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

Question

$\sup_{r \in A} r$ 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?

This can be helpful.

Answer 1

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 $$\bigcup_{r\in A}(-r,r)=(-\sup A,\sup A).$$

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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 $|x|<R$. Since $R=\sup A$, there exists $r\in A$ such that $|x|<r<R$ 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 $|x|<r$ 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.

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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 $(-\sup A,\sup A)=(\infty,-\infty)=\emptyset=\bigcup_{r\in\emptyset}(-r,r)$.

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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 $\sup A\le x$.)