$\newcommand{\br}{\\}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\N}{\mathbb{N}}$ $\newcommand{\C}{\mathbb{C}}$ $\newcommand{\P}{\mathbb{P}}$ $\newcommand{\F}{\mathbb{F}}$ $\newcommand{\L}{\mathcal{L}}$ $\newcommand{\spa}[1]{\text{span}(#1)}$ $\newcommand{\dist}[1]{\text{dist}(#1)}$ $\newcommand{\max}[1]{\text{max}(#1)}$ $\newcommand{\min}[1]{\text{min}(#1)}$ $\newcommand{\supr}[1]{\text{sup}(#1)}$ $\newcommand{\infi}[1]{\text{inf}(#1)}$ $\newcommand{\ite}[1]{\text{int}(#1)}$ $\newcommand{\ext}[1]{\text{ext}(#1)}$ $\newcommand{\bdry}[1]{\partial #1}$ $\newcommand{\argmax}[1]{\underset{#1}{\text{argmax }}}$ $\newcommand{\argmin}[1]{\underset{#1}{\text{argmin }}}$ $\newcommand{\set}[1]{\left\{#1\right\}}$ $\newcommand{\emptyset}{\varnothing}$ $\newcommand{\tilde}{\text{~}}$ $\newcommand{\otherwise}{\text{ otherwise }}$ $\newcommand{\if}{\text{ if }}$ $\newcommand{\proj}{\text{proj}}$ $\newcommand{\union}{\cup}$ $\newcommand{\intercept}{\cap}$ $\newcommand{\abs}[1]{\left| #1 \right|}$ $\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$ $\newcommand{\pare}[1]{\left(#1\right)}$ $\newcommand{\brac}[1]{\left[#1\right]}$ $\newcommand{\t}[1]{\text{ #1 }}$ $\newcommand{\head}{\text H}$ $\newcommand{\tail}{\text T}$ $\newcommand{\d}{\text d}$ $\newcommand{\limu}[2]{\underset{#1 \to #2}\lim}$ $\newcommand{\der}[2]{\frac{\d #1}{\d #2}}$ $\newcommand{\derw}[2]{\frac{\d #1^2}{\d^2 #2}}$ $\newcommand{\pder}[2]{\frac{\partial #1}{\partial #2}}$ $\newcommand{\pderw}[2]{\frac{\partial^2 #1}{\partial #2^2}}$ $\newcommand{\pderws}[3]{\frac{\partial^2 #1}{\partial #2 \partial #3}}$ $\newcommand{\inv}[1]{{#1}^{-1}}$ $\newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $\newcommand{\nullity}[1]{\text{nullity}(#1)}$ $\newcommand{\rank}[1]{\text{rank }#1}$ $\newcommand{\nullspace}[1]{\mathcal{N}\pare{#1}}$ $\newcommand{\range}[1]{\mathcal{R}\pare{#1}}$ $\newcommand{\var}[1]{\text{var}\pare{#1}}$ $\newcommand{\cov}[2]{\text{cov}(#1, #2)}$ $\newcommand{\tr}[1]{\text{tr}(#1)}$ $\newcommand{\oto}{\text{ one-to-one }}$ $\newcommand{\ot}{\text{ onto }}$ $\newcommand{\ceil}[1]{\lceil#1\rceil}$ $\newcommand{\floor}[1]{\lfloor#1\rfloor}$ $\newcommand{\Re}[1]{\text{Re}(#1)}$ $\newcommand{\Im}[1]{\text{Im}(#1)}$ $\newcommand{\dom}[1]{\text{dom}(#1)}$ $\newcommand{\fnext}[1]{\overset{\sim}{#1}}$ $\newcommand{\transpose}[1]{{#1}^{\text{T}}}$ $\newcommand{\b}[1]{\boldsymbol{#1}}$ $\newcommand{\None}[1]{}$ $\newcommand{\Vcw}[2]{\begin{bmatrix} #1 \br #2 \end{bmatrix}}$ $\newcommand{\Vce}[3]{\begin{bmatrix} #1 \br #2 \br #3 \end{bmatrix}}$ $\newcommand{\Vcr}[4]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \end{bmatrix}}$ $\newcommand{\Vct}[5]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \br #5 \end{bmatrix}}$ $\newcommand{\Vcy}[6]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \br #5 \br #6 \end{bmatrix}}$ $\newcommand{\Vcu}[7]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \br #5 \br #6 \br #7 \end{bmatrix}}$ $\newcommand{\vcw}[2]{\begin{matrix} #1 \br #2 \end{matrix}}$ $\newcommand{\vce}[3]{\begin{matrix} #1 \br #2 \br #3 \end{matrix}}$ $\newcommand{\vcr}[4]{\begin{matrix} #1 \br #2 \br #3 \br #4 \end{matrix}}$ $\newcommand{\vct}[5]{\begin{matrix} #1 \br #2 \br #3 \br #4 \br #5 \end{matrix}}$ $\newcommand{\vcy}[6]{\begin{matrix} #1 \br #2 \br #3 \br #4 \br #5 \br #6 \end{matrix}}$ $\newcommand{\vcu}[7]{\begin{matrix} #1 \br #2 \br #3 \br #4 \br #5 \br #6 \br #7 \end{matrix}}$ $\newcommand{\Mqw}[2]{\begin{bmatrix} #1 & #2 \end{bmatrix}}$ $\newcommand{\Mqe}[3]{\begin{bmatrix} #1 & #2 & #3 \end{bmatrix}}$ $\newcommand{\Mqr}[4]{\begin{bmatrix} #1 & #2 & #3 & #4 \end{bmatrix}}$ $\newcommand{\Mqt}[5]{\begin{bmatrix} #1 & #2 & #3 & #4 & #5 \end{bmatrix}}$ $\newcommand{\Mwq}[2]{\begin{bmatrix} #1 \br #2 \end{bmatrix}}$ $\newcommand{\Meq}[3]{\begin{bmatrix} #1 \br #2 \br #3 \end{bmatrix}}$ $\newcommand{\Mrq}[4]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \end{bmatrix}}$ $\newcommand{\Mtq}[5]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \br #5 \end{bmatrix}}$ $\newcommand{\Mqw}[2]{\begin{bmatrix} #1 & #2 \end{bmatrix}}$ $\newcommand{\Mwq}[2]{\begin{bmatrix} #1 \br #2 \end{bmatrix}}$ $\newcommand{\Mww}[4]{\begin{bmatrix} #1 & #2 \br #3 & #4 \end{bmatrix}}$ $\newcommand{\Mqe}[3]{\begin{bmatrix} #1 & #2 & #3 \end{bmatrix}}$ $\newcommand{\Meq}[3]{\begin{bmatrix} #1 \br #2 \br #3 \end{bmatrix}}$ $\newcommand{\Mwe}[6]{\begin{bmatrix} #1 & #2 & #3\br #4 & #5 & #6 \end{bmatrix}}$ $\newcommand{\Mew}[6]{\begin{bmatrix} #1 & #2 \br #3 & #4 \br #5 & #6 \end{bmatrix}}$ $\newcommand{\Mee}[9]{\begin{bmatrix} #1 & #2 & #3 \br #4 & #5 & #6 \br #7 & #8 & #9 \end{bmatrix}}$
Definition: Connected Spaces

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

We say that $X$ is disconnected iff
there exist disjoint non-empty open sets $V, W \subset X$ such that $V \cup W = X$.

Equivalently, $X$ is disconnected iff
$X$ contains a non-empty proper subset which is simultaneously closed and open.

We say that $X$ is connected iff
it is non-empty and not disconnected.

The property of connectedness is intrinsic.

Note: Connectedness of the Empty Set

The empty set $\emptyset $ is special, it is neither connected nor disconnected.

Example

Consider the connectedness of the space $X := [1,2] \cup [3,4]$ with the usual metric.

This set is disconnected because exist open sets $[1,2]$ and $[3,4]$ are open relative to $X$.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Let’s prove one of them is open.

$[1,2]$ is an open set with regard to $X$, because $x \in X, x \not \in [1, 2] $, then $x$ is exterior point.

Remarks

Intuitively, a disconnected set is one which can be separated into two disjoint open sets; a connected set is one which cannot be separated in this manner.

Definition: Connected Sets

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

$Y$ is connected iff the metric space $(Y, d|_{Y \times Y})$ is connected;

$Y$ is disconnected iff the metric space $(Y, d|_{Y \times Y})$ is disconnected.

Theorem

Let $X$ be a subset of the real line $\R$.

  1. $X$ is connected.
  2. $x, y \in X, x < y \implies [x, y]$ is also contained in $X$.
  3. $X$ is an interval.
Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

$(1 \implies 2)$

Assume on the contrary $\exists x, y \in X, x < y$ such that $[x, y]$ is not contained in $X$.

There exists a real number $x < z < y$ such that $z \not \in X$.

Thus the sets $(- \infty, z) \cap X $ and $(z, +\infty) \cap X $ will cover $X$.

Since these sets are non-empty (because they contain $x$ and $y$) and are open relative to $X$, and so $X$ is disconnected. CONTRADICTION.

$(2 \implies 1)$

Assume on the contradiction, $X$ is disconnected.

Then there exist disjoint non-empty sets $V, W$ which are open relative to $X$, such that $V \cup W = X$.

Since $V, W \neq \emptyset$, we choose an $x \in V, y \in W $.

Since $V$ and $W$ are disjoint, we have $x \neq y $.

Without loss of generality, we assume $x < y$.

By (2), the interval $[x, y] $ is contained in $X$.

Consider the set $[x, y] \cap V $ is both bounded, and non-empty (since it contains $x$).

Thus it has a supremum,

$$z := \supr{ [x, y] \cap V } $$

Clearly $z \in [x, y] $, hence $z \in X $. Thus either $z \in V $ or $z \in W $.

Suppose $z \in V $, then $z \neq y $.

However, $V$ is open relative to $X$, which contains $[x, y] $ so there is some ball $B_{[x, y], d} (z, r)$ which is contained in $V$.

However, this contradicts the fact that z is the supremium of $[x, y] \cap V $.

Now suppose $z \in W $, then $z \neq x $.

However $W$ is open relative to $X$, which contains $[x, y] $, so there is some ball $B_{([x, y], d)} (z, r) $ which is contained in $W$.

But this again contradicts the fact that $z$ is the supremum of $[x, y] \cap V $.

Thus in either case we obtain a contradiction.

Note: How to Prove $X$ Is Connected

  1. Look at “nice” parts of $X$(e.g. connected parts)

  2. Build up $X $ via connected parts

Proposition

If $E \subset X $ is connected, and $X \subset A \cup B $. ($A $ and $B $ disjoint and open).

Then $E \subset A $(and $E \cap B = \emptyset $)

or $E \subset B $(and $E \cap A = \emptyset $)

Example

In $\R^2 $ look at (euclidean metric) $X = B((-1, 0), 1) \cup B((1, 0), 1) \cup \set{(0,0)} $. Prove that $X $ is connected.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Use contradiction plus proposition.

Assume $X $ is disconnected, i.e. exists $A $ and $B $ open with $X \subset A \cup B $ and $A \cap B = \emptyset$(disjoint).

Homework: $B((-1, 0), 1)$ and $B((1, 0), 1)$ are connected.

without loss of generality, assume $(0, 0) \in A $, note $A $ intersect both circles. Then since these are connected.

$B((-1, 0), 1) \cup B((1, 0), 1) \subset A$.

$X \subset A$ This means $X $ is connected.

Example: Topologist sine curve in $(\R^2, d\_{l^2})$

Let $X$ be the union of two sets.

$E_1 =$ closed line segment from $(0, -1) $ to $(0 , 1) $

$E_2 =$ graph of $f(x) = \sin (\frac{ 1 }{ x }) $ from $x \in (0, 1] $

Let $X$ be the union of two sets.

Prove that $X$ is connected.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

If $X = A \cup B $ ($A \cap B = \emptyset A$ and $B$ open)

without loss of generality, $A $ contains $(0, 0) $. Then $A $ intersects

$E_1 $ and $E_2 $(which are connected)

$X = E_1 \cup E_2 \subset A $

Example

$(\R, d_{ \text{disc}})$ is disconnected.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

We choose two sets in $\R$, $A = (- \infty, 0] $ and $B = (0, \infty)$, the two sets are disjoint.

Note

In discrete metric space, every subset is both open and closed.

Theorem : Continuity Preserves Connectedness

Let $f: X \to Y$ be a function between metric spaces.

If $E \subseteq X$ is connected, and $f$ is continuous.

Then $f(E) \subseteq Y$ is connected.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Since $E$ is connected, $E$ is non-empty. Hence, $f(E)$ is also not empty.

Therefore, $f(E)$ is either connected or disconnected.

Assume $f(E)$, on the contrary, is disconnected.

There exists set $U, V \subseteq Y$ open such that
$U \cap f(E)$ and $V \cap f(E)$ non-empty, $(U \cap f(E)) \cap (V \cap f(E)) = \emptyset$.

$U \cup V \supseteq f(E)$

We have $W_1 = \inv{f}(U), W_2 = \inv{f}(V)$ such that $f(W_1) = U, f(W_2) = V$.

$W_1, W_2 $ must be disjoint.

$E \cap W_1 \neq \emptyset, E \cap W_2 \neq \emptyset $, both are open.

We also have $E \subseteq W_1 \cup W_2.$

That means $(E \cap W_1) \cup (E \cap W_2) = E $

Hence $ E $ is disconnected. CONTRADICTION.

Corollary : Intermediate Value Theorem

Let $f: X \to \R$ be a continuous map from one metric space $(X, d_X)$ to the real line. Let $E$ be any connected subset of $X$, and let $a, b$ be any two elements of $E$. Let $y$ be a real number between $f(a)$ and $f(b)$. Then there exists $c \in E$ such that $f(c) = y$.

========================================================================

Definition

Let $f: X \to Y $, a function between metric spaces $E \subseteq X $, $x_0 \in X $ , an adherent point of $E$, and $L \in Y$. (candidate for a limiting value)

We say $L$ is the limit as $X$ of $f$ tends to $x_0$ along $E$. ($\limu{ x }{ x_0 } f(x) = L, x \in E $) iff $\forall \epsilon > 0, \exists \delta > 0 $ such that if $x \in E $ with $d(x, x_0) < \delta $, then $d(f(x), L) < \epsilon $.

Observation: $f(x_0)$ may not be defined, but a limit may still exist.

Observation:

$\limu{ x }{ x_0 } f(x) \to \limu{ x }{ x_0 } f(x), x \in X \ \set{ x_0 }$

$\limu{ x }{ x_0^- } f(x) \to \limu{ x }{ x_0 } f(x), x \in \set{ X < x_0 }$

Observation:

$f$ is continuous at $x_0 $ iff

$\limu{ x }{ x_0 }, x \in X \setminus \set{ x_0 } f(x) = f(x_0) $

  1. $\lim f(x)$ exists

  2. $f$ is defined at $x_0$

  3. equality between limiting value and function value

Observation:

If condition (1) holds, then $x_0$ is a removable discontinuity.

Theorem

Let $f: X \to Y $ be a function between metric spaces. $E \subseteq X, x_0 \in \overline{ E } , L \in Y$

The following 4 statements are equivalent:

  1. $\limu{ x }{ x_0 } f(x) = L$, $x \in E $

  2. $\forall (x^{(n)})^\infty_{ n =1} $ sequences in $E$ such that $(x^{(n)})^\infty_{ n =1} \to x_0 $ we have $(f(x^{(n)}))^\infty_{ n =1} \to L$

  3. $\forall V \subseteq Y$ open containing $L $, $\exists U \subseteq X $ open such that $f(U \cap E) \subseteq V$($U \cap E$ is relatively open in $E$).

  4. $g : E \cup \set{ x_0 } \to Y$ defined by $g(x) = \begin{cases} f(x)&\text{ if }x \neq x_0 \br L&\text{ if }x = x_0 \end{cases}$

is a continuous function. Furthermore, if $x_0 \in E$, then $f(x_0) = L$.

Note

$(X, d_{disc}) $ be a metric space with the discrete metric. Let $E $ be a subset of $X$ which contains at least two elements. Then $E $ is disconnected.

TODO: proof

Note

$(X, d) $ connected. $(Y, d_{disc}) $. $f: X \to Y$.

$$f\text{ continuous} \iff f\text{ constant} $$

TODO: proof

Note

$(X,d)$, and let $(E_ \alpha)_{ \alpha \in I } $ be a collection of connected sets in $X$. Suppose also that $\bigcap_{ \alpha \in I } E_ \alpha$ is non-empty.

$\bigcup_{ \alpha \in I } E_ \alpha$ is connected.

TODO: proof