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.
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 is a subset of ordered pairs of the set ."
"A function is a relation such that for each value x in its domain there exists at most one f(x) in its image."
-Dan
Edit: I'm a tad uncomfortable about defining the function on [-1, 1] in this case. Presumably this simply means 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 typically indicates the inverse of a bijective (not merely injective) function.
-Dan
You have so formally
So if we operate on an interval, it will take the square root of all values on that interval.
Thus
We have to be a bit more careful about 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 is going to only give us the real part of the "interval" that is returned. So .
-Dan