Closeness

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relative distance

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.

“Close” is a realtive term. It doesn’t have an exact value. It depends on the context and the other measurements involved. Therefore, we cannot really talk about close as a single concept. It is a moving target. How are we going to describe a moving target algebraically?

We need moving algebra and it must take into account any and all levels of closeness.

In our case, we are concerned with a single domain number where the function suddenly has a different value than we were expecting from the surrounding values. We need our closeness to go right up against the domain number. We need a distance of $0$ . Except, the distance can’t really be $0$ , because that would be the number itself and not surrounding numbers.

$\blacktriangleright$ How do you talk about surrounding numbers within a distance of $0$ , except not $0$ ?

Let’s contemplate $0$ for a moment.

Fact: $0$ is the only nonnegative number that is less than every positive number.

This is a way to talk about $0$ by only talking about positive numbers.

If $r$ is a nonnegative real number and $r < x$ , where $x$ is any positive real number, then $r = 0$ .

The kind of “close” we are interested in is the kind that works for any positive real number.

In this way, we can begin to bring “close” under the Algebra umbrella and still be a moving target.

Let $f$ be a function. Let $d$ be a domain number.

Let $\delta$ be any small positive number.

Now consider ALL of the intervals described by $(d - \delta , d + \delta )$ .

There are an infinite number of these intervals.
They are all open intervals.
They all provide space around $d$ .
They get as small as anyone might wish for their chosen level of “closeness”.

These $\delta$ -intervals handle all of the closeness situations. If someone wants a particular closeness, then there is a $\delta$ -interval that matches their closeness.

$\blacktriangleright$ More Importantly: If we are not sure of the level of closeness, then we can talk about ALL of the $\delta$ -intervals and have ALL of the bases covered at once.

$\delta$ -Intervals

Given two real numbers, $a$ and $b$ , you can measure the distance between them, symbolized by $|b-a|$ . We use this measurement to decide if things are close. The smaller $|b-a|$ , the closer $a$ and $b$ are.

Or, we could think the other way. Given a real number $a$ and a distance $\delta$ , we could describe all of the real numbers closer to $a$ than a distance of $\delta$ . They would be the numbers inside the interval $(a-\delta , a+\delta )$ , which we could describe with absolute value.

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

Note: $\delta$ is a very popular symbol to represent small distances. So, is $\epsilon$ , which is different than set membership, $\in$ .

And, since $(a-\delta , a+\delta )$ is an open interval, we know there is a number in here, besides $a$ . In fact we know there is a number in the left half, $(a-\delta , a)$ and we know there is a number in the right half, $(a, a+\delta )$ .

And, it doesn’t matter how small $\delta$ is. There will always be a number in the left half and in the right half, because we have an open interval.

In an open interval, there is always filled space “around”. No matter what positive number $\delta$ represents, the interval $(a-\delta , a+\delta )$ surrounds $a$ . There is filled space on both sides of $a$ . This is important for comparing our expectations and actual function values.

This is not true for closed intervals.

For example, consider the number $5$ and the interval $[5, 7]$ . $5$ doesn’t have space “around” it in this interval. Of course, we could choose the interval $[4,7]$ , which has space “around” $5$ . However, there would now be the open interval $(4.5, 7)$ to use. So, why take the chance on a closed interval? If we are interested in our analysis working all of the time, then let’s just stick with open intervals. They must provide space “around” all of the numbers they contain.

$\blacktriangleright$ The situation we are contemplating is our expectations from patterns “around” a domain number not matching up with the actual function at the domain number.

Therefore, we need some language to help us jump back and forth between the domain and range or function values.

Image

We could picture $(a-\delta , a+\delta )$ inside the domain of a function, $f$ , and then investigate the image of this interval under $f$ .

Image

Let $f$ be a function with domain $Dom_f$ and range $Ran_f$ .
Let $T \subset Dom_f$ be a subset of the domain of $f$ . Then the image of $T$ is given by

$f(T) = \{ f(a) \, | \, a \in T \}$

The image of $T$ is the collection of all of the function values from the domain numbers inside $T$ .

Image

Let $f(x) = 2x + 5$ with its natural domain.

Let $I = (4-\delta , 4+\delta )$ , where $\delta > 0$ is a very small real number.

Then $I$ is an interval in the domain of $f$ .

Its image is $f(I) = \{ f(a) \, | \, a \in I \} = \{ 2a+5 \, | \, a \in (4-\delta , 4+\delta ) \}$

$f$ will stretch the interval by a factor of $2$ and shift it by $5$ .

$(2(4-\delta )+5, 2(4+\delta )+5)$

$(13-2\delta , 13+2\delta )$

What if we went the other way and started with a target in the range?

$\blacktriangleright$ Actually, a range target is often too difficult to hit EXACTLY. Instead, let’s just try to land inside the target interval.

What if, in the previous example, we wanted $f(I) \subset (6-\epsilon , 6+\epsilon )$ , where $\epsilon > 0$ is some small distance?

What are the possibilities for $I$ ?

To land near $6$ with $f(x) = 2x + 5 = 6$ , $x$ would need to be near $\frac {1}{2}$ .

$I$ would need to look like $\left ( \frac {1}{2} - \delta , \frac {1}{2}+ \delta \right )$ But this isn’t going to work if $\delta$ is too big compared to $\epsilon$ . We need to describe $\delta$ in terms of $\epsilon$ to make sure. And, we are in the business of making sure.

$\delta$ will be multiplied by $2$ and we need this to be smaller than $\epsilon$ . (Remember: closeness is relative.) Therefore, we need $\delta < \frac {\epsilon }{2}$ for the image to land inside our target interval. Let’s pick $\delta < \frac {\epsilon }{3}$ . That makes sure.

$I = \left ( \frac {1}{2} - \frac {\epsilon }{3}, \frac {1}{2} + \frac {\epsilon }{3} \right )$

When working backwards through the function, like this, we use the word preimage.

Preimage

The image of a domain interval is the set of function values that occur at those domain numbers. The preimage is the reverse.

Given an interval, $I$ , in the range (an interval of function values), the preimage is the set of of domain numbers where the function value is a member of $I$ .

Preimage

Let $f$ be a function with domain $Dom_f$ and range $Ran_f$ .
Let $S \subset Ran_f$ be a subset of the range of $f$ . Then the preimage of $S$ is given by

$f^{-1}(S) = \{ a \in Dom_f \, | \, f(a) \in S \}$

The preimage of $S$ is the set of all domain numbers whose function value is inside $S$ .

Note: The $-1$ exponent means reverse here. The reverse of $f$ . We also use a $-1$ exponent to mean reciprocal (which is the reverse of multiplication). When reading mathematics, we need to take into account the context when interpreting the symbols and notation.

A Second Note: That is NOT a negative one power. It is a signal that we are working backwards.

Note: The range is the collection of all function values. Therefore, every number in $S$ does have a corresponding domain number, possibly many.

Therefore, $f(f^{-1}(S)) = S$ .

Note: The other direction doesn’t work out so well. $f^{-1}(f(T))$ could be a larger set in the domain than $T$ .

Primage of Images can be Bigger

Let $h(y) = (y-3)(y-5)$ with is natural domain.

The set $T = \{ \, 3 \, \}$ is a subset of the domain. Its image is $h(T) = \{ \, h(3) \, \} = \{ \, 0 \, \}$ .

However, if we get the preimage of this we get

$h^{-1}(\{ 0 \}) = \{ \, 3, 5 \,\}$

Which is bigger than the original set with which we began.

Preimage

Let $G(t) = -4t - 3$ with its natural domain. The natural range is $\mathbb {R}$ .

Let $I = (10-\epsilon , 10+\epsilon )$ , where $\epsilon > 0$ is a small positive real number.

Then $I$ is a small neighborhood of $10$ in the range of $G$ .

What is the preimage of $I$ ?

$G$ will stretch a domain interval by a factor of $\answer {4}$ . It will also reflect it and shift it $3$ .

First, we need a domain interval whose image will be around $10$ .

$-4t-3 = \answer {10}$

$-4t = 13$

$t = -\frac {13}{4}$

The preimage domain interval looks like $\left ( -\frac {13}{4} - \delta , -\frac {13}{4} + \delta \right )$ . Rather than just stating that $\delta$ is small, we need to be specific about its smallness compared to $\epsilon$ , by specify $\delta$ in terms of $\epsilon$ . (Remember, closeness is relative.)

$G$ will stretch $\delta$ by a factor of $4$ , therefore, we need $\delta = \frac {\epsilon }{4}$ .

$G^{-1}((10-\epsilon , 10+\epsilon )) = \left ( -\frac {13}{4} - \frac {\epsilon }{4}, -\frac {13}{4} + \frac {\epsilon }{4} \right )$

If we are just interested in landing inside $I$ , rather than covering it exactly, then we could have used something smaller than $\frac {\epsilon }{4}$ .

$G\left ( -\frac {13}{4} - \frac {\epsilon }{5}, -\frac {13}{4} + \frac {\epsilon }{5} \right ) \subseteq (10-\epsilon , 10+\epsilon )$

$G\left ( -\frac {13}{4} - \frac {\epsilon }{11}, -\frac {13}{4} + \frac {\epsilon }{11} \right ) \subseteq (10-\epsilon , 10+\epsilon )$

$G\left ( -\frac {13}{4} - \frac {\epsilon }{32}, -\frac {13}{4} + \frac {\epsilon }{32} \right ) \subseteq (10-\epsilon , 10+\epsilon )$

Inside Preimage

Let $H(k) = 2k - 1$ with its natural domain. The natural range is $\mathbb {R}$ .

Let $I = (-5-\epsilon , -5+\epsilon )$ , where $\epsilon > 0$ is a small positive real number.

Then $I$ is a small neighborhood of $-5$ in the range of $H$ .

Identify an interval, $D$ in the domain of $H$ , such that $H(D) \subset I$ .

$H$ will stretch a domain interval by a factor of $\answer {2}$ . It will also shift it $1$ .

First, we need a domain interval whose image will be around $-5$ .

$2k - 1 = \answer {-5}$

$2k = -4$

$k = -2$

The preimage domain interval looks like $\left ( -2 - \delta , -2 + \delta \right )$ . Rather than just stating that $\delta$ is small, we need to be specific about its smallness compared to $\epsilon$ , by specify $\delta$ in terms of $\epsilon$ . (Remember, closeness is relative.)

$H$ will stretch $\delta$ by a factor of $\answer {2}$ , therefore, we need $\delta \leq \frac {\epsilon }{2}$ .

Let’s pick $\delta = \frac {\epsilon }{5}$ .

$H \left ( \left ( -2 - \frac {\epsilon }{5}, -2 + \frac {\epsilon }{5} \right ) \right ) = \left ( 2\left (-2 - \frac {\epsilon }{5}\right ) - 1, 2\left (-2 + \frac {\epsilon }{5}\right ) - 1 \right )$

$= \left ( -5 - \frac {2 \epsilon }{5} , -5 + \frac {2 \epsilon }{5} \right ) \subset \left ( -5 - \frac {2 \epsilon }{2} , -5 + \frac {2 \epsilon }{2} \right ) = (-5-\epsilon , -5+\epsilon )$

Endpoints

If our domain contains intervals with endpoints, then we have a bit of a technicality.

Let $B$ be a function with domain $(-9, -7] \cup (-2, 1) \cup [6, 10)$

Then there is no open interval inside the domain surrounding $6$ . This is because the numbers immediately less than $6$ are not included in the domain.

We may need special attention to endpoints.

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