Martin Sleziak's room

1:38 PM

4

I have original (rigorous!) mathematics and have been cold emailing some professors. Unfortunately none of them want to collaborate online with me. What can I do? I really want to get my research accepted (I don't have a graduate degree).

professors

Martin Sleziak

2:30 PM

0

A: Prove that $\cup_{r\in A}^\infty(-r,r)=(-\sup_{r\in A}r,\sup_{r\in A}r)$

From the formulation it seems that we implicitly have the assumption that $A\subseteq\mathbb R$. (After all, this is probably motivated by the discussion in the comment to this question. The discussion then continued in chat.) Also the notation in the question is rather non-standard, but looking ...

Thanks for answer @MartinSleziak.Can you please explain how you get $\bigcup_{r\in A}(-r,r)=\bigcup_{r\in A\cap(0,\infty)} (-r,r)$ and for case $A \cap (0,\infty)=\emptyset$ how you get that $R=supA \leq 0$ — unit 1991 51 mins ago

@unit1991 I have expanded on these two points a bit more in the footnotes at the end of the post. — Martin Sleziak 41 mins ago

Thank you very much and one last maybe silly question.$\cup_{r\in A}^\infty(-r,r)=(-\sup_{r \in A} r,\sup_{r \in A} r)$ if we take union up to $n$ not infinity then equality is true? — unit 1991 26 mins ago

@unit1991 In my opinion, writing something like $\bigcup_{r\in A}^\infty (-r,r)$ or $\bigcup_{r\in A}^n (-r,r)$ is a non-stardand notation with rather unclear meaning. I would suggest that we could continue the discussion in this chatroom. If you have time to come to chat, feel free to ping me there. — Martin Sleziak 11 secs ago

I will just add that the notation $\bigcup_{r\in A} M_r$ means the union of all sets $M_r$ where $r$ is from the set $A$.

$$x\in\bigcup_{r\in A} M_r \Longleftrightarrow (\exists r\in A) x\in M_r$$

If $A=\{1,2,\dots,n\}$, then we often use the notation $$\bigcup_{r\in A} M_r = \bigcup_{r=1}^n M_r.$$

If $A=\{1,2,\dots\}$ is the set of all positive integers, then we often use the notation $$\bigcup_{r\in A} M_r = \bigcup_{r=1}^\infty M_r.$$

Wikipedia: Union (set theory) § Arbitrary unions

unit 1991

@MartinSleziak Thanks for suggestion and giving me your free time.So equality is true for $\bigcup_{r\in A} M_r = \bigcup_{r=1}^n M_r.$ and $\bigcup_{r\in A} M_r = \bigcup_{r=1}^\infty M_r$?

Martin Sleziak

It's a bit unclear what you mean by this question.

$\bigcup_{r\in A} M_r $ and $\bigcup_{r=1}^n M_r.$ are just two different notation for the same thing in the case that $A=\{1,2,\dots,n\}$.

For example, $\bigcup_{r=1}^n (-r,r)=(-n,n)$. This is simply the union $(-1,1) \cup (-2,2) \cup \dots \cup (-n,n)$.

unit 1991

2:48 PM

@MartinSleziak if we take finite union of $(-r,r)$ $r \in A$ it still be $(-supA,supA)$?

Martin Sleziak

If $A$ is finite, the we still have $\bigcup_{r\in A}(-r,r)=(-\sup A,\sup A).$

In fact, for non-empty finite set we can say a bit more.

Then we actually have $\sup A=\max A$.

unit 1991

Yes thank you very much for detailed answers.

Martin Sleziak

So we get that the union is $\bigcup_{r\in A} (-r,r) = (-\max A,\max A)$. Which means that this is one of the sets $(-r,r)$. (The one with $r=\max A$.)

Ok, so is the way the proof is phrased now satisfactory for you?

Ok, I see that the answer is now accepted, so probably yes.

0

A: Prove that $\cup_{r\in A}^\infty(-r,r)=(-\sup_{r\in A}r,\sup_{r\in A}r)$

From the formulation it seems that we implicitly have the assumption that $A\subseteq\mathbb R$. (After all, this is probably motivated by the discussion in the comment to this question. The discussion then continued in chat.) Also the notation in the question is rather non-standard, but looking ...

In case it is useful for somebody, I will add link to the previous discussion in another chatroom: chat.stackexchange.com/transcript/15201/2021/11/3 chat.stackexchange.com/transcript/15201/2021/11/4

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$Mathematics$ Martin Sleziak's room