Limits Of Sets

Suppose that

(A_{n})

is a sequence of sets, then we say that

A_{n} \to A

iff \left( I \left( A _ n \right) \to I \left( A \right) \right) A sequence

(x_{n}) \subseteq ℝ

converges to

x

\forall ϵ \in ℝ^{> 0}, \exists N i n ℕ_{0} st n > N ⟹ | x_{n} - x | < ϵ

Suppose that

(f_{n})

is a sequence of functions, then we say it converges pointwise to

f

if for any

x \in dom (f)

we have that

(f_{n} (x))

converges to \left( f \left( x \right) \right)

Increasing Sequence of Sets

Let

(A_{n})

be a sequence of sets, then we say that it's increasing when

A_{0} \subseteq A_{1} \subseteq A_{2} \subseteq . . .

Decreasing Sequence of Sets

Let

(A_{n})

be a sequence of sets, then we say that it's decreasing when

A_{0} \supseteq A_{1} \supseteq A_{2} \supseteq . . .

A Sequence of Sets Increase to Another

Suppose that

(A_{n})

is an increasing sequence of sets, then we say it increases to

A

when

⋃_{n = 1}^{\infty} A_{n} = A

, and write

(A_{n}) ↗ A

A Sequence of Sets Decrease to Another

Suppose that

(A_{n})

is an decreasing sequence of sets, then we say it decreases to

A

when

⋃_{n = 1}^{\infty} A_{n} = A

, and write

(A_{n}) ↘ A

Half Closed Subset of Open for Smaller Value

Suppose that

a, b, c \in ℝ

such that

b < c

, then

(a, b] \subseteq (a, c)

Open Sets as an Increasing Sequence of Sets

Let

J : = {(x, y] \subseteq ℝ : x, y \in ℝ}

, show that there is an increasing sequence of sets that increase to the interval

(a, b)

We define the sequence $A_{n} : = (a, b - \frac{1}{n}]$ for each $n \in ℕ_{1}$ . We first show that the sequence is increasing, let $k \in ℕ_{1}$ in, and note that $k < k + 1$ , that is $\frac{1}{k} > \frac{1}{k + 1}$ , therefore $(a, b - \frac{1}{k}] \subseteq (a, b - \frac{1}{k + 1}]$ , so that $A_{k} \subseteq A_{k + 1}$ , therefore $(A_{n})$ is increasing.

We now show that $⋃_{i = 1}^{\infty} A_{n} = (a, b)$ . Let $x \in ⋃_{i = 1}^{\infty} A_{n}$ , by definition this means that there is some $j \in ℕ_{1}$ such that $x \in A_{j}$ , in other words $x \in (a, b - \frac{1}{j}]$ since $b - \frac{1}{j} < b$ then we know that $(a, b - \frac{1}{j}] \subseteq (a, b)$ (by the above lemma), therefore we know that $x \in (a, b)$ .

We now work on the opposite inclusion, so assume that $x \in (a, b)$ , note that $x < b$ , so that $b - x > 0$ , and define $N : = ⌊ \frac{1}{b - x} ⌋ + 1$ , note that $N > 0$ then we have the following: $\frac{1}{b - x} < N$ therefore by multiplying each side by $\frac{1}{N}$ and then $b - x$ , we obtain that $\frac{1}{N} < b - x$ .

Now notice that $\frac{1}{N} < b - x$ is equivalent to $- \frac{1}{N} > x - b$ which is the same as $b - \frac{1}{N} > x$ thus we can also say that $x < b - \frac{1}{N} < b$ and we we clearly see that $x \in (a, x]$ therefore we can conclude that $x \in (a, b - \frac{1}{N}]$ , but $(a, b - \frac{1}{N}]$ is exactly $A_{N}$ , so we've shown there is some $k \in ℕ_{1}$ such that $x \in A_{k}$ (namely $k = N$ ) therefore by definition $x \in ⋃_{i = 1}^{\infty} A_{i}$ as needed.