Convergence and Completeness

Converges in Rn

A sequence

(x_{n}) : N_{1} \to R^{n}

is said to converge to

a \in R^{n}

if for every

ϵ \in R^{+}

there is an integer

N

such that for all

k \geq N

we have

‖ x_{k} - a ‖ < ϵ

and in this case we write

lim_{k \to \infty} x_{k} = a

Convergence iff Limit gets arbitrarily Close to a Point

lim_{k \to \infty} x_{k} = a ⟺ lim_{k \to \infty} ‖ x_{k} - a ‖ = 0

A Sequence in Rn Converges to a Point iff Each Component Converges

Suppose that we have the sequence

(x_{k}) : N_{1} \to R^{n}

, then

lim_{k \to \infty} x_{k} = a ⟺ lim_{k \to \infty} x_{k, i} = a_{i}

for

i \in [1, \dots, n]

Cauchy in Rn

We say that a sequence

(x_{k}) : N_{1} \to R^{n}

is cauchy if for every

ϵ \in R^{+}

there is an

N \in N_{1}

such that for all

k, l \geq n

we have

‖ x_{l} - x_{k} ‖ < ϵ

Complete Set in

R^{n}

Given

S \subseteq R^{n}

is complete if every cauchy sequence of points in

S

converges to a point in

S

Less Than or Equal to Iff Less Than Or Equal to With Epsilon

For any

x, y \in R

the statement

x \leq y

is equivalent to

\forall ϵ \in R^{+}, x < y + ϵ

$⟹$ Suppose that $x \leq y$ then let $ϵ \in R^{+}$ thus we know that $x \leq y < y + ϵ$ so then $x < y + ϵ$ .

$⟸$ Suppose the latter, we'll prove the former, but for the sake of contradiction, assume that $x > y$ , but then we know that $y \leq y + (\frac{x - y}{2}) \leq x$ but on the other hand by considering $ϵ = \frac{x - y}{2}$ we see that $x < y + (\frac{x - y}{2})$ this is a contradiction because we cannot have $α \leq x$ and $α > x$ at the same time. Thus we must have $x \leq y$

Note that the same holds for strict inequalities because $x < y ⟹ x \leq y$

Closed Balls are Complete

Show that for any

p \in R^{n}

and

r \in R^{+}

, that

\bar{B (p, r)}

is complete.

Suppose that $(a_{n}) : N_{1} \to \bar{B (p, r)}$ converges to some point $c$ , we'll prove that $c \in \bar{B (p, r)}$ . To show that $c \in \bar{B (p, r)}$ one has to prove that $‖ p - c ‖ \leq r$ .

Since $(a_{n}) \to c$ , we know: $\forall ϵ \in R^{+}, \exists N \in N_{1}, st \forall n \in N_{1}, n \geq N ⟹ ‖ a_{n} - c ‖ < ϵ$

Focusing on the last inequality we also have that $‖ p - c ‖ \leq ‖ p - c ‖ + ‖ a_{n} - c ‖ < ϵ + r$ Thus we can conclude that for any $ϵ \in R^{+}$ we have $‖ p - c ‖ < ϵ + r$ , therefore we have $‖ p - c ‖ < r$ as needed.

R^{n}

is Complete

Every cauchy sequence in

R^{n}

converges to a point in

R^{n}

If a Sequence in

R^{n}

Converges, Then their Norms Converge

lim_{n \to \infty} x_{n} = a ⟹ lim_{n \to \infty} ‖ x_{n} ‖ = ‖ a ‖

Suppose the former, let

ϵ \in R^{+}

therefore we obtain an

N^{'} \in N_{1}

such that

‖ x_{n} - a ‖ < ϵ

for any

n \geq N^{'}

, take

N = N^{'}

recall that we need to show that for any

k \geq N

we have that

| ‖ x_{n} ‖ - ‖ a ‖ | < ϵ

, but due to the reverse triangle inequality we have

| ‖ x_{n} ‖ - ‖ a ‖ | \leq ‖ x_{n} - a ‖ < ϵ

we get exactly what we needed.

We can observe that the converse is false, simply by taking $x_{n} = e_{i}$ for all $n \in N_{1}$ and then for any $e_{k}$ where $i \neq k$ we have that $lim_{n \to \infty} ‖ e_{i} ‖ = ‖ e_{k} ‖$ simply because the norm of the one-hot vectors is always one, but clearly $e_{i} \neq e_{k}$ .