(h(D))\lt I$, contradicting the fact that $I$ is the infimum. – joriki Apr 23 '20 at 15:11
  • @Arjihad: By the way, you could do all this without the functions and just say that for any $D$ you can find $D'$ with $s(D')\lt s(D)$ and for any sequence of $D_n$ with $s(D_n)$ converging to $\sigma$ you can find $\Sigma$ with $s(\Sigma)=\sigma$. – joriki Apr 23 '20 at 15:13
  • Thank you. Sorry that this wasnt made clear by me right on. – Arji Apr 23 '20 at 16:18
  • @Arjihad: No worries; in fact with hindsight I thought that I should have seen the possibility that this was what you meant. – joriki Apr 23 '20 at 16:20