Question

let A = power set of ℝ. Define f: ℝ->A by formula f(x) = {y∈ℝ | y^2 < x}.

Is one-to-one? is it onto?

Answer 1

suppose f(x1) = f(x2). I'm trying to determine whether x1 = x2. So far I got,
f(x1) = {y∈ℝ | y^2 < x1}
f(x2) = {y∈ℝ | y^2 < x2}

the fact that f(x) isn't a real number throws me off, thus, I don't really know how to solve for x1 and x2 and show they are (or are not) equal

Answer 2

Well, \(f(x)\) is a just a set of real numbers.

Answer 3

right, f(x) is a set, which isn't a number

Answer 4

Consider \(x=0\) and consider \(x=-1\). That will give away the answer.

Answer 5

x = 0 then f(0) is a set of y that y^2 < 0, which is a empty set.
x = 1 gives -1 < y < 1

Answer 6

ok?

Answer 7

I said \(x=-1\) not \(x=1\).

Answer 8

oh... f(-1) also gives an empty set.

Ah so f isn't one-to-one

Answer 9

The onto part should be pretty obvious as well.

Answer 10

f is onto right?

Answer 11

Think about it for a few seconds.

Answer 12

Consider a function: \[
g(x) = \frac{1}{x-3}
\]Would it's domain be in \(A\)? If it is, then \(f(x)\) has to map to it.

Answer 13

its^

Answer 14

{3} isn't in A

Answer 15

First of all, the domain would be \(\mathbb R \setminus \{3\}\).

Answer 16

Is \(\mathbb R \setminus \{3\}\subseteq \mathbb R\)?

Answer 17

yeah

Answer 18

For that matter is \(\{3\}\subseteq \mathbb R\)?

Answer 19

no

Answer 20

oh wait yeah

Answer 21

How are you defining the power set?

Answer 22

set of all subsets

Answer 23

Okay, so why would you think it isn't in \(A\)?

Answer 24

You said \(A = \mathcal P(\mathbb R)\)

Answer 25

oh {3} is indeed in A

Answer 26

Does \(f(x)\) ever map to \(\{3\}\)?

Answer 27

huhm.. so I need to find an x such that f(x) = {3}, which is impossible. So f is not onto

Answer 28

I see now. Thank you!

Answer 29

Can you prove that \(\forall x\quad \{3\}\notin f(x)\)?

Answer 30

huhm.... suppose x in R, then -sqrt(x) < y < sqrt(x)

Answer 31

What?

Answer 32

i just solved for y

Answer 33

ok

Answer 34

The way I see it is given, x > 0, f(x) will be a set that has an equal length, and {3} doesn't have a length. So, x > 0 works.

Now I need to consider x <= 0, which will gives an empty set

Answer 35

I see where you are going.
Another idea: \(3\in f(x) \implies 3^2<x \implies 2^2<x \implies 2\in f(x)\).
So \(\{3\}\notin f(x)\) since \(2\notin \{3\}\).

Answer 36

you meant to say {3} ∈ ℝ right?

Answer 37

wait wait typo

Answer 38

you meant to say {3} ∈ f(x) ?

Answer 39

I mean to say \(\{3\}\notin \text{Im} [{ f(x)}]\)

Answer 40

how does {3} ∈ f(x) => 3^2 < x?

Answer 41

No, it is \(3\in f(x)\).

Answer 42

It returns a set of real numbers, not a set of sets

Answer 43

but f(x) is a set...

Answer 44

Yes, \(f(x)\) is a set. I'm saying suppose that \(3\) is in that set.

Answer 45

ohh.... :o

Answer 46

If \(3\) is in the set, then \(3^2<x\). We know that \(2^2<x\) based on this fact.
This means \(2\) is also in the set.

Answer 47

I see

Answer 48

Well, your proof that \(0\) must be in \(f(x)\) or \(f(x)\) must be empty also works.

Answer 49

you mean the proof I gave that f(x) is either empty or a set that has a length?

Answer 50

Whoops, I didn't read it thoroughly, I thought you were going to say something like either it is empty or \(0\) is in it.

Answer 51

ok :D

Answer 52

well, I did say that -sqrt(x) < y < sqrt(x), so yeah either f(x) is empty or 0 is in it. I didn't thought of it this way when I said it. All I did was merely solving for y. But now I see the connection to what I did

Answer 53

once again, thank you for your time ^.^