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Math 127: Functions

Mary Radclie

1 Basics

We begin this discussion of functions with the basic denitions needed to talk about functions. Denition 1.LetXandYbe sets. AfunctionffromXtoYis an object that, for each elementx2X, assigns an elementy2Y. We use the notationf:X!Yto denote a function as described. We write f(x) =yorf:x7!yto denote that the element inYassigned toxisy. We callXthedomainoff, and we callYthecodomainoff. Iff(x) =y, we say thatxmaps toyunderf.

In general, we will often talk about functions from this perspective of \mapping;" we see the role of

a function as taking things fromX, and sending them over toY, where the functionfis the map that

explains where each element inXis supposed to go. This is visualized in Figure 1.; fis represented by the red arrows here, so that ifx2Xis assigned y2Y, we have a red arrow fromxtoy. Notice that it is permitted to have more than one element inXmap to the samey2Y, and it is also permitted to have elements inythat are not mapped to by anyx2X.

Let's think about this notion with a function we probably feel a little more comfortable with. Consider

the functionf:Z!Zdened byf(n) = 2n, as shown in Figure 2. Practically, what this function is

asking us to do is map elements inZover to new elements inZ, where the directions on the map, described

by the function, are to multiply something by 2. So if you ask the function \what do I do with the number

1

3?" the function will answer \Send it over to 6." In this way, the function provides us with a mechanism

to transform the number 3 according to some rule, which is the denition of the function. Figure 2: A representation of the functionf:Z!Zdened byf(n) = 2n.

Now, in order for this map to make sense, there are a few basic properties that we need to ensure are

satised. In particular, we want to make sure that every element ofXcan be mapped to somewhere, and that the \somewhere" is appropriate. Together these properties constitute what is known as well- denedness. Denition 2.An assignment of valuesyto elementsx2Xis said to be awell-denedfunctionf:X!Y if it satises the following three properties: Totality: For everyx2X, there existsysuch thatf(x) =y.

Existence: For everyx2X,f(x)2Y.

Uniqueness: For everyx2X, there is only oney2Ysuch thatf(x) =y.

While these properties are trivially true in the denition of a function, it's useful to think about the

failure cases to think about what kinds of assignments we might make that are in fact NOT functions. In

general, if you are to dene a function yourself, it's worth thinking about these things to ensure that the

function you are creating is a well-dened function.Example 1.Deneg:R!Rbyg(x) =1x1. This is NOT a well-dened function, becauseg(1)

is not an element ofR. There is noysuch thatg(1) =y, and hence this function fails the totality property.Example 2.Denef:N!Nbyf(n) =n1. This is NOT a well-dened function, because f(1) = 0 is not an element ofN. This assignment of values fails the existence property.2 Example 3.Denef: (0;1)! f0;1;2;:::;9g(where (0;1) =fx2Rj0< x <1g) byf(x) =d, wheredis the rst digit after the decimal point in the base-10 expansion forx. This is NOT a well-dened function, because some numbers have more than one possible base-10 expansion.

Indeed

a, 0:5 = 0:4999999:::, and hence we have more than oney2 f0;1;:::;9gwithf(12 ) =y. This assignment of values fails the uniqueness property.a

Just to be sure, let's prove that 0:9999= 1, which in turn clearly implies that 0:4999= 0:5. First, leta=

0:9999:::, which is some real number. Then 10a= 9:9999:::, and hence 9a= 10aa= 9:99990:9999= 9.

Since 9a= 9, we therefore have thata= 1. Thus, 1 =a= 0:99999:QED.When you are dening a function, be on the lookout for these kinds of problems. Some things to think

about: If you were handed a value ofx2X, would you always know what to do with it? If you're using a formula: can EVERY value ofx2Xbe plugged in to your formula? If not, what are you doing with the other values? Can the formula ever give you numbers that live outside your codomain? Is there more than one way to represent elements in the domain? If so, do those representations give the same output under this function?

As with several of our previous topics, we have some special functions that will show up from time to

time, and will be quite useful. Denition 3.Theidentity functionon a setX, denoted byX, is the functionX:X!Xsuch that

8x2X; f(x) =x.

Theempty functionis any functionf:; !X. Note that there is no need, in the empty function, to dene any values for elements in the domain, as there are none! Finally, we have to address the question of what it means for two functions to be the same. Denition 4.LetX;Y;A;Bbe sets, and letf:X!Yandg:A!Bbe functions. We say thatfis identically equal tog, denoted byfg, if the following conditions are met: X=A Y=B

8 x2X,f(x) =g(x).

There is a reason for distinguishing equality of functions withrather than =. There are often times, for example, when it is important to nd a specic choice ofx2Xfor whichf(x) =g(x). However, this is quite dierent fromf(x) =g(x) for every singlex2X. Moreover, we can have functions that are the same on all their common domain elements (that is, f(x) =g(x)8x2X\A), but are not identically equal because perhaps they have dierent domains. An example might be as follows: denef:N!Nbyf(x) = 3x, and forAthe set of even integers, dene g:A!Zbyg(a) = 3a. Then on anyxthat is both a natural number and an even integer,fandg

produce the same value. However, we cannot say that these are really the same function, because there

are numbers thatfcan map andgcannot, and vice versa. Similar qualications, of course, apply to the codomain.

For these reasons, we distinguish between identical equality of functions and equality on elements. This

is similar to the use ofin the world of propositional logic: we use the symbolto mean that the two objects are the same, all the time, regardless of how we set the corresponding parameters. 3

1.1 Functions and Subsets

We have, already, some language and notation to discuss things like the domain or codomain of a function.

However, what if we wish to consider subsets of the domain or codomain? We need to develop concepts to deal with these structures. Denition 5.LetX;Ybe sets, and letf:X!Ybe a function. Given a setUX, we dene the imageofUinY, denotedf(U), to be the subset ofYgiven by f(U) =fy2Yj 9x2Uwithf(x) =yg: IfU=X, the image ofUinYis also called therangeoff. This is to say, if we are interested in a subsetUofX, we might want to ask ourselves: what values in

Yare assigned to elements in this subset? This is precisely what the image ofUis getting at.Example 4.Denef:Z!Nbyf(x) =jxj+1. To illustrate the concept of image, let's consider

a few subsets ofZhere. IfU=f0;1;2g, then the image ofUis whatever function values are assigned to these 3 numbers. As f(0) = 1; f(1) = 2; f(2) = 3; we have thatf(U) =f1;2;3g. Notice that asUhas 3 elements in it, we cannot have more than 3 elements inf(U), due to the uniqueness property of well-dened functions. IfU=f2;1;0;1;2g, then the image ofUis whatever function values are assigned to these ve numbers. Since f(2) = 3; f(1) = 2; f(0) = 1; f(1) = 2; f(2) = 3; we have thatf(U) =f1;2;3g. Notice that despite the fact thatUhas 5 elements in it, we still only have 3 elements inf(U). This is due to the fact that some elements off(U) are assigned to more than one member of U. If we wish to consider the range off, we think aboutU=Z. In this case, we have that every element ofNis a member of the range; ifn2N, then by takingz=n1, we havef(z) =n. Hence, for everyn2N,9z2Zsuch thatf(z) =n, so everyn2Nis a member of the range off. IfU=f0g, then the image ofUis justf(U) =f1g. In general, this is true: for any function f:X!Y, we have thatf(fxg) =ff(x)gfor anyx2X; that is, the image of the set

containing only one elementxis the set containing only one element,f(x).In the case of the function shown in Example 4, we have that the range of the functionfis precisely

the same as the codomain off. This is because for this function, every element in the codomain is the

image of some element in the domain. Of course, this is not true for all functions; if we takeg:Z!Z

dened byg(z) = 2z, we clearly have that the codomain is all ofZ, but the range is just the even integers.

Of course, we may be interested in more than just what the function values are on a subset ofX. Indeed, there are circumstances where we might wish only to consider a subset ofX, and throw away super uous elements. In that case, we can dene a function on a subset ofXas follows. Denition 6.LetX;Ybe sets, and letf:X!Ybe a function. Given a setUX, therestriction of ftoU, denoted asfjU, is the functionfjU:U!Ydened byfjU(x) =f(x) for allx2U. 4 The restriction offto a subset will occasionally appear as a useful tool. Now, for the other side of the coin, suppose we have a subset ofY, and wish to think about what

elements inXmight map to that subset. In that case, we can dene a similar type of object to the image,

called the preimage. Denition 7.LetX;Ybe sets, and letf:X!Ybe a function. Given a subsetVY, we dene the preimageofVunderf, denoted byf1(V), to be the subset ofXgiven by f

1(V) =fx2Xjf(x)2Vg:

This is to say, the preimage ofVis the set of all elements inXwhose image is a member ofV. To

ensure we understand the concept, let's consider an example.Example 5.Letf:Z!Zwithf(z) =j2zjfor allz2Z. Let's consider the preimage of a few

setsV. IfV=f2g, then the preimage ofVis all those elements inZthat map to 2; that is, it is all choices ofzfor whichf(z) =j2zj= 2. There are two such elements, namely1. Hence f

1(V) =f1;1g.

IfV=f1g, then the preimage ofVis all those elements inZthat map to 1; that is, it is all choices ofzfor whichf(z) =j2zj= 1. There are no such elements! Hence,f1(V) =;. IfV=f0g, then the preimage ofVis all those elements ofZthat map to 0, which is clearly justz= 0. Hencef1(V) =f0g. IfV=f0;1;2;3;4g, then by repeating above ideas, we have that the elements inZfor which f(z)2Vare exactlyf1(V) =f2;1;0;1;2g. If we takeV=fx2Zjx0 andxis eveng, then every element ofZhas its image inV.

Hence, we havef1(V) =Z.Unlike with the image, as we see above the number of elements in the preimage of a setVis not

necessarily tied to the number of elements inV. We can have empty preimages, which we cannot have with images, and we can have many elements that map to the samey2Vcontribute to the preimage. In general, the preimage and image play well with unions and intersections of sets. Proposition 1.LetX;Ybe sets, and letf:X!Ybe a function. 1.

If U1;U2X, thenf(U1[U2) =f(U1)[f(U2).

2.

If U1;U2X, thenf(U1\U2)f(U1)\f(U2).

3.

If V1;V2Y, thenf1(V1[V2) =f1(V1)[f1(V2).

4.

If V1;V2Y, thenf1(V1\V2) =f1(V1)\f1(V2).

We shall not prove all these properties here; those that we do not prove will be left as an exercise. In

particular, we shall prove items 2 and 3. Proof.[Partial Proof.] LetX;Y;fbe as in the statement of Proposition 1. 2. Let U1;U2X. Suppose thaty2f(U1\U2). Then by denition, there existsx2U1\U2having f(x) =y. Sincex2U1\U2, we thus have thatx2U1andx2U2, soy2f(U1) andy2f(U2). But theny2f(U1)\f(U2), and thusf(U1\U2)f(U1)\f(U2). 5

3.W epro vethe set e qualityb ydouble con tainment.

First, suppose thatx2f1(V1[V2). Then by denition,f(x)2V1[V2, so eitherf(x)2V1orf(x)2 V

2. Wolog, suppose thatf(x)2V1. Then we havex2f1(V1), and hencex2f1(V1)[f1(V2).

For the other direction, suppose thatx2f1(V1)[f1(V2). Then we havex2f1(V1) orx2 f

1(V2); wolog suppose thatx2f1(V1). Then by denition, we havef(x)2V1, and thus

f(x)2V1[V2. But thenx2f1(V1[V2). Therefore, we have thatf1(V1[V2) =f1(V1)[f1(V2) by double containment. In addition, we have similar properties for set complements. Proposition 2.LetX;Ybe sets, and letf:X!Ybe a function. 1.

If UX, thenf(XnU)f(X)nf(U).

2.

If VY, thenf1(YnV) =Xnf1(V)

The proofs of these properties are left as an exercise.

1.2 Composition

Denition 8.LetX;Y;Zbe sets, and letf:X!Yandg:Y!Zbe functions. We dene the compositiongf:X!Zto be the function dened bygf(x) =g(f(x)). We can imagine a composition as follows. The functionfgives us a rule for how to map the setXinto the setY, and the functionggives us a rule for how to map the setYinto the setZ. By chaining these rules together, rst by doing thefrule, then thegrule, we can get fromXall the way toZ. This new

function can disregard the setYcompletely, as shown in Figure 3. The notation for function composition

is set up so that the rst function you perform is closest to thex;gfindicates thatfhas to be performed

rst, because when you putxnext to the function,fis closest to it. (a) (b) Figure 3: To construct a composition of functionsfandg, we can imagine that we have two mappings, as shown in (a). To get the compositegf, we start fromX, and then follow both the red and blue arrows to get toZ, resulting in the purple arrows shown in (b). Once we have these purple arrows established, we can then disregardYand the original functionsf andg, and treatgfas its own function fromXtoZ. 6 To ensure that we understand this, let's take a look at an example. Example 6.Denef:N!Zbyf(x) = 2x, and deneg:Z!Qbyg(x) =x12 . The composition gfis a function fromNtoQ, that looks as follows:gf(x) =g(f(x)) =g(2x) =2x12 =x6 . Once we have this expression ofgf, we can simply treat it as a function fromNtoQ, ignoring anythingquotesdbs_dbs20.pdfusesText_26