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 $\R_{>0}$, we ought 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 $k \in \N$.

If $k = 1$, 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 $k \ge 2$:

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

$\displaystyle \bigcup_{k \mathop \in \N} S_k = \R_{>0}$
Let $x \in \R_{>0}$ be arbitrary.

Case 1: $1 < x$
If $1 < x$, let $k = \left\lceil{ x }\right\rceil$.

From Real Number between Ceiling Functions, $x < k$.

From Ordering of Reciprocals:


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

So $\dfrac 1 k < x < k$, and $x \in S_k$.

Case 2: $x = 1$
If $x = 1$, let $k = 2$.

Then:


 * $\dfrac 1 2 < 1 < 2$

and hence $x \in S_2$.

Case 3: $0 < x < 1$
If $0 < x < 1$, let $k = \left\lceil{ \dfrac 1 x }\right\rceil$.

From Ordering of Reciprocals:


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

From Real Number between Ceiling Functions, $1 < \dfrac 1 x < k$.

From Ordering of Reciprocals, $\dfrac 1 k < x < 1$.

So $\dfrac 1 k < x < k$, and $x \in S_k$.

Hence:


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

Hence the result, from the definition of exhausting sequence of sets.