Lecture 27 - IVT

Recall:

Separated, Disconnected

$A, B$ are separated if both:
$\overset{―}{A} \cap B = \emptyset, A \cap \overset{―}{B} = \emptyset$
If $E \subseteq R$ can be written as $E = A \cup B$ where $A, B \neq \emptyset$ and $A, B$ are separated, then $E$ is disconnected.

This is an implication definition: Suppose

E = A \cup B

where

A, B

nomempty and separated implies that

E

is disconnected.

The idea here is that the limit points of $A$ aren't in $B$ , and vice versa.

connected

A set is connected if it is not disconnected. Namely, $E$ is connected for any $A, B \neq \emptyset$ and $A, B$ are not separated ( $\overset{―}{A} \cap B \neq \emptyset$ or $A \cap \overset{―}{B} \neq \emptyset$ ) then $E \neq A \cup B$ .

Examples

$A = (0, 1), B (1, 2)$ then these two sets are separated (the closure of $A$ throws on $0, 1 \notin B$ and the closure of $B$ with the limit points $1, 2$ are not in $A$ neither). Thus $E = A \cup B = (0, 1) \cup (1, 2)$ must be a disconnected set (by construction).

Note that

A, B

being disjoint is not equivalent. For example

A = (0, 1), B = [1, 2)

are not separated since the limit point

1

from

A

is in

B

. Namely:

$\overset{―}{A} \cap B = {1} \neq \emptyset$

Let $E = Q$ . Is $E$ disconnected or connected? Since $Q$ is dense in $R$ , then it is disconnected via the following construction. Let $A = {x \in Q | x < \sqrt{2}}$ and $B = {x \in Q : x > \sqrt{2}}$ . Here $A, B \neq \emptyset$ and $A, B$ are indeed separated. However, $\overset{―}{A} = (- \infty, \sqrt{2}]$ and $\overset{―}{B} = [\sqrt{2}, \infty)$ but $A \cap \overset{―}{B} = \overset{―}{A} \cap B = \emptyset$ showing that $E = A \cup B$ must be disconnected.

We also talked about the condition for connectivity:

Condition for Connected

Let $E \subseteq R$ . $E$ is connected iff $\forall a < b \in E$ and $\forall c \in R$ if $a < c < b$ then $c \in E$ .

Really this is saying that $E$ is an 'interval', in the more traditional sense.

Proof

#todo/school : Read this proof up 📅 2024-11-05 ✅ 2024-11-06

( $\Rightarrow$ ): Suppose $E$ is connected, and let $a, b \in E$ and $a < c < b$ . Set:
$A = (- \infty, c) \cap E; B = (c, \infty) \cap E$
Because $a \in A$ and $b \in B$ , then neither set is empty and neither set contains a limit point of the other (see Separated, Disconnected, Connected Sets#Examples to see how that works, namely example 2). If $E = A \cup B$ , then we would have that $E$ is disconnected, which it is not (by our given). It must then be that $A \cup B$ is missing some element of $E$ and $c$ is the only possibility. Thus, $c \in E$ .

( $\Leftarrow$ ): Suppose $\forall a < b \in E$ and $\forall c \in R$ that $(a < c < b \to c \in E)$ . To show that $E$ is connected, assume $E = A \cup B$ for contradiction, where $A, B$ are nonempty and disjoint (being separated allows us to get disjoint for free). We need to show that one of these sets contains a limit point of the other. Pick $a_{0} \in A$ and $b_{0} \in B$ and, suppose (for the sake of argument) that $a_{0} < b_{0}$ . Because $E$ itself is an interval, then interval $I_{0} = [a_{0}, b_{0}] \subseteq E$ . Now, bisect $I_{0}$ into two equal halves. The midpoint of $I_{0}$ must either be in $A$ or $B$ , so choose $I_{1} = [a_{1}, b_{1}] \subseteq E$ to be the half that allows us to have $a_{1} \in A$ and $b_{1} \in B$ . Continue this process, where each $I_{n} = [a_{n}, b_{n}] \subseteq E$ where $a_{n} \in A$ and $b_{n} \in B$ and the length $(b_{n} - a_{n}) \to 0$ . By the Nested Interval Property then $\exists x$ where:
$x \in ⋂_{n = 0}^{\infty} I_{n}$
It is straightforward to see that $a_{n} \to x$ and $b_{n} \to x$ . But now $x \in E$ must belong to either $A$ or $B$ , thus making it a limit point of the other, contradicting $A, B$ being separated. As such, then $E \neq A \cup B$ so then $E$ must be connected.

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We'll use this to make easy showing of whether or not a set is connected. This will help to get into a useful theorem that will get to the IVT:

Theorem

The continuous image of a connected set is connected.

Proof

Suppose $C \subseteq R$ be a connected set. Let $f : C \to R$ which is continuous (on $C$ ).

(We want to show here that $f (C)$ is connected, using definition of connectedness).

Suppose $\exists A, B \subseteq R$ , where $A, B \neq \emptyset$ , such that $f (C) = A \cup B$ . We want to show that a limit point of $A$ is in $B$ , or vice versa.

Let $A^{'} = {x \in C : f (x) \in A}$ . Similarly $B^{'} = {x \in C : f (x) \in B}$ .

Sometimes we say that

f^{- 1} (A)

is the a shorthand to

A^{'}

, even though

f

may not be bijective (and thus non-invertible).

First off since $A \neq \emptyset$ then $\exists a \in A$ , so then $A^{'}$ is nonempty. Thus then $\exists x_{a}$ where $f (x_{a}) \in A$ and thus $x_{a} \in A^{'}$ . A similar argument shows that $\exists x_{b} \in B^{'}$ . Further notice that $C = A^{'} \cup B^{'}$ , because $f (C) = A \cup B$ .

Now because $C$ is connected, then (without loss of generality) then $\exists (x_{n}) \subseteq A^{'}$ such that $x_{n} \to x$ where $x \in B^{'}$ .

Now because $f$ is continuous at $x \in B^{'} \subseteq C$ then $f (x_{n}) \to f (x) \in B$ by definition. Now because each $f (x_{n}) \in A$ then clearly $B$ contains the limit points of the set $A$ . This shows then that $\overset{―}{A} \cap B \neq \emptyset$ . The same argument shows that $A \cap \overset{―}{B} \neq \emptyset$ .

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As an immediate corollary we get the Intermediate Value Theorem.

Intermediate Value Theorem (IVT)

Let $a < b \in R$ where $f : [a, b] \to R$ be continuous. Let $y \in R$ satisfies either $f (a) > y > f (b)$ or $f (a) < y < f (b)$ then $y \in f ([a, b])$ .

Proof
Using Continuous Image of Connected Set is Connected, since $[a, b]$ is a connected set, then $f ([a, b])$ is connected. Using Condition for Connected (which we get immediately from the givens), then $y \in f ([a, b])$ immediately (use $a = f (a) < y < f (b) = b$ or vice versa, and since $y \in R$ and in between the $f

Starting Derivatives

We'll want to talk about one-sided limits before getting to our talk on derivatives.

Let $f : (a, b) \to R$ .

Let $a < c \leq b$ . We claim that we can define a left-handed limit (a limit from the left) as:
$lim_{x \to c^{-}} f (x) = L$
iff:
$\forall (x_{n}) \subseteq (a, c) where x_{n} \to c, f (x_{n}) \to L$

We may write

f (c^{-})

to denote this limit rather than

L

Similarly if $a \leq c < b$ , we write the right-handed limit (a limit from the right) as:
$lim_{x \to c^{+}} f (x) = L$
iff:
$\forall (x_{n}) \subseteq (c, b) where x_{n} \to c, f (x_{n}) \to L$

We may write

f (c^{+})

to denote this limit rather than

L

c \in (a, b)

here, then

lim_{x \to c} f (x)

exists iff both one-sided limits

lim_{x \to c^{-}} f (x), lim_{x \to c^{+}} f (x)

exist and are equal.

Proof

( $\Rightarrow$ ): If $lim_{x \to c} f (x) = L$ exists then by the Characterization of Continuity then both the sequences we can construct for $(x_{n}) \subseteq (a, c)$ and $(y_{n}) \subseteq (c, b)$ where $x_{n}, y_{n} \to L$ then $f (x_{n}), f (y_{n}) \to f (c)$ , thus showing both one-sided limits exist.

( $\Leftarrow$ ): Suppose both sided limits equal and exist. Let $(x_{n}) \subseteq (a, c)$ where $x_{n} \to c$ then by the left, sided limit existing, then $f (x_{n}) \to f (c)$ . Similarly if $(y_{n}) \subseteq (c, b)$ where $y_{n} \to c$ then $f (y_{n}) \to f (c)$ . This implies $lim_{x \to c^{-}} f (x) = lim_{x \to c^{+}} f (x) = f (c)$ so then the limit must exist by the Existence of Functional Limits.

#todo/school : Show the reverse implication 📅 2024-11-12 ✅ 2024-11-14

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This brings us to a good discussion on discontinuity:

Now let $f : (a, b) \to R$ where $c \in (a, b)$ . Assume first that $f$ is not continuous at $c$ . We say that $c$ is a discontinuity of $f$ . Note that since $c$ is a Limit Point, then:

Note

$f$ is continuous at $c$ iff $lim_{x \to c} f (x) = f (c)$

So then discontinuity comes from 3 possible cases:

The limit $lim_{x \to c} f (x)$ exists, but is not equal to $f (c)$ .

Then here is a removable discontinuity.

$lim_{x \to c} f (x)$ does not exist. Now we could either have both One-Sided Limits exist (they just don't equal each other)

We call this a jump discontinuity.

$lim_{x \to c} f (x)$ does not exist, and at least one of the two one-sided limits don't exist.

For example the function:
$f (x) = {\begin{cases} \frac{1}{x - c} & x \neq c \\ 0 & x = c \end{cases}$
is this type of discontinuity. Another example would be:

---
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yLabel: 
bounds: [0,3,-1,1]
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---
f(x) = sin(1/x)

Where the function above is:
$f (x) = {\begin{cases} 0 & x \leq 0 \\ \sin (\frac{1}{x}) & x > 0 \end{cases}$
In this case then $c$ is an essential discontinuity.

s then $y \in f([a,b])$).

#todo/school : Do this problem (in the HW) 📅 2024-11-12 ✅ 2024-11-13

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Starting Derivatives

We'll want to talk about one-sided limits before getting to our talk on derivatives.

!One-Sided Limits

This brings us to a good discussion on discontinuity:

!Discontinuity (and its types)