Exhausting Sequence of Sets on the Strictly Positive Real Numbers

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


From Real Number is between Ceiling Functions:

$x < k$

From Ordering of Reciprocals:

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

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


From Ordering of Reciprocals:

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

From Real Number is between Ceiling Functions:

$1 < \dfrac 1 x < k$

From Ordering of Reciprocals:

$\dfrac 1 k < x < 1$


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.

$\blacksquare$