# Functions acting on sets

1. Jan 26, 2017

### sa1988

1. The problem statement, all variables and given/known data

ONLY QUESTION 2

2. Relevant equations

3. The attempt at a solution

Not sure what's going on here. I think the issue is in my own flawed understanding of the notation used in sets generally. So the question states:

$f : R \rightarrow R$ such that $f(x) = x^{2}$

My understanding thus far is that the cartesian product of two sets X and Y is:

$X \times Y = \{(x,y) : x\in X, y\in Y\}$

So in the case of $f(x) = x^2$, we have for part a):

a) $f((-1,2)) = (-1,2) \times (-1,2) = \big((-1,-1),(-1,2),(2,-1),(2,2)\big)$

but then part of me wonders if I've got it all wrong and it should really just be $f((-1,2)) = ((1,4))$ ..??

And then for part b:

b) $f((-1,2]) = ...?$

I don't really understand this at all since it has a square bracket which I'm led to believe means it represents a continuous interval of numbers not including that which is on the side of the curled bracket (according to this - interval notation). If that's the case, I don't know how to perform $X \times Y$ in the way I defined above.

And then parts c) and d) we have stuff to do with $f^{-1}$ which is a whole other thing entirely.

(Just to check - am I right in saying that $f^{-1}$ on a set $Y$ is all the elements $x \in X$ such that $f(x) \in Y$ ??)

Or in other words: $f^{-1}(Y) = \{ x \in X : f(x) \in Y \}$ - right?

Hints much appreciated, thanks.

2. Jan 26, 2017

### Staff: Mentor

I thought it to be elements of the Cartesian product, too, at first glance. But it's not.

$(-1,2)$ is the open interval $I_1 := \{x \in \mathbb{R}\,\vert \, -1 < x < 2\}\, ,$
$(-1,2]$ is the half-open interval $I_2:= \{x \in \mathbb{R}\,\vert \, -1 < x \leq 2\}\, ,$
$(-\infty,0)$ is the open interval $I_3 :=\{x \in \mathbb{R}\,\vert \, -\infty < x < 0\}\, ,$ and
$(-\infty,0]$ is the closed interval $I_4:=\{x \in \mathbb{R}\,\vert \, -\infty < x \leq 0\}\, .$
Thus the task is to write $f^\varepsilon(I_i) = \{f(x) \in \mathbb{R}\,\vert \,x \in I_i\}$ with $\varepsilon \in \{\pm 1\}\,$.

I don't know where this ambiguous use of parenthesis came from, but it seems to be wide spread. Personally, I prefer to write them as $I_1=]-1,2[\, , \,I_2=]-1,2]\, , \,I_3=]-\infty,0[\, , \,I_4=]-\infty,0]$ but I'm pretty alone with that. Guess we have to live with this bad habit to write open intervals $(a,b)$.

3. Jan 26, 2017

### Math_QED

No, you are using the wrong definitions. I will give you the definition.

Let $f: A \rightarrow B$ be a function. Let $V \subset A$ Then we define $f(V) = \{f(v) | v \in V \}$

Although you are thinking right here, you should be careful: $f((-1,2)) \neq (1,4)$

Hint: What is $f(0)$?

4. Jan 26, 2017

### sa1988

Great stuff, so that's that cleared up.

I'm still a little fuzzy on what the $f(x) = x^2$ part means though. I think my weakness is in knowing how to 'read' definitions such as yours and the one below.

In words I'm interpreting it as, "A set of all the values resulting from the function f(v), which are derived from all elements v in the set V."

So is this correct?

For $f(x) = x^2$

$f((-1,2)) = (1,4)$
$f((-1,2]) = (1,4]$
$f^{-1}((-\infty,0)) = R_{<0}$
$f^{-1}(-\infty,0]) = R_{\leq 0}$

5. Jan 26, 2017

### Staff: Mentor

Take @Math_QED 's hint! What happened to $f(0)\,$? $0 \in ]-1,2[$ so shouldn't $f(0)$ be an element of $f(]-1,2[)\,$? And for which $x$ is $f(x) = -1\in ]\infty,0[\, ?$

6. Jan 26, 2017

### sa1988

Heh, yes indeed it should! Bloody obvious too, actually.

So I'll modify the answers to:
$f((-1,2) = (0,4)$
$f((-1,2]) = (0,4]$

The answer is essentially a number line representing all the possible outcomes of running a given set (a "number line", in this case) through the $x^2$ function. Unless I'm still missing something somewhere...

7. Jan 26, 2017

### Staff: Mentor

You still left out zero in both intervals!
And remember: $f^{-1}(y) = \{x \in \mathbb{R}\,\vert \,f(x)=y\}$. So how do you get negative $y$?

8. Jan 26, 2017

### sa1988

Damn those brackets and negative signs. Second attempt...

$f((-1,2)) = [0,4)$
$f((-1,2]) = [0,4]$
$f^{-1}((-\infty,0)) = i R_{<0}$
$f^{-1}(-\infty,0]) = i R_{\leq 0}$

9. Jan 26, 2017

### Math_QED

An interesting additional question would be:

what is $f((-1,\frac{1}{2}))$?

10. Jan 26, 2017

### sa1988

$[0,1)$ ?

which is interesting because it infers that $f((-1,\frac{1}{n})) = [0,1)$ $\forall$ $n \in R_+$

if I'm not mistaken?

11. Jan 26, 2017

### Staff: Mentor

Yep.
$i\mathbb{R}$ isn't an option here as $f(x)\, : \, \mathbb{R} \rightarrow \mathbb{R}$. You can only choose among real numbers!

12. Jan 26, 2017

### Staff: Mentor

Yes.
Almost. If you allow $n$ to be positive but arbitrary close to $0$, then you get a lot of big numbers $\frac{1}{n^2}$.

13. Jan 26, 2017

### sa1988

Ah, true. Haha I think I may be doomed. This is the very start of a 3rd year module on topology and I can't even get my head around the basics of working with sets. Funny really because the information theory module I just did was a breeze! It was loosely similar to all this.

In regard to those two which I got wrong, I really can't think how a non-complex real number can be squared with itself to form negative infinity.

14. Jan 26, 2017

### Staff: Mentor

I wonder how you'd get any negative number here. (But don't forget the zero again in the last case.)
As to topology, I think a good advise is: Don't take anything out of intuition for granted. Topology is full of mysterious examples of all kind. E.g. you can fill a square by just a line. However, here it's about sets, even if empty.

15. Jan 26, 2017

### sa1988

I think I've cracked it:

$f^{-1}((-\infty,0)) = \emptyset$
$f^{-1}(-\infty,0]) = (0)$

16. Jan 26, 2017

### Staff: Mentor

Yes, although it might be better to write $(0)$ as $\{0\}$ or to make a kind of joke $[0,0]$.

17. Jan 26, 2017

### sa1988

Duly noted. Many thanks!