Question

Let the set T be given by T = { x in the reals | x^2 < 2}. Describe maximum, minimum, supremum (least upper bound), and infimum (greatest lower bound) for T (they may not exist).

Answer 1

Answer:

• The supremum is $\sqrt{2}$.
• The infimum is $-\sqrt{2}$.
• There is no maximum.
• There is no minimum.

Step-by-step explanation:

We have the set $T=\{x\in\mathbb{R}:x^2<2\}$. Now, let us recall that $\sqrt{x^2}=|x|$, and the inequality $x^2<2$ is equivalent to $|x|<\sqrt{2}$, so our set can be written as

$T=\{x\in\mathbb{R}:|x|<\sqrt{2}\}$.

The inequality $|x|<\sqrt{2}$ is equivalent to $-\sqrt{2}<x<\sqrt{2}$. So,

$T=\{x\in\mathbb{R}:-\sqrt{2}<x<\sqrt{2}\}$. Thus, $T=(-2,2)$.

Now, we have that $T$ is the open interval (-2,2). From this we can extract all the information we need:

• The supremum is $\sqrt{2}$.
• The infimum is $-\sqrt{2}$.
• There is no maximum, because the interval is open.
• There is no minimum, because the interval is open.