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unit 1991
unit 1991
5:32 PM
Cab anyone help with this question math.stackexchange.com/questions/4295551/…?
 
Martin Sleziak
Martin Sleziak
5:46 PM
To make the transcript easier to read, the question is: Prove that $\cup_{r\in A}^\infty(-r,r)=(-\sup_{r\in A}r,\sup_{r\in A}r)$
BTW more usual notation would be `\bigcup, i.e. $\bigcup_{r\in A}^\infty(-r,r)=(-\sup_{r\in A}r,\sup_{r\in A}r)$
Let us denote $R=\sup_{r\in A} r$.
We know that $R$ is an upper bound, i.e., for each $r\in A$ we have $r<R$.
@unit1991 Do you see that you get $(-r,r)\subseteq (-R,R)$ for each $r\in A$? And can you get one inclusion from this?
So far, we have only used the fat that $R$ is an upper bound. The fact that it is the least upper bound will be useful in the opposite inclusion.
A reminder that info on MathJax in chat can be found in this post on meta or the bookmarklet can be obtained directly from robjohn's website.
2
Oh, I see - you probably already managed that part.
> Now let's $x\in(-\sup r,\sup r)$ have difficulties for proving this part and proof of first part is right?
So it remains to show that $(-R,R) \subseteq \bigcup_{r\in A} (-r,r)$.
You have $x\in(-R,R)$ which means $|x|<R$.
That means $|x|<R-\varepsilon$ for some $\varepsilon>0$.
Can you show that there is $r\in A$ somewhere between $R-\varepsilon$ and $R$?
This was already mentioned in a comment:
hint: if $x\in(-\sup A,\sup A)$ then there exists $\epsilon > 0$ such that $-\sup A + \epsilon < x < \sup A - \epsilon$ — podiki 4 hours ago
And it was also pointed out that the case $R=+\infty$ has to be treated separately.
@podiki the sup equals infinity case need a little different treatment. — Henno Brandsma 4 hours ago
 
unit 1991
unit 1991
@MartinSleziak Thanks for replying.But why $sup$ exists?I gave example of a set that does not have a supremum.And can you give hints or say how to handle $\infty$ case.
 
 
4 hours later…
Martin Sleziak
Martin Sleziak
10:06 PM
@unit1991 I suppose you mean this set: $\{x\in\mathbb{Q}:x\geq 0, \ x^{2}<2\}$.
Actually, that set has a supremum. And it is equal to $\sqrt2$.
BTW you should clarify in the question whether A contains only positive numbers.
If $\sup A=+\infty$, then for each $n\in\mathbb N$ there is an $r\in A$ such that $r>n$.
Both links above go to a comment. So here is one more try to actually link to the question: Prove that $\cup_{r\in A}^\infty(-r,r)=(-\sup_{r\in A}r,\sup_{r\in A}r)$.
 

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