Let A⊆R.

A function f: A→A is called an embedding of A.

A function f: A→A is called a compression of A if (∀a∈A)[f(a)≤a].

A function f: A→A is decreasing if (∀x∈A)(∀y∈A)[x < y ⇒ f(x)≥f(y)].

(1)  There are 27 embeddings f:{1,2,3} → {1,2,3}.  Write them all down.  (Hint:  Find a way to representeach embedding using a three digit number.)

(2)  Make a conjecture about the number of embeddings of the set {1,2,3, . . . ,2020} has. Prove your conjecture.

(3)  The set A={1,2,3} has 6 compressions.  Indicate them on your list from part 1.

(4)  Make a conjecture about the number of compressions the set {1,2,3, . . . ,2020} has. Prove your conjecture.

(5)  The set A={1,2,3} has 10 decreasing functions.  Indicate them on your list from part 1.

(6)  Find a counterexample to the claim that every compression on {1,2,3} is a decreasing function.

(7)  Find a counterexample to the claim that every decreasing function on {1,2,3} is a compression.
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(1) 27 embeddings:

{1,2,3}→

{(1,1,1),(1,1,2),(1,1,3),(2,1,1),(2,1,2),(2,1,3),(3,1,1),(3,1,2),(3,1,3),

((1,2,1),(1,2,2),(1,2,3),(2,2,1),(2,2,2),(2,2,3),(3,2,1),(3,2,2),(3,2,3),

(1,3,1),(1,3,2),(1,3,3),(2,3,1),(2,3,2),(2,3,3),(3,3,1),(3,3,2),(3,3,3)}

(2) Number of embeddings=nⁿ where n is the set size=2020.

Each element of the set can map to itself and to each of the other elements (n mappings for each element) and the mapping of each of the n elements is independent of the others, giving nⁿ combinations in total.

(3) Compressions (f(a)≤a): 

f(1)=1⇒f(a)=a, f(2)=1⇒f(a)<a, f(3)=1⇒f(a)<a

f(1)=2⇒f(a)>a, f(2)=2⇒f(a)=a, f(3)=2⇒f(a)<a

f(1)=3⇒f(a)>a, f(2)=3⇒f(a)>a, f(3)=3⇒f(a)=a

If we exclude all cases where f(a)>a we are left with:

f(1)=1, f(2)=1, f(3)=1, f(2)=2, f(3)=2, f(3)=3 (6 functions)

(4) For {1,2,...,2020}, f(1)=1, f(1)<2,...2020 (2020 for f(1)); f(2)=2, f(2)<3,... (2019 for f(2)); ...; f(2020)=2020 (1 for f(2020)). The sum of natural numbers between 1 and 2020 is 2020×2021/2=2041210 compressions.

(5) Consider the cases where x<y:

Case A: (x,y)=(1,2)

Case B: (x,y)=(1,3)

Case C: (x,y)=(2,3)

Implications of f(x)≥f(y):

Case A: f(1)=1⇒f(2)=1; f(1)=2⇒f(2)=1 or 2; f(1)=3⇒f(2)=1, 2 or 3

Case B: f(1)=1⇒f(3)=1; f(1)=2⇒f(3)=1 or 2; f(1)=3⇒f(3)=1, 2 or 3

Case C: f(2)=1⇒f(3)=1; f(2)=2⇒f(3)=1 or 2; f(2)=3⇒f(3)=1, 2 or 3

We now need to combine these cases:

f(1)=1f(2)=1f(3)=1 (1 function)

f(1)=2[f(2)=1f(3)=1] or [f(2)=2f(3)=1 or 2] (3 functions)

f(1)=3[f(2)=1f(3)=1] or [f(2)=2f(3)=1 or 2] or [f(2)=3f(3)=1, 2 or 3] (6 functions). This gives us 10 decreasing functions in all.

(6) Where we have one of the 6 functions in (3) not included in the functions in (5), we have a counterexample: f(1)=1 and f(3)=3 (because, if x=1 and y=3, x<y but f(x)<f(y)—not decreasing).

(7) Where we have one of the 10 functions in (5) not included in (3), we have a counterexample: f(1)=f(2)=f(3)=3 (because f(a)=a only when a=3, but f(a)>a otherwise—so not compression).

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