Boundedness of real numbers.

Question

If $A,B\subset\mathbb{R}$ are bounded, show that $A\cup B$ is also bounded. Here is my proof. Suppose $x\in A$ is an arbitrary element in $A$. Since $A$ is bounded, we know that any arbitrary $x$ is bounded such that $a\leq x\leq b$. By similar argument, we know an arbitrary element $y\in B$ is bounded such that $c\leq y\leq d$. Since $A\cup B$ consists of only arbitrary elements of $x$ and $y$ only, this means every arbitrary elements of $A\cup B$ is also bounded, and hence the set $A\cup B$ is bounded. ( We can visualize it by letting $\min\{a,c\}$ be the lower bound and $\max\{b,d\}$ be the upper bound of the set $A\cup B$.)

Is this proof legit?

Answer 1

What you have written is not false but to me it seems a bit sloppy. Usually in such exercises the "safest" way is to find a bound and prove that this is indeed a bound. You did exactly that, but you described why $\max\{b,d\},\min\{a,c\}$ are bounds, you haven't proven it. Because this is an easy exercise it is almost obvious, but I would prefer to see something like :

$x\in A\cup B \Rightarrow x\in A $ or $x\in B\Rightarrow x\leq b $ or $x\leq d \Rightarrow x\leq \max\{b,d\}$

Answer 2

An option:

$A,B \subset \mathbb{R}$ are bounded.

Show that $A\cup B$ is bounded.

Since $A,B$ bounded:

There exist $M,N$ such that:

$|a| \lt M$, $M$ real, positive, for $a \in A$.

$|b| \lt N$, $N$ , real positive for $b \in B.$

Let $R=\max(M,N).$

Let $x \in A\cup B$, then:

1) $x \in A$, or 2) $x \in B$:

1)If $x \in A:$ $|x| \lt M \le R.$

2)If $x \in B$: $|x| \lt N \le R.$

Hence : $A\cup B$us bounded .