# Increasing Sequence of Sets forms Nest

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## Theorem

Let $\sequence {a_n}$ be an increasing sequence of sets:

- $\forall k \in \N: S_k \subseteq S_{k + 1}$

Let $c = \set {a_1, a_1, \ldots, a_n, a_{n + 1}, \ldots}$ be the class of all terms of $\sequence {a_n}$.

Then $c$ is a nest.

## Proof

Recall the definition of nest:

$c$ is a **nest** if and only if:

- $\forall x, y \in c: x \subseteq y$ or $y \subseteq x$

Let $a_i$ and $a_j$ be arbitrary elements of $c$.

From Ordering on Natural Numbers is Trichotomy, either $i < j$ or $i = j$ or $i > j$.

- Case $(1)$

Let $i = j$.

Then we have $a_i = a_j$ and so both $a_i \subseteq a_j$ and $a_j \subseteq a_i$.

- Case $(2)$

Let $i \ne j$.

Without loss of generality, suppose $i < j$.

Then as Subset Relation is Transitive:

- $a_i \subseteq a_j$

$\blacksquare$

Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: I believe the object of this exercise was to use theorems specifically introduced as part of the exposition by S&F in order to construct the minimally inductive class, but right now I have a day job to attendYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 8$ Definition by finite recursion: Exercise $8.5$