Function examples

Question

I have the following: Give example of sets F, G, R, S so that: F $\subseteq $RxS and G={(y,x): (x,y) $\in$ F) and :

1. F and G are not Functions.
2. F is a Function but G is not.
3. F is not a Function but G is a Function.
4. F and G are both Functions.

First of all G={(y,x): (x,y) $\in$ F) really confuses me. What does it mean exactly?

Secondly could you help with

Answer 1

Okay:

G={(y,x): (x,y) ∈ F) what does it mean?

F, G, R, S are all sets. R is going to have a bunch of elements, say {r, s, t}. S is going to have a bunch of elements, say {a, b}

Then the notation R x S is a set whose elements are all ordered pairs where the first term of the pair is in R and the second term is in S. We write this as RxS = {(x, y)| x in S; y in R}. We read that as "RxS is the set of pairs (x, y) where x is in S and y is in R. And it means the set:

{(r, a) (r,b) (s, a) (s, b) (t, a) (t, b) }

That's all the pairs where first term is in R and the second term is in S. Of course, this is only for my choice of R and S. R and S can be anything we want. For example, if R is the real numbers and S is also the real numbers RxS is the two dimensional Euclidean plane of all ordered pairs of numbers.

So F is a subset of RxS. Say just the pairs (r, a) and (s, b) and (s, a).

So what does G={(y,x): (x,y) ∈ F) mean?

G is the set of ordered pairs (y, x) where (x, y) are in F.

In other words, G is all the pairs of F but with the orders reversed.

So in my case, as F = {(r, a) (s,b) (s,a)} then G = {(a, r) (b, s) (a, s)}.

=====

So let's get to the problem:

F and G are not Functions.
F is a Function but G is not.
F is not a Function but G is a Function.
F and G are both Functions.

A function in this case means a set of pairs with the condition that for any first term there is only one pair with that first term. You can't have more than one pair with a specific first term. In my example:F = {(r, a) (s,b) (s,a)} is not a function because s is the first term of two pairs (s, b) and (s, a). A function can only have one pair starting with an s and F has two. So it's not a function.

[Note: the restriction is only on the first term. There can be multiple occurrences of the second term. Example: {(1, a) (2, b) (3, a) (4, c)} is a function. The first terms are all different. The second terms are not but that doesn't matter.]

[We often think of functions as "rules" that turn one term into another. For example f(x) = x + 5. This is actually a set of pairing where each input is paired with an out put. So the function f(x) can be thought of as {(0, 5) (1, 6) (2, 7) .....}. It's the set of all pairs (x, x+5) where the first term is the "input" and the second term is the "output". As each input has only one output, each first term must be different. That's why we have the condition."]

Answer 2

Essential here is the definition of function. A function $F:R\rightarrow S$ is actually a subset of $R\times S$ that satisfies the following condition:

$$\text{ for every }x\in R\text{ there is exactly one }y\in S\text{ such that } \langle x,y\rangle\in F$$

Now let $R$ and $S$ both be sets with more than one element.

1) Can you find a subset of $R\times S$ that does not satisfy this condition?

How about $R\times S$ itself?

And if $F=R\times S$ then what can be said about $G:=\{\langle y,x\rangle\mid \langle x,y\rangle\in F\}$?

2) How does $G$ look like if $F$ is a constant function?

3) Think of 2) again and start with $G$ being a constant function.

4) I leave that to you.