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Definition: Balls

Let $(X, d) $ be a metric space, $x_0 \in X, r> 0 $.

The (open) ball of radius $r $ centered at $x_0 $ relative to $d $ in $X $ is the set $B_{(X, d)}(x_0, r) = \set{ x \in X : d (x_0, x) < r } $

Example

$X = \R, f: \R \to \R f(x) = \begin{cases} 1 & x = 0 \br 0 & if x = 1 \br x & \otherwise \end{cases}$

Define $d: \R \times \R \to [0, \infty)$ by $d(x, y) = \abs{ f(x) - f(y)} $

Does $(x_n)^ \infty _{n = 1 } $ defined by $x_n = \frac{ 1 }{ n } $ converge? If so, to what value.

Claim: $(x_n)_{n = 1} ^ \infty \to 1$

Check!

Note

$\R^n = \set{ xo = (x_1, …, x_n): x_i \in \R } $

If $(_k)^\infty_{k=1} $ is a sequence in $\R^n$ then we write

$x^{(k)} = (x^{(k)}_1, x^{(k)}_2, …, x^{(k)}_n) $

Notice that if you fix an index $i $, then $(x_i^{(k)})^\infty_{k=1} $ is a sequence.

Theorem 1.1.18

o Let $X = \R^n $, $(x^{(k)})^\infty_{k=1} $ is a sequence in $\R^n$ and $x_0 = (x_1, x_2, …, x_n) $ a point in $\R^n $

The following statements are equivalent.

  1. $(x^{(k)})^\infty_{k=1} \to x_0 $ in $l^1 $ metric.
  2. $(x^{(k)})^\infty_{k=1} \to x_0 $ in $l^2$ metric.
  3. $(x^{(k)})^\infty_{k=1} \to x_0$ in $l^\infty $ metric.
  4. for every $j \in \set{ 1, 2, …, n } $, $(x^{(k)}_j)^\infty_{k=1} \to x_j $.
Proof 
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<span class="proof__expand"><a>[expand]</a></span>

$(1 \implies 2)$

Assume $(x^{(k)})^\infty_{k=1} \to x_0 $ in $d_{l^1}$,

$$\limu{ k }{ \infty } d_{l^1}(x^{(k)}, x_0) \to 0 $$

Since $\forall i, d_{l^1}(x^{(k)}_i, x_0) >= d_{l^2} (x^{(k)}_i, x_0) $, by squeeze theorem,

$$\limu{ k }{ \infty } d_{l^2}(x^{(k)}, x_0)= 0$$

$\therefore (x^{(k)})^ \infty _{k = 0} \to x_0 $ in $d_{l^2}$.

$(2 \implies 1) $

Assume $(x^{(k)})^\infty_{k=1} \to x_0 $ in $d_{l^2} $,

$$\limu{ k }{ \infty } d_{l^2}(x^{(k)}, x_0) = 0$$

Since $d_{l^1}(x^{(k)}, x_0) \leq \sqrt[ ]{ n } d_{l^2}(x ^{(k)}, x_0) $, by squeeze theorem,

$$\limu{ k }{ \infty } d_{l^1} (x^{(k)}, x_0) = 0 $$

$\therefore (x^{(k)})^\infty_{k=1} \to x_0 $ in $d_{l^1} $.

$(1 \implies 3, 3 \implies 1) $ follow from a similar computation.

$(1 \implies 4) $

Fix $i \in \set{ 1, 2, …, n } $ to show $x^{(k)}_i \to x_i$; we prove $\limu{ k }{ \infty } \abs{ x_i^{(k)} \to x_i } = 0 $.

Since $0 \leq \abs{ x_i^{(k)} - x_i } \leq \sum_{ i = 1 }^{ n }\abs{ x_i^{(k)} - x_i } $, by squeeze theorem,

$$\limu{ k }{ \infty } \abs{ x^{(k)}_i - x_i } = 0 $$

$\therefore (x_i^{(k)}) \to x_i, \forall i$.

$(4 \implies 1)$

Assume $(x_i^{(k)}) \to x_i \forall i$,

$$\limu{ k }{ \infty } \abs{ x_i^{(k)} -x_i } =0, \forall i$$

$d_{l^1} (x^{(k)}, x_0) = \sum_{ i = 1 }^{ n } \abs{ x_i^{(k)} - x_i }$

$\therefore (x^{(k)})^\infty_{k=1} \to (x_0)$ in $d_{l^1} $.

Definition: Interior, Exterior and Boundary

Let $(X, d)$ a metric space, let $E \subseteq X$.

$x \in X$ is an interior point of $E$, if $$\exists r > 0 \text{ s.t. } B(x, r) \cap E = E\text{ i.e. }B(x, r) \subseteq E$$

$x \in X $ is an exterior point of $E$ if $$\exists r > 0 \text{ s.t. } B(x, r) \cap E = \emptyset$$

$x \in X $ is a boundary point if $x$ is neither an interior point nor an exterior point.

Note

$(X, d) $ metric space $E \subseteq X, x_0 \in X $.

  1. $x_0 \in \ite{E} \iff \exists r >0$ such that $B(x_0, r) \subseteq E$
  2. $\ite{E} \cap \ext{E} = \emptyset$
  3. $\ite{E} \cap \bdry{E} = \emptyset$
  4. $\ext{E} \cap \bdry{E} = \emptyset$
  5. $X = \ite{E} \cup \ext{E} \cup \bdry{E}$
Example

$X = \R, d_{l^2}, E = (0, 1]. $

$\ite{E} = (0, 1)$

$\ext{E} = (-\infty, 0) \cup (1, +\infty) $

$\bdry{E} = \set{0, 1} $

Example

$X= \R, E = [0, 1] d_{l^1}. \forall x \in X$, If $0 < r \leq 1, B(x, r) = \set{ x } $. Why?

$B(x,r) = \set{ y \in R: d(x, y) < r } $

$\ite{E} = (0, 1]$

$\ext{E} = (-\infty, 0] \cup [1, \infty)$

$\bdry{E} = \emptyset$

Example

$0 \not \in E = (0, 1], $ but $0 $ is “close” to $E $.

Use language of open balls.

Definition: Adherent Point and Closure

Let $(X,d)$ be a metric space, let $E \subseteq X$, and let $x_0 $ be a point in $X $.

$x_0 $ is an adherent point of $E$ if $$\forall r > 0, B(x_0, r) \cap E \neq \emptyset$$

The set of all adherent points of $E$ is called the closure of $E$, and is denoted $\overline{ E }$.

$E $ is the collection of all adherent points of $E_0$. This set is called the closure of $E$.

Theorem

Let $(X, d)$ metric space, let $E \subset X$, and let $x_0 \in X$.

The following statements are logically equivalent.

  1. $x_0$ is an adherent point.
  2. $x_0$ is $\ite{E} \cup \bdry{E}$.
  3. there exists a sequence in $E$, $(x^{(n)})^\infty_{n=1}$ such that $(x^{(n)})^\infty_{n=1} \to x_0$.
Proof 
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<span class="proof__expand"><a>[expand]</a></span>

(1 -> 3)

Since $x_0 $ is an adherent point, $\forall r > 0, B(x_0, r) \cap E \neq \emptyset$.

Choose $x^{(1)} \in B(x_0, 1) \cap E$, since $B(x_0, 1) \cap E \neq \emptyset$.

Choose $x^{(2)} \in B(x, \frac{ 1 }{ 2 }) \cap E \neq \emptyset$.

Similarly we choose $(x^{(n)})^\infty_{n=1}$ such that

1. $x^{(n)} \in E, \forall n \in N$

2. $d(x^{(n)},x_0) < \frac{ 1 }{ n } $

Hence, $x^{(n)} \in B(x_0, \frac{ 1 }{ n })$ i.e. $d(x^{(n)}, x_0) < \frac{ 1 }{ n } $.

By squeeze theorem, $\limu{ n }{ \infty } d(x^{(n)}, x_0) = 0$.

(3 -> 1)

Suppose $(x^{(n)})^\infty_{n=1} \to x_0$.

Need to show $\forall r > 0, B(x_0, r) \cap E \neq \emptyset$.

Since $\limu{n}{\infty} d(x^{(n)}, x_0) = 0, \exists N $ such that if $n > N \implies d(x^{(n)}, x_i) < r$.

Pick some $n_0 >N $, $d(x^{(n_0)}, x_0) < r $.

Hence $x^{(n_0)} \in B(x_0, r), x^{(n_0)} \in E$.

$x^{(n_0)} \in B(x_0, r) \cap E$.

Definition: Open and Closed Sets

Let $(X,d)$ be a metric space, and let $E \subseteq X$.

$E$ is closed if $E \cap \bdry{E} = \bdry{E}$, i.e. $\bdry{ E } \subseteq E$.

$E$ is open if $E \cap \bdry{E}= \emptyset$.

$E$ is neither open nor closed if $E \cap \bdry{E} \neq \emptyset$

Note

Is $\bdry{ \R } \subseteq \R $?

Yes, $\emptyset \subseteq \R$

So $\R$ is closed?

Now, let $E = \emptyset $, $\bdry{ \emptyset } = \emptyset$

$\bdry{ \emptyset } = \emptyset $

$\bdry{ \emptyset } \cap \emptyset = \emptyset$

$\emptyset $ is both open and closed.

Note

For every metric space $(X, d)$.

$X $ is open and closed.

$\emptyset $ is open and closed.

Note

$\ite{ X } = X $

Since $\forall E, \ite{ E } \subseteq E $

so $\ite{ X } \subseteq X $.

Let $x_0 \in X$, pick any $r > 0 $

$B(x_0, r) = \set{ x \in X: d(x_0, x) < r } \subseteq X$

Thus $x_0 $ is an interior point of $X $ and hence $X \subseteq \ite{ X } $.

Note

Let $X $ be a set and $d $ the discrete metric on $X_0 $.

Every subset $E \subseteq X $ is always open and closed.

Lemma

Let $(X, d) $ be a metric space.

  1. Let $E \subseteq X $. $E$ is open $\iff E = \ite{ E }$.
  2. Let $E \subseteq X$. $E $ is closed $\iff E = \overline{ E }$.
  3. Let $x_0 \in X, r > 0$.
    The open ball $B(x_0, r) = \set{x \in X : d(x_0, x) \leq r}$ is an open set.
    The closed ball $\set{x \in X : d(x_0, x) \leq r}$ is a closed set.
  4. $\forall x_0 \in X, E = \set{x_0}$ is closed.
  5. Let $E \subseteq X $. $E $ is open $\iff X \setminus E $ is closed.
  6. Let $E_1, E_2, …, E_n \subseteq X$ be a finite number of open sets. $E_1 \cap E_2 \cap … \cap E_n $ is still open.
    Let $F_1, F_2, …, F_n \subseteq X$ be a finite number of closed sets. $F_1 \cup F_2 \cup … \cup F_n$ is still closed.
  7. Let $I$ be an indexing set. $I$ can be infinite or uncountable.
    If $\set{ E_ \alpha }_ { \alpha \in I } $, $E_ \alpha $ open $\forall \alpha \in I, \bigcup_ { \alpha \in I } E_ \alpha $ is still open.
    If $\set{ F_ \alpha }_ { \alpha \in I }$, $F_ \alpha$ closed $\forall \alpha \in I, \bigcap_ { \alpha \in I } F_ \alpha$ is still closed.
Proof 
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<span class="proof__expand"><a>[expand]</a></span>

$(1) $

Assume $E $ is open $\bdry{ E } \cap E = \emptyset$. Pick a point in $E $,

$x_0 $ need to find $r > 0 $.

$B(x_0, r) \subseteq E, x_0 \in \ite{ E } $ or $x_0 \in \bdry{ E }$ or $x_0 \in \ext{ E } $.

Since $\bdry{ E } \cap E = \emptyset, x_0 \in E$ is nnot in $\bdry{ E } $.

$x_0 \in E \implies x_0 \not \in \ext{ E }$

$\forall r > 0, B(x_0, r) contains x_0$

so $B(x, r) \cap E \forall r $

Thus $x_0 $ must be in $\ite{ E } $

To show $E = \ite{ E } $ implies $E $ is open.

$(2)$

suppose $E = \overline{ E } $, $\overline{ E } = E \cup \bdry{ E } $

$\bdry{ E } \subseteq \overline{ E } = E $.

Now suppose $E $ is closed. $\bdry{ E } \subseteq E $

$\overline{ E } = E \subseteq E $.