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1) Let $M : =\mathbb{N} \cup \{-1,0\} $. We define the relation $R \subset M \times M$, whereas $\lfloor y \rfloor := max \{m \in \mathbb{Z} : m \leq y\}$ \

$ R :=\{ (x,y) \in M \times M: x < \lfloor \frac{y}{2} \rfloor \}$

1) Is R well-founded?

2) How does the answer of 1) change if we set $M : = \mathbb{Z}$?

To 1) I would say yes we know that $\mathbb{N}$ is well founded and $\{-1,0\}$ is also well founded so, $\mathbb{N} \cup \{-1,0\}$ should also be well-founded? To 2) I would say no, since $\mathbb{Z}$ is not well-founded.

Is this correct? Please correct me if not, any additional info would be helpful.

Parinn
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  • The usual order $\le$ is a well-founded order on $\Bbb N$, but $R$ is a different order; how do you know that it is well-founded on $\Bbb N$? Similarly, you know that $\le$ is not well-founded on $\Bbb Z$, but how do you know that $R$ is not well-founded on $\Bbb Z$? You still have some work to do here. – Brian M. Scott May 02 '20 at 20:35
  • You are right, now after thinking more about it, it is not that obvious that these assumptions hold. Could you please push me into the right direction as to how I can show this? – Parinn May 02 '20 at 23:44

1 Answers1

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Here are some pointers.

To show that $R$ is well-founded on $M=\Bbb N\cup\{-1,0\}$, you need to show that every non-empty subset of $M$ has an $R$-minimal element. In other words, you need to show that if $\varnothing\ne S\subseteq M$, there is an $s_0\in S$ such that there is no $s\in S$ for which $s\,R\,s_0$, meaning that there is no $s\in S$ such that $s<\left\lfloor\frac{s_0}2\right\rfloor$. Try proving that for any $m,n\in M$, if $m\,R\,n$, then $m<n$; once you realize this fact, it’s not hard to describe an $R$-minimal element of any non-empty $S\subseteq M$.

As for the case $M=\Bbb Z$, does $M$ itself have an $R$-minimal element? Or can you prove that no matter what $m\in M$ you take, there is some $n\in M$ such that $n\,R\,m$? (Be a little careful here: $-3\,R\,-4$ even though $-3>-4$, because $-3<-2=\left\lfloor\frac{-4}2\right\rfloor$.)

Brian M. Scott
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  • Thanks prof. Scott. Prof can you help us with this problem: https://math.stackexchange.com/questions/3656033/a-product-inequality-with-binomial-coefficients – James May 03 '20 at 03:38
  • Thank you. I am understanding now what I have to do, however I am still struggling (this topic is new and because of Corona we have no lectures and have to study it at our own). For the first one can/should I use the characteristics of the floor function, i.e $\lfloor \frac{s_0}{2} \rfloor \geq \frac{s_0-2+1}{2}$ ? Or which way would be best to show this? To the second question: we know that if M=$\mathbb{Z}$ , then it does not have a R-minimal element, right? – Parinn May 03 '20 at 13:53
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    @Parinn: I think that you’re making the first problem too hard. If $\varnothing\ne S\subseteq M$, let $s_0=\min S$, the ordinary smallest element of $S$. If $s$ is any other element of $S$, then $\left\lfloor\frac{s_0}2\right\rfloor\le s_0<s$, so $s_0,R,s$; this shows that $s_0$ is $R$-minimal in $S$ as well as minimal in the usual sense. (You do have to be careful to check that $\left\lfloor\frac{s_0}2\right\rfloor\le s_0$ even when $s_0=-1$, since $\frac{-1}2>-1$.) For the second question it’s true that $\Bbb Z$ has no $R$-minimal element, but we don’t know that yet: the point of the ... – Brian M. Scott May 03 '20 at 16:20
  • ... exercise is to prove it. You need to show that for any $n\in\Bbb Z$ there is some $m\in\Bbb Z$ such that $m,R,n$, i.e., such that $m<\left\lfloor\frac{n}2\right\rfloor$. If $n\ge 1$, you can take $m=0$. If $n=0$, $m=-1$ will work. After that it gets a little trickier. If $m=-2$, say, doubling it and setting $n=-4$ doesn’t quite work: $\left\lfloor\frac{-4}2\right\rfloor=\lfloor-1\rfloor=-2$, which is not less than $-2$. But you can check that $n=-5$ works, and that $n=-11$ works for $n=-5$. Can you formulate and prove a general recipe for getting $m$ from $n$? – Brian M. Scott May 03 '20 at 16:24
  • @Brian M.Scott : Thank you for your explanations. As you mention, I am having problems with -1 in the first case, since as you say $\lfloor \frac{-1}{2} \rfloor > -1$. I do not know what to do from here. Because $\lfloor -1/2 \rfloor$ > -1, so for ${s_0}=1$ we could not get a R-minimal element? – Parinn May 05 '20 at 10:10
  • @BrianM.Scott: in the second case, so for m,n $\in {\mathbb{Z}}$(-negative) could m be 2*n-1? – Parinn May 05 '20 at 10:15
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    @Parinn: What I said is that $\frac{-1}2>-1$ which is true, but $\left\lfloor\frac{-1}2\right\rfloor=-1$, so $-1$ really is $R$-minimal in any subset of the first $M$ that contains it. Yes, in the second case $2n-1$ will work if $n$ is negative. – Brian M. Scott May 05 '20 at 14:10
  • Thank you so much – Parinn May 06 '20 at 17:26
  • @Parinn: You’re very welcome. – Brian M. Scott May 06 '20 at 17:27