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

$$\sup_{r \in A} r$$ I am stuck at proof of $$\cup_{r\in A}^\infty(-r,r)=(-\sup_{r \in A} r,\sup_{r \in A} r)$$

First question is how we know that $$\sup$$ of given set exists?

Let's take $$\{x\in\mathbb{Q}:x\geq 0, \ x^{2}<2\}$$ this set does not have a supremum.

Is this equality true for all sets $$A$$?

For proof.

Let $$x \in \cup_{r\in A}^\infty(-r,r)$$ that means there exists at least one $$(-r,r)$$ s.t.$$x\in(-r,r)$$ and if we know that $$\sup$$ exists then it will belong to $$(-\sup r,\sup r)$$

Now let's $$x\in(-\sup r,\sup r)$$ have difficulties for proving this part and proof of first part is right?

• Your example does not have a supremum in $\mathbb Q$, but it does have one in $\mathbb R$. In fact every bounded subset of $\mathbb R$ has a supremum. Nov 3, 2021 at 13:38
• Please, change $supr_{r \in A}$ with $\sup_{r \in A} r$ Nov 3, 2021 at 13:39
• What is the definition of $A$?
– user987907
Nov 3, 2021 at 13:41
• hint: if $x\in(-\sup A,\sup A)$ then there exists $\epsilon > 0$ such that $-\sup A + \epsilon < x < \sup A - \epsilon$ Nov 3, 2021 at 13:59
• @podiki the sup equals infinity case need a little different treatment. Nov 3, 2021 at 14:22

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 at the comments in chat it is clear that we're trying to show that $$\bigcup_{r\in A}(-r,r)=(-\sup A,\sup A).$$

First let's have a look at the case $$A\subseteq(0,\infty)$$ and $$A\ne\emptyset$$. This is the case needed in the linked question - but we can check later what happens if we allow negative values or if we consider the empty set. (Although such cases are much less relevant.)

Let us denote $$R=\sup A$$. We know that either $$A$$ is bounded (and $$R\in\mathbb R$$) or $$A$$ is unbounded (and $$R=+\infty$$).

In either case, we have $$(-r,r)\subseteq(-R,R)$$ for each $$r\in A$$, and thus $$\bigcup_{r\in A}(-r,r) \subseteq (-R,R).$$

The relevant part of the proof is the opposite inclusion.

If $$R$$ is finite. Let us suppose that $$R$$ is finite. Let us take arbitrary $$x\in(-R,R)$$, i.e., we have $$|x|. Since $$R=\sup A$$, there exists $$r\in A$$ such that $$|x| and $$x\in(-r,r).$$ So we see that for every $$x\in(-R,R)$$ we also have $$x\in\bigcup_{r\in A}(-r,r).$$

If $$R=+\infty$$. In this case $$(-R,R)=(-\infty,\infty)=\mathbb R$$. For any $$x\in\mathbb R$$ there exists an $$r\in A$$ such that $$|x| and $$x\in(-r,r)$$. Consequently, we get $$x\in\bigcup_{r\in A}(-r,r).$$

In both cases we have shown $$(-R,R)\subseteq \bigcup_{r\in A}(-r,r)$$ and we have the desired inequality.

What if $$A$$ is empty? For $$A=\emptyset$$ we have $$\sup A=\sup\emptyset=-\infty.$$ (If we want to define $$\sup\emptyset$$ in the context of sets of real numbers, this is the only reasonable value. The other option is to leave this supremum undefined.) So we have $$(-\sup A,\sup A)=(\infty,-\infty)=\emptyset=\bigcup_{r\in\emptyset}(-r,r)$$.

What if $$A$$ contains some $$r\le0$$? Whenever $$r\le 0$$, we have $$(-r,r)=\emptyset$$. So adding such $$r$$'s does not change the union. So we have1 $$\bigcup_{r\in A}(-r,r)=\bigcup_{r\in A\cap(0,\infty)} (-r,r).$$ How taking the intersection influences the supremum? We either have $$A\cap(0,\infty)\ne\emptyset$$, an in this case $$\sup A=\sup(A\cap(0,\infty))$$. And in the case that $$A\cap(0,\infty)=\emptyset$$ we get that2 $$R=\sup A\le 0$$ and thus $$(-R,R)=\emptyset$$.

So if we were able to solve the cases $$A\subseteq(0,\infty)$$ and $$A=\emptyset$$, we get that the same is true for any $$A\subseteq\mathbb R$$.

1Just notice that the only difference is that on the LHS we have added some empty sets. I.e., we have $$\bigcup_{r\in A}(-r,r)=\left(\bigcup_{r\in A\cap(-\infty,0]} (-r,r)\right) \cup \left(\bigcup_{r\in A\cap(0,\infty)} (-r,r)\right) = \emptyset \cup \bigcup_{r\in A\cap(0,\infty)} (-r,r) = \bigcup_{r\in A\cap(0,\infty)} (-r,r).$$

2If $$A\subseteq (-\infty,0]$$ then $$0$$ is an upper bound for the set $$A$$. So for the least upper bound we have $$\sup A \le 0$$. (This is a general property of supremum. The assumption that $$a\le x$$ for each $$a\in A$$ implies that $$\sup A\le x$$.)

• 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 $R=supA \leq 0$ Nov 7, 2021 at 13:39
• @unit1991 I have expanded on these two points a bit more in the footnotes at the end of the post. Nov 7, 2021 at 13:49
• 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? Nov 7, 2021 at 14:04
• @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. Nov 7, 2021 at 14:34