function between sets

**seerTneerGevoLI** · September 5th 2007, 05:26 PM

I decided to take a Real Analysis course, and I'd appreciate some help throughout this semester on questions I have if possible. I'm alrdy having trouble w/ one of the 1st problems..

1.) Given a function such that f : D -> R and given that a subset B "is a subset of" R, let f^(-1)(B) be the set of all the points from domain D that will get mapped into B;

IE: f^(-1)(B) = {x "is an element of" D : f(x) "is an element of B}. Note that this is referred to as the preimage of B.

a.) Let f(x) = x^2. Suppose A is the interval [0, 4] and B is the interval [-1, 1] (note both of these are closed), what is f^(-1)(A) and f^(-1)(B).

So this seems elementary, and perhaps I'm overlooking it.

We know the domain D maps to the reals..

So for A, we'd have [0,0], [1,1], [2,4], [3,9], [4,16]
And for B, we'd have [-1,1], [0,0], [1,1]

I'm not quite sure what they're wanting.

b.) Does f^(-1)(A ^ B) = f^(-1)(A) ^ f^(-1)(B) in this case? Does f^(-1)(A v B) = f^(-1)(A) v f^(-1)(B)?

(**The v and ^ by itself means union and intersection, respectively)

Well, A ^ B would be [0,0],[1,1] if im not mistaken, and
A v B would be [-1,1], {A} => [-1,1], [0,0], [1,1], [2,4], [3,9], [4,16]

But if that's f^(-1)(A ^ B) and f^(-1)(A v B), what is it by component (ex. f^(-1)(A) ^ f^(-1)(B)

c.) The good behavior, as showed above, of these preimages is general. Show that for some arbitrary function g : R -> R, its always true that g^(-1)(A ^ B) = g^(-1)(A) ^ g^(-1)(B) and g^(-1)(A v B) = g^(-1)(A) v g^(-1)(B) for all sets A, B "is a subset of" R

Not sure how to show this if I can't even show it for a specific function.

Thanks for your help and clarification.

**topsquark** · September 5th 2007, 05:45 PM

Originally Posted by seerTneerGevoLI

1.) Given a function such that f : D -> R and given that a subset B "is a subset of" R, let f^(-1)(B) be the set of all the points from domain D that will get mapped into B;

IE: f^(-1)(B) = {x "is an element of" D : f(x) "is an element of B}. Note that this is referred to as the preimage of B.

a.) Let f(x) = x^2. Suppose A is the interval [0, 4] and B is the interval [-1, 1] (note both of these are closed), what is f^(-1)(A) and f^(-1)(B).

So this seems elementary, and perhaps I'm overlooking it.

We know the domain D maps to the reals..

So for A, we'd have [0,0], [1,1], [2,4], [3,9], [4,16]
And for B, we'd have [-1,1], [0,0], [1,1]

I'm not quite sure what they're wanting.

Bear in mind that [0, 4] is an interval. It is the entire real line segment from 0 to 4, not just 0, 1, 2, 3, and 4.

So
$f^{-1}( [0, 4] ) = [0, 2]$
and
$f^{-1}( [-1, 1] ) = [0, 1]$

Note in this last one $f^{-1}( [-1, 1] ) = f^{-1}( [0, 1] )$ because the first expression is not real on the set [-1, 0).

-Dan

**topsquark** · September 5th 2007, 05:51 PM

Originally Posted by seerTneerGevoLI

a.) Let f(x) = x^2. Suppose A is the interval [0, 4] and B is the interval [-1, 1].

b.) Does f^(-1)(A ^ B) = f^(-1)(A) ^ f^(-1)(B) in this case? Does f^(-1)(A v B) = f^(-1)(A) v f^(-1)(B)?

(**The v and ^ by itself means union and intersection, respectively)

Well, A ^ B would be [0,0],[1,1] if im not mistaken, and
A v B would be [-1,1], {A} => [-1,1], [0,0], [1,1], [2,4], [3,9], [4,16]

But if that's f^(-1)(A ^ B) and f^(-1)(A v B), what is it by component (ex. f^(-1)(A) ^ f^(-1)(B)

$A \cap B = [0, 4] \cap [-1, 1] = [0, 1]$

$A \cup B = [0, 4] \cup [-1, 1] = [-1, 4]$

So
$f^{-1}(A \cap B) = f^{-1}( [0, 1] ) = [0, 1]$
and
$f^{-1}(A) \cap f^{-1}(B) = f^{-1}( [0, 4] ) \cap f^{-1}( [-1, 1] ) = [0, 2] \cap [0, 1] = [0, 1]$

So the two are, indeed equal.

I leave it to you to do the intersection case.

-Dan

**topsquark** · September 5th 2007, 05:57 PM

Originally Posted by seerTneerGevoLI

c.) The good behavior, as showed above, of these preimages is general. Show that for some arbitrary function g : R -> R, its always true that g^(-1)(A ^ B) = g^(-1)(A) ^ g^(-1)(B) and g^(-1)(A v B) = g^(-1)(A) v g^(-1)(B) for all sets A, B "is a subset of" R

Not sure how to show this if I can't even show it for a specific function.

I can't help you too much with this one off the top of my head. However I can remind you of one piece of information that will likely be helpful:
"A relation $f: \mathbb{R} \to \mathbb{R}$ is a subset of ordered pairs of the set $\mathbb{R} \times \mathbb{R}$ ."

"A function $f: \mathbb{R} \to \mathbb{R}: x \mapsto f(x)$ is a relation such that for each value x in its domain there exists at most one f(x) in its image."

-Dan

**topsquark** · September 5th 2007, 06:03 PM

Originally Posted by topsquark

Bear in mind that [0, 4] is an interval. It is the entire real line segment from 0 to 4, not just 0, 1, 2, 3, and 4.

So
$f^{-1}( [0, 4] ) = [0, 2]$
and
$f^{-1}( [-1, 1] ) = [0, 1]$

Note in this last one $f^{-1}( [-1, 1] ) = f^{-1}( [0, 1] )$ because the first expression is not real on the set [-1, 0).

-Dan

Edit: I'm a tad uncomfortable about defining the function $f^{-1}$ on [-1, 1] in this case. Presumably this simply means $f(x) = x^2$ is injective (since I would presume that [-1, 1] is a subset of the image of f, perhaps range would indeed be the better word here), but I'm just not in tune with it since $f^{-1}$ typically indicates the inverse of a bijective (not merely injective) function.

-Dan

**seerTneerGevoLI** · September 5th 2007, 06:45 PM

Originally Posted by topsquark

Bear in mind that [0, 4] is an interval. It is the entire real line segment from 0 to 4, not just 0, 1, 2, 3, and 4.

So
$f^{-1}( [0, 4] ) = [0, 2]$
and
$f^{-1}( [-1, 1] ) = [0, 1]$

Note in this last one $f^{-1}( [-1, 1] ) = f^{-1}( [0, 1] )$ because the first expression is not real on the set [-1, 0).

-Dan

How exactly is f^(-1)([0,4]) = [0,2] ?

I don't quite see what's going on. Sorry if this is very elementary.

**topsquark** · September 6th 2007, 05:34 AM

Originally Posted by seerTneerGevoLI

How exactly is f^(-1)([0,4]) = [0,2] ?

I don't quite see what's going on. Sorry if this is very elementary.

You have $f(x) = x^2$ so formally $f^{-1}(x) = \sqrt{x}$

So if we operate $f^{-1}$ on an interval, it will take the square root of all values on that interval.

Thus
$f^{-1}([0,4]) = [ \sqrt{0} , \sqrt{4} ] = [0, 2]$

We have to be a bit more careful about $f^{-1}( [-1, 1] )$ since some of the values returned by the square root are imaginary. (This is why I mentioned I was uncomfortable in the other post.) However we may ignore this problem in the sense that $f^{-1}(A)$ is going to only give us the real part of the "interval" that is returned. So $f^{-1}( [-1, 1] ) = [0 , 1]$ .

-Dan

**Plato** · September 6th 2007, 05:58 AM

Originally Posted by topsquark

Edit: I'm a tad uncomfortable about defining the function $f^{-1}$ on [-1, 1] in this case.

There is no need to worry about that.
$\begin{array}{rcl} x \in f^{ - 1} (A \cap B)\quad & \Leftrightarrow & \quad f(x) \in A \cap B \\ & \Leftrightarrow & \quad f(x) \in A \wedge f(x) \in B \\ & \Leftrightarrow & \quad x \in f^{ - 1} (A) \wedge x \in f^{ - 1} (B) \\ & \Leftrightarrow & \quad x \in \left[ {f^{ - 1} (A) \cap f^{ - 1} (B)} \right] \\ \end{array}$

**seerTneerGevoLI** · September 12th 2007, 06:30 AM

Originally Posted by topsquark

Bear in mind that [0, 4] is an interval. It is the entire real line segment from 0 to 4, not just 0, 1, 2, 3, and 4.

So
$f^{-1}( [0, 4] ) = [0, 2]$
and
$f^{-1}( [-1, 1] ) = [0, 1]$

Note in this last one $f^{-1}( [-1, 1] ) = f^{-1}( [0, 1] )$ because the first expression is not real on the set [-1, 0).

-Dan

Apparently, according to my professor, f^{-1}( [-1,1] ) = [-1,1]

And, f^{-1}( [0,4] ) = [-2,2]

**Plato** · September 12th 2007, 06:44 AM

It really does depend of what the D in the original is. I would agree with your instructor if D includes those two intervals.

**red_dog** · September 12th 2007, 07:06 AM

Originally Posted by seerTneerGevoLI

c.) The good behavior, as showed above, of these preimages is general. Show that for some arbitrary function g : R -> R, its always true that g^(-1)(A ^ B) = g^(-1)(A) ^ g^(-1)(B) and g^(-1)(A v B) = g^(-1)(A) v g^(-1)(B) for all sets A, B "is a subset of" R

$x\in g^{-1}(A\cap B)\Leftrightarrow g(x)\in A\cap B\Leftrightarrow (g(x)\in A)\wedge (g(x)\in B)\Leftrightarrow$
$\Leftrightarrow (x\in g^{-1}(A))\wedge (x\in g^{-1}(B))\Leftrightarrow x\in g^{-1}(A)\cap g^{-1}(B)$

Prove the second part in the same way.

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