examples bijections

7/23/2019 Examples Bijections

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Further examples of bijections

These examples are meant to aid you in understanding bijections. One has to identify

first the sets Aand B that we are trying to find a bijection between, then define a map

f : A B, and then show that f is a bijection to conclude |A| = |B|. To show f is

a bijection, one either writes down an inverse for the function f, or one shows in two

steps that (i) f is injective and (ii) fis surjective. If two sets Aand B do not have the

same size, then there exists no bijection between them. It is therefore often convenient

to think of a bijection as a pairing up of the elements ofA with the elements ofB . In

fact, in a general situation, if|A|= |B|= n, there are n! bijections between A and B.

Example. Give a bijection f from the set of subsets of [5] of size two to the set of

subsets of [5] of size three.

Solution. In this example, A is the set of subsets of [5] of size two and B is the set of

subsets of [5] of size three. The subsets of size two and three are listed below:

A 12 13 14 15 23 24 25 34 35 45

B 123 124 125 134 135 145 234 235 245 345

Here we are writing the sets {1, 2}, {1, 3}, . . ., as 12, 13, . . .to make the notation easier.

So we could define the bijection f : A B byf(12) = 123, f(13) = 124, f(14) = 125,

and so on following the table above. Note there are many bijections betweenA and B

many ways of pairing up sets of size two to sets of size three: in fact there are 10!

bijections.

Example. Is it possible to tile a chessboard with dominoes (here each domino covers

two squares of a chessboard)? Is it possible to tile a chessboard with diagonally opposite

squares removed with dominoes?

Solution. The answer is yes to the first question, because we can for example tile each

column with four vertical dominoes in each column. The answer to the second question

is no: if we could tile the chessboard with corners removed with dominoes, then each

domino would cover one white square and one black square. That means there would be

a bijection from the set of white squares to the black squares. But then there should be

as many white squares as black squares, and that is false because the removed corners

had the same color. What if the chessboard was a 9 9 chessboard? Then what would

the answers to the two above questions be?

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Example. Find a bijection from the positive integers N = {1, 2, 3, . . .} to the integers

Z= {. . . , 2, 1, 0, 1, 2, . . .}.

Solution. In this example, A= Nand B = Z. Define f : N Zbyf(x) = (1)xx2

,

wherey

denotes the largest integer less than or equal toy

(i.e. roundy

down). Thenf(1) = 0, f(2) =1, f(3) = 1, f(4) = 2, f(5) = 2, and so on. Now f is a bijection,

since iff(x) =f(y), then (1)x = x2

= (1)yy2

, and so (1)x = (1)y which means

x and y must differ by an even number. But then we needx2

= y2

, and so the only

possibility is x= y . So fis injective. To see that f is surjective, given an integer z, we

must find an x N such that f(x) =z. Well ifzis positive, then f(2z) =z, and ifz is

negative then f(2|z|+ 1) =z, as required.

Example. Prove that ifn is odd, there are as many subsets of [n] of size more than n2

as subsets of [n] of size less than n2 .

Solution. Let A be the set of subsets of [n] of size more than n2

and let B be the set of

subsets of [n] of size less than n2

. For each set a A, define f(a) = [n]\a, i.e. fmaps a

set a to the complement ofa. Then fis its own inverse, and ifa A and f(a) B , so

f :A B is a bijection.

The next example is slightly harder, since we find a bijection from a set of sequences to

a set of subsets.

Example. Prove that the number of sequences (x1, x2, . . . , xk) of lengthk wherexi [n]

and xi > xi1+ 1 for i= 2, 3, . . . , kequalsnk+1

k

.

Solution. We define a bijection f from A, the set of given sequences, to B, the set of

subsets of size k in [nk+ 1]. Given (x1, x2, . . . , xk) A, define

f(x1, x2, . . . , xk) ={x1, x21, x32, . . . , xkk+ 1}.

This maps a sequence in A to a set inB . Note that the set {x1, x2 1, . . . , xk k + 1}is

a subset of [n k + 1] becausexk

n and because no two of the elementsx1, x2 1, x3

2, . . . , xk k + 1 are equal by the condition xi > xi1+ 1. For instance ifk = 3 then

the sequence (1, 3, 5) gets mapped to {1, 2, 3} and the sequence (1, 4, 6) gets mapped

to {1, 3, 4}. We now have to check that f is a bijection. Suppose (x1, x2, . . . , xk) and

(y1, y2, . . . , yk) are two sequences in A. Iff(x1, x2, . . . , xk) =f(y1, y2, . . . , yk), then

{x1, x21, x32, . . . , xk k+ 1}= {y1, y21, y32, . . . , ykk+ 1}

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and this means x1 = y1, x2 1 = y2 1, x3 2 = y3 2, and so on. Therefore

x1 = y1, x2 = y2, . . . , xk = yk which means (x1, x2, . . . , xk) = (y1, y2, . . . , yk) i.e. f is

injective. To check that f is surjective, for each set {b1, b2, . . . , bk} B where b1 bi1+ i1 = xi1+ 1. So we have checkedf is

surjective.

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