Question

If f:S--> T is a function from S into T, show that the following are equivalent
a) f is one-to-one
b) For every y in T, the set \(f^{-1({y})\) contains at most one point.
c) \(f(D1\cap D2)= f(D1)\cap f(D2)\) for all subsets D1, D2 of S
Please, help

Answer 1

@perl

Answer 2

@IMStuck

Answer 3

is it all equivalent or choose 2 from the 3 ?

Answer 4

each is equivalent to other two

Answer 5

so prove
a -> b , b - > c , and c - > a
that will show statements are equivalent

Answer 6

It shows me nothing @perl

Answer 7

Did you got the question you posted yesterday solved?
i looked similar to this no?

Answer 8

it*

Answer 9

No, this is the other one. I didn't solve the yesterday problem yet. I have a bunch of homework to to. I just need a hint or an outline on the problems I think I get stuck. I will go over all of them later.

Answer 10

I'm not good at this
but i think to show that \(f(D1~ intersect ~ D2)=f(D1)~intersect~f(D2)\)
we must show that an element y belongs to both

Answer 11

But we don't have to prove it, all we need is proving if f is one to one (1) , then (3)
not prove (3) itself.

Answer 12

how would you prove that it is one to one?
kind of difficult i see that proving c is much better to start with?

Answer 13

oh, I got what you mean

Answer 14

We don't prove it is one to one.
We prove: "if f is one to one, then f(D1 and D2)= f(D1 and f(D2)

Answer 15

and how would justify that?
we don't know if f in one to one
see we need to work the converse?

Answer 16

Ok, let finish it today (hihihi, I am lazy!!)
f(D1 and D2) = {x such that image of x is in (D1 and D2)}

Answer 17

yeah! that's the definition
don't know how to proceed further lol.
what is this course?

Answer 18

Advanced calculus 1

Answer 19

I still have abstract algebra and probability and statistics reference course. hehehhe.. I am ready to die.

Answer 20

hehe

Answer 21

this looks to me more like abstract algebra
were are dealing with

Answer 22

@perl, at least, you give me step2, right?

Answer 23

we are*

Answer 24

which one are you proving ?

Answer 25

hihihi Actually, the problem is quite obvious, so that the hardest part is how to present it professionally.

Answer 26

what is that part 2? f inverse?

Answer 27

@xapproachesinfinity are you good at abstract?? hhehehe, let me become your fan, I will nag you when I get stuck

Answer 28

no man, not good at it hehe.
i'm just sticking my noise in this stuff! i'm just become a calculus man hehe

Answer 29

for part2) we have definition of inverse function which says : f^- = {x in S such that f(x) in T} and f is one to one, which leads us to f(x) is unique for any x in s

Answer 30

that is not one to one @perl

Answer 31

it is still 1 though?
why are you taking onto case?

Answer 32

i know, im trying to satisfy the hypotheses

Answer 33

That is not onto, either @xapproachesinfinity

Answer 34

oh yeah my bad! i thought 2 was linked

Answer 35

so knowing that f(D1 and D2)=f(D1)andf(D2)
doesn't necessarily mean it is one to one
as perl showed?

Answer 36

Hey, you guys are testing me?? I am a diligent student. I am not lazy to just ask for answer without knowing anything. hehehehe

Answer 37

dealing with this stuff means you are pretty good hehe