Write down this very logical theorem about the absolute value

Question

This statement sounds pretty straightforward but writing it down correctly is a bit harder than expected (probably because it is so logically)

Using the following metric : $(\mathbb{R},|.|)$ proof that:

$A \subseteq \mathbb{R}$ is bounded $\Leftrightarrow \exists m,M \in \mathbb{R} : m \leq a \leq M , \forall a\in A$

Note: A sunset is bounded $\Leftrightarrow \exists a \in V, \exists r > 0 so that A \subset B(a,r)$

It seems so self explanatory that it becomes hard to prove.

Answer 1

Notice that a metric space (say $\mathbb R^n$ or $\mathbb C$) that does not have an order, [ $(a,b),(c,d) \in \mathbb R^2$ so that $(a,b) < (c,d)$ has no universal or clear meaning], has not concept of being bounded above or bounded below.

Instead the definition of as set $A \subset $ of metric space $X$ being bounded has a strict definition that there is a $K \in \mathbb R^+$ so that for any $x,y \in A$ then $d(x,y) \le K$.

Now if $X$ is also an ordered set then we have a very similar but slightly definition of bounded above and bounded below as $A$ is bounded above (below) if there exist an $M\in X$ ($m \in X$) so that for all $a \in A$ then $a \le M$ ($a \ge m$).

We can't have this definition is $X$ is not an ordered set because $a \le M$ for $a, M \in X$ simply is not defined and makes no sense.

However this definition is very useful and it is easy, almost trivial to prove that for an ordered metric space $X$ then $A$ is bounded if and only if $A$ is both bounded above and below.

SO:

Proposition: If $A\subset \mathbb R$ is bounded then then $A$ is bounded above and below.

Pf: $A$ is bounded so there exists a $K$ so that for any $a,b \in A$ then $|a-b| \le K$. Let $x$ be an arbitrary point in $A$. Let $m = x - K$ and let $m = x + K$.

If $a \in A$ then if $a < m$ and $x > m$ so $|x - a| = |x-m| + |m-a| = K + |m-a| > K$. But that contradicts that $|x-a| \le K$. SO $m \le a$ for all $a \in A$ and $A$ is bounded below by $m$.

Likewise if $a > M$ then $|a-x| = |a-M| + |M-x| = |a-M| + K > K$ which contradicts that $|x-a| \le K$. So $M \ge a$ for all $a \in A$ and $A$ is bouded below by $M$.

ANd that's it. But lets be thorough.

Proposition: If $A\subset \mathbb R$ is bounded below and above the $A$ is bounded.

Pf: Let $m$ be a number so that $m \le a$ for all $a \in A$ and let $M$ be a number so that $M \ge a$ for all $a \in A$ then let $a,b \in A$ so that $a \le b$.

Then $m \le a \le b \le M$. Then $|M-m| = |a-m| + |b-a| + |M-b| \ge |b-a|$. So for all $a,b \in A$ then $|a- b| \le |M-m|$ so $A$ is bounded.

That's it. We are done. Its a subtle difference but is important to talk of metric spaces without order to be bounded.

Answer 2

Here a short version:

Note that $$\exists a \in \mathbb{R}, \exists r > 0:\; A \subset B(a,r) \Leftrightarrow \exists R > 0:\; A \subset B(\color{red}{0},R)$$ This is so, because, if $A$ is bounded ($A\subset B(a,r)$), we have for all $x \in A$ $$|x| = |x-a+a|\leq|x-a|+|a|\leq r+|a| = R$$ Now, you are almost done.

Case $\Rightarrow$: $$\forall x \in A: |x| \leq R \Rightarrow -R \leq x \leq R$$ Case $\Leftarrow$: $$\exists m,M \in \mathbb{R} : m \leq x \leq M , \forall x\in A \Rightarrow \mbox{ Set }R := \max(|m|,|M|)\Rightarrow -R \leq m \leq x \leq M \leq R$$