Open Intervals

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always space

As we study functions we have expectations of their values or feelings about “nice” behavior and sometimes functions just don’t follow our ideas of “niceness”. We have two types of situations where this appears.

There is a domain number. The function has a value at this domain number. When other domain numbers are close to this domain number, their function values are not close to this function value.
There is a real number, which is not a domain number. So, the function has no value at this real number. When domain numbers are close to this real number, their function values are not close to each other.

We have two concepts we need to get control over. We need algebraic language that allows us to describe exactly the concepts of “around” and “close”. These ideas are vital to Calculus and the study of functions. And we will never get anywhere with vague or graphical language.

$\blacktriangleright$ First up is “around”.

We want an exact notion of “around”, because the function values surrounding our target are generating our expectations. Their values are making the target function value suspicious.

So, we need precise algebraic language for “around” and open intervals will provide it.

An interval of real numbers, or on the real line, is a whole piece of the real line.

Open Intervals

A finite open interval contains all of the real numbers between two distinct real numbers, but excludes the two distinct numbers themselves.

We symbolize finite open intervals like $(a, b)$ .

$(a,b)$ includes all real numbers strictly between $a$ and $b$ .
$(a,b)$ includes all real numbers that are simultaneously greater than $a$ and also less than $b$ .
$(a,b)$ includes all real numbers, $r$ , such that $a < r < b$ .
An infinite open interval contains either (1) all of the real numbers less than some specific real number, or (2) all of the real numbers greater than some specific real number.

We symbolize infinite open intervals like $(-\infty , b)$ or $(a, \infty )$ .

$(-\infty , b) = \{ \, r \in \mathbb {R} \, | \, r < b \, \}$
$(a, \infty ) = \{ \, r \in \mathbb {R} \, | \, a < r \, \}$

Closed Intervals

A finite closed interval contains all of the real numbers between two distinct real numbers, including the two distinct numbers themselves.

We symbolize finite open intervals like $[a, b]$ .
$[a,b]$ includes all real numbers that are simultaneously greater than or equal $a$ and also less than or equal $b$ .
$[a,b]$ includes all real numbers, $r$ , such that $a \leq r \leq b$ .
An infinite closed interval contains either (1) all of the real numbers less than or equal to some specific real number, or (2) all of the real numbers greater than or equal to some specific real number.
We symbolize infinite open intervals like $(-\infty , b]$ or $[a, \infty )$ .
$(-\infty , b] = \{ \, r \in \mathbb {R} \, | \, r \leq b \, \}$
$[a, \infty ) = \{ \, r \in \mathbb {R} \, | \, a \leq r \, \}$

Distinct means different or not equal.

You cannot tell from the definitions, but open intervals are weird, which means this is where we learn a lot about real numbers and functions.

$\blacktriangleright$ Open intervals always contain a real number, for instance the average of the two endpoint numbers.

$\frac {a+b}{2} \in (a,b)$

$\blacktriangleright$ Open intervals do not contain a maximum real number.

Note: $b$ is not a member of $(a,b)$ , so it cannot be the maximum number in the interval.

Suppose $(a,b)$ does contain a maximum number. Then the maximum would be some number, $c \in (a,b)$ . Why can’t that happen?

Suppose $c \in (a,b)$ is the maximum number in the interval $(a,b)$ . Then, consider the number $\frac {c+b}{2}$ .

$\frac {c+b}{2} \in (a,b) \, \text { and } \, c < \frac {c+b}{2}$

If we have the maximum number in the interval, $c$ , then we can find a greater number also in the interval. Therefore, $c$ cannot be the maximum number in the interval. As a result, there is no maximum number in an open interval.

This type of argument is called a proof by contradiction.

You assume a fact is true, like $c \in (a,b)$ is the maximum number in the open interval.

Then you use logic to discover other facts that also must be true based on the truth of your assumed fact. You do this until you discover something that must also be true, but you know it is not.

At this point, you know that your original asummption cannot be true.

An explanation that includes both true and false statements is called a contradiction.

The conclusion is that your original assumption must be false, because your assumption has generated a contradiction, like a number greater than the maximum value in an interval.

$\blacktriangleright$ Similarly, open intervals do not contain a minimum real number.

$\blacktriangleright$ We can make similar observations about infinite open intervals.

Space

Let $(a,b)$ be an open interval.

Let $c \in (a,b)$ .

Then there is an open proper subinterval inside $(a,b)$ that also contains $c$ .

Consider $\frac {a+c}{2}$ and $\frac {c+b}{2}$

$a < \frac {a+c}{2} < c$ and $c < \frac {c+b}{2} < b$

$c \in \left (\frac {a+c}{2} , \frac {c+b}{2} \right ) \subset (a,b)$

There is always space around any number inside an open interval. This space will help us describe our expectations. The space means that an open interval cannot just touch a number, like an endpoint. Any number that an open interval touches is swallowed up completely by the open interval.

We are only talking about intervals here, but there is also a larger notion of an open set, which may not be an interval. The observation above is the defining characteristic of open sets. If there is an open interval around every member of the set, then the set is an open set.

Open Set

Let $S$ be a set of real numbers.

$S$ is an open set provided for each number in $S$ there exists an open interval containing the number and this open interval is totally inside the set $S$ .

For each $c \in S$ , there exists $(a,b) \subset S$ such that $c \in (a,b)$

Set Arithmetic

We have several types of intervals of real numbers.

Finite Open: $(a, b)$
Finite Closed: $[a, b]$
Finite Half-Opened, Half-Closed: $[a, b)$ or $(a,b]$
Infinite Open: $(-\infty , b)$ or $(a, \infty )$
Infinite Closed: $(-\infty , b]$ or $[a, \infty )$

These are special types of sets.

Sets have an arithmetic. Three of the operations are intersection and union and complement.

Set Operations

Given two sets $S$ and $T$ , the union is $S \cup T = \{ \, r \in \mathbb {R} \, | \, r \in S \, \text { or } \, r \in T \, \}$

Given two sets $S$ and $T$ , the intersection is $S \cap T = \{ \, r \in \mathbb {R} \, | \, r \in S \, \text { and } \, r \in T \, \}$

Given a set $S$ , the complement of is $S^{c} = \{ \, r \in \mathbb {R} \, | \, r \not \in S \, \}$

We can use this arithmetic to connect open and closed sets.

$\blacktriangleright$ The complement of a finite open interval is the union of two closed infinite intervals.

$(a,b)^{c} = (-\infty , a] \cup [b, \infty )$

$\blacktriangleright$ The complement of an infinite open interval is an infinite closed interval.

$(-\infty , a)^{c} = [a, \infty )$

$(a, \infty )^{c} = (-\infty , a]$

$\blacktriangleright$ The complement of a finite closed interval is the union of two open intervals.

$[a,b]^{c} = (-\infty , a) \cup (b, \infty )$

$\blacktriangleright$ The complement of an infinite closed interval is an infinite open interval.

$(-\infty , a]^{c} = (a, \infty )$

$[a, \infty )^{c} = (-\infty , a)$

Intervals and their arithmetic are part of a big idea of sets. Intervals are special sets. There are other types of sets and they all use this arithmetic.

There are two special sets.

$\mathbb {R}$ is an open set, because any real number, $r$ , has an open interval around it - namely $(r-1, r+1) \subset \mathbb {R}$ .
The empty set, $\emptyset$ , is an open set. This is true, because it is not false. You cannot find a number inside $\emptyset$ such that there is no open interval containing it - because there are no numbers in $\emptyset$ . We say that this is vacuously true, because there is nothing to prove it wrong.

Sets also follow this idea that complements of open sets are closed and vice versa.

$\mathbb {R} = \emptyset ^c$ , which makes $\mathbb {R}$ a closed set.
$\mathbb {R}^c = \emptyset$ , which makes $\emptyset$ a closed set.

Open and Closed

$\mathbb {R}$ and $\emptyset$ are both open and closed sets at the same time.

Intersections

$(-\infty , 3] \cap [2, \infty ) = \left [\answer {2}, \answer {3}\right ]$

$(-\infty , 3] \cap (-\infty , 2] = \left (-\infty , \answer {2}\right ]$

$[6, \infty ) \cap [4, \infty ) = \left [\answer {6}, \infty \right )$

$(-\infty , 3] \cap [4, \infty ) = \emptyset$

$[1, 7] \cap [4, 9] = \left [\answer {4}, \answer {7}\right ]$

$[1, 7] \cap [2, 6] = \left [\answer {2}, \answer {6}\right ]$

$[1, 7] \cap [9, 11] = \emptyset$

The observation above gives the idea of why intersections of closed sets are closed sets.

Nested Intervals

An interval is nested inside another interval if it is completely contained inside the other interal.

Nested Closed Intervals

If $a \leq b < c \leq d$ , then $[b,c]$ is nested inside $[a,d]$ .

$[b,c]$ is a subset of $[a,d]$ , i.e. $[b,c] \subseteq [a,d]$ .

$\blacktriangleright$ The intersection of two nested closed intervals equals the inner closed interval.

$\blacktriangleright$ The intersection of a finite number of nested closed intervals equals the closed inner most interval.

$\blacktriangleright$ What about an infinite number of nested closed intervals?

Nested

$\bigcap _{n=1}^{\infty } \left [3 - \frac {1}{n}, 5 + \frac {1}{n}\right ]$

The intersection would be all real numbers that are in every one of those nested closed intervals.

$\bigcap _{n=1}^{\infty } \left [3 - \frac {1}{n}, 5 + \frac {1}{n}\right ] = \left [\answer {3}, \answer {5}\right ]$

An infinite intersection of nested closed intervals is again a closed interval.

Nested

$\bigcap _{n=1}^{\infty } \left [4 - \frac {1}{n}, 4 + \frac {1}{n}\right ]$

The intersection would be all real numbers that are in every one of those nested closed intervals.

$\bigcap _{n=1}^{\infty } \left [4 - \frac {1}{n}, 4 + \frac {1}{n}\right ] = [4, 4] = \left \{ \answer {4} \right \}$

An infinite intersection of nested closed intervals is again a closed interval, which might just contain a single number.

Infinite intersections of open intervals are not neccessarily open, because open intervals are weird.

Nested

$\bigcap _{n=1}^{\infty } \left (3 - \frac {1}{n}, 5 + \frac {1}{n}\right )$

The intersection would be all real numbers that are in every one of those nested open intervals.

$\bigcap _{n=1}^{\infty } \left (3 - \frac {1}{n}, 5 + \frac {1}{n}\right ) = [3, 5]$

An infinite intersection of open intervals can be a closed interval, because open intervals are weird.

It could even be a singleton set.

nested

$\bigcap _{n=1}^{\infty } \left (4 - \frac {1}{n}, 4 + \frac {1}{n}\right )$

The intersection would be all real numbers that are in every one of those nested open intervals.

$\bigcap _{n=1}^{\infty } \left (4 - \frac {1}{n}, 4 + \frac {1}{n}\right ) = [4, 4] = \{ 4 \}$

It could even be the empty set.

nested

$\bigcap _{n=1}^{\infty } \left (0, \frac {1}{n}\right )$

The intersection would be all real numbers that are in every one of those nested open intervals.

$0$ is not in any of the nested intervals, so $0$ is not in the intersection.

Suppose $r>0$ is some real number. Then there exists some natural number $N$ , such that $\frac {1}{N} < r$ .

Then $r$ is not in all of the nested intervals inside $\left (0, \frac {1}{N}\right )$ . So, $r$ is not in the intersection.

No real number is in the intersection.

$\bigcap _{n=1}^{\infty } \left (0, \frac {1}{n}\right ) = \emptyset$

Open intervals are weird. Closed intervals are nice. They are so nice that even an intersection of an infinite number of nonempty closed intervals has to contain something. It cannot be empty.

Intersection Theorem

Suppose we have a infinitely decreasing, nested, sequence of non-empty closed intervals.

$I_0 \supset I_1 \supset \cdots \supset I_{k} \supset I_{k+1} \cdots$

Then

$\bigcap _{k=0}^{\infty } I_{k} \ne \emptyset$

This turns out to be extremely important and useful as we will see in Calculus.

And, the reason this is true is because open intervals must contain space.

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