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:

[itex] f : R \rightarrow R[/itex] such that [itex]f(x) = x^{2}[/itex]

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

[itex]X \times Y = \{(x,y) : x\in X, y\in Y\}[/itex]

So in the case of [itex]f(x) = x^2[/itex], we have for part a):

a) [itex]f((-1,2)) = (-1,2) \times (-1,2) = \big((-1,-1),(-1,2),(2,-1),(2,2)\big)[/itex]

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

And then for part b:

b) [itex]f((-1,2]) = ...? [/itex]

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 [itex]X \times Y[/itex] in the way I defined above.

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

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

Or in other words: [itex]f^{-1}(Y) = \{ x \in X : f(x) \in Y \}[/itex] - 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 [itex]f(x) = x^2[/itex], we have for part a):

a) [itex]f((-1,2)) = (-1,2) \times (-1,2) = \big((-1,-1),(-1,2),(2,-1),(2,2)\big)[/itex]

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

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!