Functions and Sets: Understanding Notation and Inverse Functions

[sa1988]

Homework Statement

ONLY QUESTION 2[/B]

Homework Equations

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.

 

[fresh_42]

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 $(a,b)$.

 

[member 587159]

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 $f(V) = \{f(v) | v \in V \}$

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)) ..??

Hints much appreciated, thanks.

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

Hint: What is $f(0)$?

 

[sa1988]

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 $(a,b)$.

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.

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 $f(V) = \{f(v) | v \in V \}$

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 $f(x) = x^2$

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

 

[fresh_42]

So is this correct?

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

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 $f(x) = -1\in ]\infty,0[\, ?$

 

[sa1988]

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

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...

 

[fresh_42]

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 $y$?

 

[sa1988]

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 $y$?

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

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

 

[member 587159]

An interesting additional question would be:

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

 

[sa1988]

An interesting additional question would be:

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

$[0,1)$ ?

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

if I'm not mistaken?

 

[fresh_42]

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

$f((-1,2)) = [0,4)$
$f((-1,2]) = [0,4]$

Yep.

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

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

 

[fresh_42]

$[0,1)$ ?

Yes.

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

if I'm not mistaken?

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

 

[sa1988]

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

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.

 

[fresh_42]

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.

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.

 

[sa1988]

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.

I think I've cracked it:

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

 

[fresh_42]

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

 

[sa1988]

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

Duly noted. Many thanks!