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 $k \ge 2$:
| \(\ds \) | \(\) | \(\, \ds x \, \) | \(\, \ds \in \, \) | \(\ds S_k\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 1 k\) | \(<\) | \(\, \ds x \, \) | \(\, \ds < \, \) | \(\ds k\) | Definition of Open Real Interval | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 1 {k + 1}\) | \(<\) | \(\, \ds x \, \) | \(\, \ds < \, \) | \(\ds k\) | Ordering of Reciprocals | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 1 {k + 1}\) | \(<\) | \(\, \ds x \, \) | \(\, \ds < \, \) | \(\ds k + 1\) | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds x \, \) | \(\, \ds \in \, \) | \(\ds S_{k + 1}\) | Definition of Open Real Interval |
It follows that $\sequence {S_k}_k$ is increasing.
$\Box$
$\ds \bigcup_{k \mathop \in \N} S_k = \R_{>0}$
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:
- $\forall x \in \R_{>0} : \exists k \in \N : x \in S_k$
$\Box$
The result follows from the definition of exhausting sequence of sets.
$\blacksquare$