Exhausting Sequence of Sets on the Strictly Positive Real Numbers

Theorem
For each $k \in \N$, let $S_k = \openint {\dfrac 1 k} k$.

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

Proof
To prove that $\sequence {S_k}_k$ is exhausting $\R_{>0}$, it is sufficient to show:


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

$\sequence {S_k}_k$ is increasing
Let $k \in \N$.

Let $k = 1$.

Then:
 * $S_k = \openint {\dfrac 1 k} k = \O$

Thus, by Empty Set is Subset of All Sets:


 * $\openint {\dfrac 1 k} k \subseteq \openint {\dfrac 1 {k + 1} } {k + 1}$

If $k \ge 2$:

It follows that $\sequence {S_k}_k$ is increasing.

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

Case 1: $1 < x$
Let $1 < x$.

Let $k = \ceiling x$.

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 $k = \ceiling {\dfrac 1 x}$.

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:


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

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