Question

note for up here in the question e is the belongs to symbol
Suppose A and B are sets, and let F = {f l f : A => B}. Also suppose R is a relation on B, and define a relation S on F as follows:
S = { (f,g) e F x F l for all x e A((f(x),g(x)) e R}

Answer 1

a) if R is reflexive, must it be the case that S is reflexive?

Answer 2

\[S = \left\{ (f,g) \in F \times F l \forall x \in A((f(x),g(x))) \in R \right\}\]

Answer 3

the l there is the such that line

Answer 4

dam i hated this doing stuff like this when i was taking the class

i didnt fully understood it but i give you some ideas

F is defined to be a functions that takes in elts of set to and spits out elts of set B

R is defined to be a relation on B implies R is a subset of BxB

a relation S on F implies S is a subset of FxF

Answer 5

F is actually the set of all functions that go from A to B

Answer 6

reads as the set of all f such that f is a function from A to B

Answer 7

ah yea sorry it takes the set of functions from A to B

Answer 8

so we know R is reflexive then that means for some a in R, aRa

Answer 9

actually my textbook defines it as
\[\forall a \in A(aRa)\]

Answer 10

hm.. i guess i do it differently when i took the class

what does A mean?

Answer 11

o in that definition it is just the set relation R was on.

Answer 12

we'll just go with your definition then

since we know its reflexsive and its a relation on B then a is also a subset of BxB

Answer 13

wait a minute i believe it should be like this:

\[\forall a \in B(aRa) \in R \]
sine R is a relation on B and not A

so \[a \in B x B\]
so B is reflexive since R is a relation on B

Answer 14

B is a set though not a relation....

Answer 15

R is a relation on B so for some a in R then aRa implies a in an elt of BxB

Answer 16

it implies (a,a) is an element of B yes

Answer 17

yea my notation is prob off since we did different notation from you

the part im really stuck on is after knowing (a,a) is in B
how to show it is in the entire F

Answer 18

actually I think this statement is false

Answer 19

why do you say that?

Answer 20

cause in for for S to be reflexive I need to have that
\[\forall f \in F( (f,f) \in S)\]

Answer 21

i see what you are trying to say

if so then we can find a counter example to this problem where f in F and not in S

Answer 22

yes

Answer 23

let try with simple functions first

f(x) = x
g(x) = x^2

Answer 24

so these 2 functions are in S and they are not reflexive

Answer 25

You can't tell that unless you define A, B an R first

Answer 26

actually I think I just proved it to be true...

Answer 27

hm.. i couldnt get any counter example since i kept getting both R and S were reflexive

Answer 28

what did you write to make you think its true?

Answer 29

i wanted to say that since (a,a) is in B then (a,a) is also in A but i wasnt sure if i was allowed to do that then (a,b) exist in F so F is reflexive and since S is a relation on F the S must my reflexive

Answer 30

(a,a) in F*

Answer 31

Let h be an arbitrary function that belongs F then (h, h) belongs F cross with F. Since (h, h) belongs to F cross with F, it belongs to S if for x that belong to A, (h(x),h(x)) belongs to R Since h belongs to F we know it is a function from A to B. Thus h(x) belongs to B. Since R is reflexive we know that for b that belong to B (b,b) belongs to R. Since for all b that belong to B (b,b) belongs to R and h(x) belongs to B we have that (h(x),h(x)) belongs to R. Thus
h,h) belongs to S. Therefore S is reflexive since h was arbitrary

Answer 32

I think this proof is okay.. but im not too sure about this part:

Since for all b that belong to B (b,b) belongs to R and h(x) belongs to B we have that (h(x),h(x)) belongs to R.

Answer 33

think of it this way if all balls are red and I have a ball then my ball has to be red

Answer 34

yea it could work actually
its just that i usually see proofs proving it in one direction or the other directions thats all

Answer 35

that is proving it in one direction

Answer 36

actually yea i think its fine

Answer 37

you should probably do a bit more explanation on that sentence i mentioned

Answer 38

other than that i think its fine. I gotta go now some maybe someone else can take a look at it and see if its good or not. Gnight