Showing $\sup A\ge 2$

Question

I am struggling to understand the proof in the textbook.

Let $A$ contain elements $x$, and $x$ is real number which satisfies $x^2 < 2$. Let $\sup A = r$, and show that $r^2 \ge 2$.

In the textbook, the proof:

Let $s$ be a positive number $s$ such that $s^2 < 2$ and try to find that $(s+h)^2 < 2$.

So assuming $0 < h < \dfrac{2-s^2}{2s+1}$, and also $h < 1$, it shows that $(s+h)^2 < 2$. $s+h$ is also an element of $A$, since it satisfies the condition. (I understand up to here).

But, I don't understand the conclusion:

As $s$ was an arbitrary positive number such that $s^2 < 2$, it follows that $r^2 \ge 2$.

Why is it necessarily true?? Thanks for helping me in advance.

Answer 1

Because, if $r$ is the sup, you can't find a bigger number verifying also this condition.

Here, if a number $s$ is such that $s^2<2$, he cannot be the sup, since you can find $s+h$ (hence strictly bigger then $s$) verifying the same condition. So $r$ cannot be such that $r^2<2$ : consequently $r^2\geq 2$.

But it is true that using a proof by contradiction would be more easy to understand.

Answer 2

A "less formal" way of thinking about it:

$$\ A=\{x\in\mathbb{R}:x^{2}<2\} \ $$

The set A will be infinite and made up of all the real numbers in the open interval: $\ (-\sqrt{2},\sqrt{2}). \ $

Will there be an element in this set, A, which is greater than (or equal to) all of the other elements of A? Nope. Thus, this is an example of the importance of understanding the difference between a supremum, or least upper bound, and a maximum. A maximum is an element of the set, and also greater than (or equal to) every element in the set; therefore, when a maximum exists, it is also the supremum. However, a supremum does not have to be an element of the set. Our supremum, in this case, is not an element of the set. Our supremum will be $\ \sqrt{2} \ $ as we can safely say that $\ \sqrt{2} \ $ is greater than every element in the set A, and that out of all the upper bounds for the set A (A is a proper subset of $\ \mathbb{R}\ $so 3, 100, 650404, 56050505.404 - these are all upper bounds for A),$\ \sqrt{2} \ $is the least upper bound. Thus, it follows that if we square our least upper bound, which was defined as $ \ r \ $here, so $ \ r^{2}, \ $it will be greater than or equal to 2; in other words, if we square$\ \sqrt{2}, \ $the following holds: $\sqrt{2}^{2}\ge2. \ $