User:Keith.U/Sandbox/Lemma

Theorem
For each $k \in \N$, let $S_k = \left\langle{ \left({ \frac{1}{k}, \,.\,.\, k }\right) }\right\rangle$.

Then $\left\langle{ S_k }\right\rangle$ is an exhausting sequence of sets on $\R_{>0}$.

$(1) : \quad \left\langle{ S_k }\right\rangle$ is increasing
Suppose $k \in \N$.

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

If $\left({ \dfrac{1}{k}, \,.\,.\, k }\right) = \varnothing$, then $\left({ \dfrac{1}{k}, \,.\,.\, k }\right) \subseteq \left({ \dfrac{1}{k+1}, \,.\,.\, k+1 }\right)$ from Empty Set is Subset of All Sets.

If $k \geq 2$:

So $\left\langle{ \left({ \dfrac{1}{k}, \,.\,.\, k }\right) }\right\rangle$ is increasing.

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 $\dfrac{1}{k} < x < k$.

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

Then:
 * $\dfrac{1}{2} < 1 < 2$

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 $\dfrac{1}{k} < x < k$.

Hence:
 * $\forall x \in \R_{>0} : \exists k \in \N : x \in \left({ \dfrac{1}{k}, \,.\,.\, k }\right)$

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