# 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.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 Proof Wiki 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 $8.5$