Equivalence of Definitions of Bounded Subset of Real Numbers

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Theorem

The following definitions of the concept of Bounded Subset of Real Numbers are equivalent:

Definition 1

Let $T \subseteq \R$ be both bounded below and bounded above in $\R$.


Then $T$ is bounded in $\R$.

Definition 2

Let $T \subseteq \R$ be a subset of $\R$ such that:

$\exists K \in \R: \forall x \in T: \size x \le K$

where $\size x$ denotes the absolute value of $x$.


Then $T$ is bounded in $\R$.


Proof

Definition 1 implies Definition 2

Let $S$ be a bounded subset of the real numbers by definition 1.

Then by definition $S$ is bounded both above and below.

As $S$ is bounded below:

$\exists L \in \R: \forall x \in S: L \le x$

As $S$ is bounded above:

$\exists H \in \R: \forall x \in S: H \ge x$

Let:

$K = \max \set {\size L, \size H}$

Then:

$K \ge \size L$ and $K \ge \size H$

It follows from Negative of Absolute Value: Corollary 2 that $-K \le L \le K$ and $-K \le H \le K$.

In particular:

$-K \le L$ and so $\forall x \in S: -K \le x$
$H \le K$ and so $\forall x \in S: x \le K$

Thus:

$\forall x \in S: -K \le x \le K$

and, by Negative of Absolute Value: Corollary 2:

$\forall x \in S: \size x \le K$

Thus $S$ is a bounded subset of the real numbers by definition 2.

$\Box$


Definition 2 implies Definition 1

Let $S$ be a bounded subset of the real numbers by definition 2.

Thus:

$\exists K \in \R: \forall x \in S: \size x \le K$

Then by Negative of Absolute Value: Corollary 2:

$\forall x \in S: -K \le x \le K$

Thus by definition, $S$ is both bounded above (by $K$) and bounded below (by $-K$).

Thus $S$ is a bounded subset of the real numbers by definition 1.

$\blacksquare$


Sources

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