# Exhausting Sequence of Sets on the Strictly Positive Real Numbers

## Theorem

For each $k \in \N$, let $S_k = \left({\dfrac 1 k \,.\,.\, k }\right)$.

Then $\left({S_k}\right)_k$ is an exhausting sequence of sets on $\R_{>0}$.

## Proof

To prove that $\left({S_k}\right)_k$ is exhausting $\R_{>0}$, it is sufficient to show:

$(1):\quad \forall k \in \N: S_k \subseteq S_{k + 1}$
$(2):\quad \displaystyle \bigcup_{k \mathop \in \N} S_k = \R_{>0}$

### $\left({S_k}\right)_k$ is increasing

Let $k \in \N$.

Let $k = 1$.

Then:

$S_k = \left({\dfrac 1 k \,.\,.\, k}\right) = \varnothing$

Thus, by Empty Set is Subset of All Sets:

$\left({\dfrac 1 k \,.\,.\, k}\right) \subseteq \left({\dfrac 1 {k + 1} \,.\,.\, k + 1}\right)$

If $k \ge 2$:

 $\displaystyle$  $\, \displaystyle x \,$ $\, \displaystyle \in\,$ $\displaystyle S_k$ $\displaystyle \implies \ \$ $\displaystyle \frac 1 k$ $<$ $\, \displaystyle x \,$ $\, \displaystyle <\,$ $\displaystyle k$ Definition of Open Real Interval $\displaystyle \implies \ \$ $\displaystyle \frac 1 {k + 1}$ $<$ $\, \displaystyle x \,$ $\, \displaystyle <\,$ $\displaystyle k$ Ordering of Reciprocals $\displaystyle \implies \ \$ $\displaystyle \frac 1 {k + 1}$ $<$ $\, \displaystyle x \,$ $\, \displaystyle <\,$ $\displaystyle k + 1$ $\displaystyle \implies \ \$ $\displaystyle$  $\, \displaystyle x \,$ $\, \displaystyle \in\,$ $\displaystyle S_{k + 1}$ Definition of Open Real Interval

It follows that $\left({S_k}\right)_k$ is increasing.

$\Box$

### $\displaystyle \bigcup_{k \mathop \in \N} S_k = \R_{>0}$

Let $x \in \R_{>0}$.

#### Case 1: $1 < x$

Let $1 < x$.

Let $k = \left\lceil{x}\right\rceil$.

$x < k$
$1 < k \implies \dfrac 1 k < 1$

So:

$\dfrac 1 k < x < k$

and so $x \in S_k$.

#### Case 2: $x = 1$

Let $x = 1$.

Let $k = 2$.

Then:

$\dfrac 1 2 < 1 < 2$

and hence $x \in S_2$.

#### Case 3: $0 < x < 1$

Let $0 < x < 1$.

Let $k = \left\lceil{\dfrac 1 x}\right\rceil$.

$0 < x < 1 \implies 1 < \dfrac 1 x$
$1 < \dfrac 1 x < k$
$\dfrac 1 k < x < 1$

So:

$\dfrac 1 k < x < k$

and so $x \in S_k$.

Hence:

$\forall x \in \R_{>0} : \exists k \in \N : x \in S_k$

$\Box$

The result follows from the definition of exhausting sequence of sets.

$\blacksquare$