Question

Show that the set B = {x | x^(2) > 2} is non-empty and bounded below. Is it bounded
above? Justify.

Answer 1

Answer on question #69109 – Math – Real Analysis

Question

Show that the set $B = \{x \mid x^{\wedge}(2) > 2\}$ is non-empty and bounded below. Is it bounded above? Justify

Solution

The set $B = \{x \mid x^{\wedge}(2) > 2\}$ contains elements from the interval $(\sqrt{2}; +\infty)$, hence it is not empty.

This set is not bounded below because the statement $x^2 > 2$ means $(x < -\sqrt{2})$ or $(x > \sqrt{2})$, hence $B$ is not bounded below because $\inf_{x \in B} B = -\infty$.

The same reason gives that $B$ is not bounded above because $\sup_{x \in B} B = +\infty$, so there is no number $C$ for which it is true that $\forall x \in B: x < C$.

The set $D = \{x > 0 \mid x^{\wedge}(2) > 2\}$ contains elements from the interval $(\sqrt{2}; +\infty)$, hence it is not empty.

This set is bounded below by $\sqrt{2}$, because

$$\forall x \in D: x > \sqrt{2}$$

that follows from the definition of the set $D$.

$D$ is not bounded above because $\sup_{x \in D} D = +\infty$, so there is no number $C$ for which it is true that $\forall x \in D: x < C$.

**Answer**: $B$ is not empty. $B$ is unbounded. $D$ is bounded below by $\sqrt{2}$. $D$ is not bounded above.

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