I suppose I shouldn't have said that...
But since I did...Since we are going to the trouble of stating that a,b are integers and since we know that the set of integers are an integral domain, having all of the properties of an integral domain, AND since we know that by definition, sqrt(-1)=i is an element of the complex numbers and NOT an integer...
I dunno...I guess I might have assumed too much...
I was just thinking that since he bothered to ask the question by mentioning that a,b were integers, we ought to be able to use that fact in order to define the numbers and their properties, right?
You can do it in cases...
Let a,b∈ Z and suppose that:
case 1: a<0 and b>0...
Then a=(-1)g ( g=abs(a) )
Then ab = (-1)gb = -(1g)b = -(gb) [negative]
case 2: a<0, b<0
Then a=(-1)g and b=(-1)h ( g=abs(a) h=abs(b) )
Then ab=(-1)g(-1)h=(-1)(-1)gh=(1)gh=gh [positive]
case 3: a>0, b>0
then ab>0 [obviously positive]
I was using those theorems...
By my definition, a set is open iff its complement is closed.
The empty set was both open and CLOSED, per our earlier "proof"
Then the set R is open since its complement (R\R...aka the empty set) is closed.
That's what I was trying to say...And I just realized that your definition is actually the same as mine, but you are using that every point in an open set (you called them 'a') are interior points.
As for your kicker...I have no idea how to do it...I'll need to think it over for awhile...(Are we certain that it can be done?)
hmmm...I believe the exact same argument would work for the entire set of real numbers, wouldn't it?
And I just re-read your explanation, Ricky, and I'm having a little bit of trouble with your statement of the definition of an open set as
"A set O is open if for all points a∈O there exists an ∈-neighborhood of a."
What did you mean by that?...You can create an epsilon neighborhood around any point, whether the set is open or closed. I think that you meant to say something about the intersection of that neighborhood and the set...
The definitions of open and closed that I remember have to do with inclusion of boundary points.
A boundary point of a set S is defined as a point, say x (in R), such that for any epsilon>0, the neighborhood N(x;epsilon)
intersected with S is nonempty, and that same neighborhood N intersected with R\S is nonempty...
Now the definition that I know for a closed set is a set that contains all of those boundary points. (And thus, an open set is one whose complement is closed)
For example, consider the set of integers Z.
the set of boundary points of Z (denoted bd Z) is actually all of the integers. [Does that make sense?...because any neighborhood that you take about an integer includes both an integer (in Z), and all of the other real numbers between integers (in R\Z)]
So since Z contains bd Z (in fact, it IS bd Z), the set Z of integers is closed.
I hope I'm making sense...
So then, for this particular question, I would say (R="the set of real numbers")
bd R = empty set (because there are no points in R that would fit the definition of a boundary point of R)
then, bd R is a subset of R ----> R is closed
but bd R is also a subset of R\R (which is the empty set) -----> R is open.
Thus, we have that R is both open and closed....
I DO love the title...
And I believe that the empty set is the set that is both open and closed. It is sort of vacuously true that it contains its boundary points (also the empty set), so its closed. But the boundary points (empty set) are also contained in the complement of the empty set, so it is open...
Whew...that was tricky to say....
who's on first?
ATTORNEY: Doctor, before you performed the autopsy, did you check for a pulse? WITNESS: No.
ATTORNEY: Did you check for blood pressure?
ATTORNEY: Did you check for breathing?
ATTORNEY: So, then it is possible that the patient was alive when you began the autopsy?
ATTORNEY: How can you be so sure, Doctor?
WITNESS: Because his brain was sitting on my desk in a jar.
ATTORNEY: But could the patient have still been alive, nevertheless?
WITNESS: Yes, it is possible that he could have been alive and practicing law.
In that case....give us the proof please, krassi!!!!
It's been bugging me since you posted this and I'm having the same problem...I've gone through all my notes to try to find some long lost number theory theorem that would make it all come together, but I can't....and I'm a bit obsessive so I need to see the proof...and I need to see it fast!
|Im f| = |A|.
|A| ≤ |B|
When you use the absolute value symbols, you mean the order of those sets (number of elements in the set), right?
If so, then I'm not sure that you need to prove that they are subsets both ways, because actually each of those terms would just represent a number.