🍈 Zettelkasten

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Apr 19, 20252 min read

Theorem

If ${a_{n}}$ is Bounded
If ${a_{n}}$ is Monotone
In particular, it is either:
1. Strictly Increasing and Bounded Above
2. Strictly Decreasing and Bounded Below Then, ${a_{n}}$ Converges

Suppose ${a_{n}}$ is Bounded Above
Suppose ${a_{n}}$ is Strictly Increasing
Then, $\exists c \in R$ s.t $\forall a \in {a_{n}}, a \leq c$
Then, $\forall x_{1}, x_{2} \in N, x_{1} < x_{2} ⟹ a_{x_{1}} < a_{x_{2}}$
We want to show $\exists l \in R, \forall ϵ > 0, \exists N > 0$ s.t $n > N ⟹ ∣ a_{n} - l ∣ < ϵ$
Consider $A = {a_{n} ∣ n \in N} \subset R$
Note that $a_{1} \in A$ so $A$ is non-empty
Note $A$ is bounded above, as we know ${a_{n}}$ is bounded above
By the Completeness Axiom, the Supremum of the set $A$ exists
Choose $l = s u p (A)$
Then, choose arbitrary $ϵ > 0$
Choose $N \in N$ s.t $l - ϵ < a_{N}$ by definition of supremum
Suppose $n > N$
Note that $l - ϵ < a_{N} < a_{n}$ as ${a_{n}}$ is strictly increasing
Note by definition that since $l = s u p (A), \forall a \in A, a < l$
Note $a_{n} \in A$ by definition
Then, $a_{n} \leq l < l + ϵ$
Then, by transitivity of $<$ , it follows that $l - ϵ < a_{n} < l + ϵ$
$⟹ - ϵ < a_{n} - l < ϵ$
$∣ a_{n} - l ∣ < ϵ$ by $∣ \cdot ∣$ defn
Thus, we have shown convergence
$□$