1. ## function proof

Challenge problem

Given, the following definition of a function from a set A to a set B
denoted as: :

1)

2) for all , ,there exists a unique ,such that

Prove that the following f:

f={(1,a),(2,b)} is a function from A to B,where :

A = {1,2} and

B = {a,b}

Moderator approved CB

2. Originally Posted by alexandros
Challenge problem

Given, the following definition of a function from a set A to a set B
denoted as: :

1)

2) for all , ,there exists a unique ,such that

Prove that the following f:

f={(1,a),(2,b)} is a function from A to B,where :

A = {1,2} and

B = {a,b}

Moderator approved CB
I don't understand why this is a challenge problem; it's just a simple definition. Usually the formal definition is that f is an ordered triple $\displaystyle (A,B,F)$ with $\displaystyle F \subseteq (A\times B)$ as you defined above.

Anyway it's easy to check that {(1,a),(2,b)}$\displaystyle \subseteq(A\times B)$={(1,a),(1,b),(2,a),(2,b)} and that for x=1, the unique y is a, and for x=2, the unique y is b.

(Fixed wording.)

3. Originally Posted by undefined
I don't understand why this is a challenge problem; it's just a simple definition. Usually the formal definition is that f is an ordered triple $\displaystyle (A,B,F)$ with $\displaystyle F \subseteq (A\times B)$ as you defined above.

Anyway it's easy to check that {(1,a),(2,b)}$\displaystyle \subseteq(A\times B)$={(1,a),(1,b),(2,a),(2,b)} and that for x=1, the unique y is a, and for x=2, the unique y is b.

(Fixed wording.)
I think there is a difference between ""check" and "prove",don't you?

4. Originally Posted by alexandros
I think there is a difference between ""check" and "prove",don't you?
There is no difference in any finite case.
I absolutely agree, I cannot see why this is a challenge problem.

5. If you want me to reveal my proof so you can see the difference ,i can do so.

6. Originally Posted by alexandros
If you want me to reveal my proof so you can see the difference ,i can do so.

Please do. Without putting any shame on you, I think that including questions like this one in the challenge problems section demerit and defile the intention of it.

Tonio

7. proof:

Let :$\displaystyle \vee$ denote "or" ,and

let :$\displaystyle \wedge$ denote "and"

1)Let (x,y)εf =>

$\displaystyle [(x,y)=(1,a)\vee (x,y)=(2,b)]\Longrightarrow [(x,y)=(1,a)\vee (x,y)=(1,b)\vee (x,y) = (2,b)\vee (x,y)=(2,a)]\Longrightarrow (x,y)\ A\times B\Longrightarrow f\subset A\times B$

2) Let xεΑ => χ=1v x=2.

For x=1

put $\displaystyle y = a\Longrightarrow [y = a\vee y = b]\Longrightarrow y\in B$

Also

$\displaystyle (x,y) = (1,a)\Longrightarrow [(x,y) = (1,a)\vee (x,y) = (2,b)]\Longrightarrow (x,y)\in f$

Thus there exists y such that,$\displaystyle y\in B$ and $\displaystyle (x,y)\in f$

So much for the existence part.

Now for the uniqueness part.For that we must prove:

for all ,y, z:$\displaystyle [( y\in B\wedge (x,y)\in f\wedge z\in B\wedge (x,z)\in f)\Longrightarrow y = z]$

So let :

$\displaystyle (x,y)\in f\wedge(x,z)\in f\Longrightarrow[(x,y) = (1,a)\vee (x,y) = (2,b)]\wedge[(x,z) = (1,a)\vee(x,z) = (2,b)]$

BUT $\displaystyle 1\neq 2$ (an assumption needed in the initial hypothesis for f to be a function),thus $\displaystyle 1\neq 2\Longrightarrow 1\neq 2\vee y\neq b\Longrightarrow (1,y)\neq (2,b)\Longrightarrow (x,y)\neq (2,b)$

,since x=1.

In the same way .$\displaystyle (x,z)\neq (2,b)$

Hence (x,y)=(1,a) and (x,z) =(1,a) and so y=z

So:

for x= 1 there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$

in the same way:

for x=2 there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$

Therefor.:

For all ,x:$\displaystyle x\in A$ there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$ .

Thus :$\displaystyle f:A\to B$

8. Originally Posted by alexandros
proof:

Let :$\displaystyle \vee$ denote "or" ,and

let :$\displaystyle \wedge$ denote "and"

1)Let (x,y)εf =>

$\displaystyle [(x,y)=(1,a)\vee (x,y)=(2,b)]\Longrightarrow [(x,y)=(1,a)\vee (x,y)=(1,b)\vee (x,y) = (2,b)\vee (x,y)=(2,a)]\Longrightarrow (x,y)\ A\times B\Longrightarrow f\subset A\times B$

2) Let xεΑ => χ=1v x=2.

For x=1

put $\displaystyle y = a\Longrightarrow [y = a\vee y = b]\Longrightarrow y\in B$

Also

$\displaystyle (x,y) = (1,a)\Longrightarrow [(x,y) = (1,a)\vee (x,y) = (2,b)]\Longrightarrow (x,y)\in f$

Thus there exists y such that,$\displaystyle y\in B$ and $\displaystyle (x,y)\in f$

So much for the existence part.

Now for the uniqueness part.For that we must prove:

for all ,y, z:$\displaystyle [( y\in B\wedge (x,y)\in f\wedge z\in B\wedge (x,z)\in f)\Longrightarrow y = z]$

So let :

$\displaystyle (x,y)\in f\wedge(x,z)\in f\Longrightarrow[(x,y) = (1,a)\vee (x,y) = (2,b)]\wedge[(x,z) = (1,a)\vee(x,z) = (2,b)]$

BUT $\displaystyle 1\neq 2$ (an assumption needed in the initial hypothesis for f to be a function),thus $\displaystyle 1\neq 2\Longrightarrow 1\neq 2\vee y\neq b\Longrightarrow (1,y)\neq (2,b)\Longrightarrow (x,y)\neq (2,b)$

,since x=1.

In the same way .$\displaystyle (x,z)\neq (2,b)$

Hence (x,y)=(1,a) and (x,z) =(1,a) and so y=z

So:

for x= 1 there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$

in the same way:

for x=2 there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$

Therefor.:

For all ,x:$\displaystyle x\in A$ there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$ .

Thus :$\displaystyle f:A\to B$

With all due respect the above is completely absurd and if I may add, even ridiculous: nobody not out of his mind would ever present such a cumbersome, long and ridiculously messy proof of such a basic, easy-to-check and straighforward fact in such a basic case, UNLESS he's been said to do so in order to check his/her command in this or that logic-related or definition-related matters.

Anyway, and as has already been pointed out by some other members of the forum, unless some really interesting, unexpected and/or brilliant insight is produced I cannot understand how this question has been aproved as a challenge one, and even less can I understand how the above horrible, awful proof could possibly be accepted by any teacher/lecturer or professor in any university.

Tonio