04 Jun 08
Originally posted by FabianFnasWell, you can use the Riemann sphere, although it's not really a "plane", being compact, having different Euler characteristic, etc.
Well, it was a wild guess, the closest I can get to an answer.
Is it possible to construct a plane where inf really is included? As you can with R, i.e. R* ?
04 Jun 08
'Originally posted by JirakonJust because all points have finite coordinates does not mean there is a 'y-max'
A point is an ordered pair of two REAL numbers. No infinity.
If that's the case, then there is no solution:
Suppose such a set exists in which all points have finite coordinates. Take the point(s) with the highest y-value (y_max). Now imagine the line y = y_max + 1. This line does not intersect any of the points. Therefore no such set exists.
Otherwise your "proof" could be used to prove that the set of points on x=0 is finite.
04 Jun 08
1 edit
Originally posted by PalynkaWell, it would have to be something complicated - discrete but dense. Think of those functions that have different values depending on whether x is rational or irrational. You can get some very strange behaviour.
I don't think it can be discrete points. Imagine if you have a set of discrete points and pick a line that intersects only two points. If you pick one of those points as center and keep rotating that line infinitesimally, then I don't see how it's possible to cover all possibilities with any mesh of discrete points.
I agree, I don't think there is such as set, but I do know that you can construct sets with properties that intuition really doesn't work well on. So I'll need a proper proof to be sure.
04 Jun 08
Originally posted by JirakonI don't see why what I said implies the set must be bounded. Take the Euclidean axis. No line that you can draw or represent as a function intersects a point of those axes only at infinity.
A point is an ordered pair of two REAL numbers. No infinity.
If that's the case, then there is no solution:
Suppose such a set exists in which all points have finite coordinates. Take the point(s) with the highest y-value (y_max). Now imagine the line y = y_max + 1. This line does not intersect any of the points. Therefore no such set exists.
04 Jun 08
Originally posted by mtthwBut if it's discrete, you'd always find a rotation of an infinitesimal epsilon that slips through a discontinuity. Note that every two points in the set define a unique line and all possible lines must be defined by that set.
Well, it would have to be something complicated - discrete but dense. Think of those functions that have different values depending on whether x is rational or irrational. You can get some very strange behaviour.
I agree, I don't think there is such as set, but I do know that you can construct sets with properties that intuition really doesn't work well on. So I'll need a proper proof to be sure.
04 Jun 08
Originally posted by PalynkaOK try S = {x, y: s.t. x and y are irrational}
But if it's discrete, you'd always find a rotation of an infinitesimal epsilon that slips through a discontinuity. Note that every two points in the set define a unique line and all possible lines must be defined by that set.
Even an infinitesimal rotation is going to sweep through an infinite number of points.
Yes, I know that that set breaks lots of the other conditions - I'm just using it to illustrate the point. Dense sets - and I mean a set such that any interval is going to contain at least one value - can have some weird properties.
04 Jun 08
Originally posted by mtthwBut there are still holes in that set. The irrational set is uncountable and therefore would have holes.
OK try S = {x, y: s.t. x and y are irrational}
Even an infinitesimal rotation is going to sweep through an infinite number of points.
Yes, I know that that set breaks lots of the other conditions - I'm just using it to illustrate the point. Dense sets - and I mean a set such that any interval is going to contain at least one value - can have some weird properties.
04 Jun 08
1 edit
Originally posted by GastelYes, it would. That's why I called it a discrete set. It's designed to illustrate the problems with one specific objection.
But there are still holes in that set. The irrational set is uncountable and therefore would have holes.
Oh, and if I was going to be picky 🙂 - it's uncountable and there are holes - but it's not got holes because it's uncountable.
04 Jun 08
Originally posted by mtthwNice example. I did add my note that every set of two points in the S set defines a unique line, but I see your point about proving it formally.
OK try S = {x, y: s.t. x and y are irrational}
Even an infinitesimal rotation is going to sweep through an infinite number of points.
Yes, I know that that set breaks lots of the other conditions - I'm just using it to illustrate the point. Dense sets - and I mean a set such that any interval is going to contain at least one value - can have some weird properties.
15 Jun 08
Originally posted by joe shmoNo, not at all pejorative.
by "lower grade" do you mean lower caliber, or lower amount of time attended?
because I can not prove it....
From when you enter the math classes in University, you have to wait quite a while before you get the hint why 1+2=3. The reason is that there is so many important things to go through before you go into the very essence of the number systems. You take it for obvious that 1+2=3 and you don't have to care much about it.
So, who can come up with the water strong proof that 1+2=3?
15 Jun 08
Originally posted by FabianFnasIt's a postulate, just like the one that goes 1+1=2. We have no conclusive way of proving it, yet it exists. It's a bit like "Between every two points there is only one line." Can we prove it? No. Why not? Because we cannot demonstrate using points or lines because in the real world they do not exist. Same thing with 1+2=3 or 1+1=2. If 1+2=3 is a theorem, then I actually don't know because I'm not out of high school yet.
No, not at all pejorative.
From when you enter the math classes in University, you have to wait quite a while before you get the hint why 1+2=3. The reason is that there is so many important things to go through before you go into the very essence of the number systems. You take it for obvious that 1+2=3 and you don't have to care much about it.
So, who can come up with the water strong proof that 1+2=3?
15 Jun 08
Originally posted by scherzoNo no, 1+2=3 is certainly provable. Just use the axioms. The axioms, on the other hand, is not provable. If they were they are not axioms. And now Gödel comes into my mind.
It's a postulate, just like the one that goes 1+1=2. We have no conclusive way of proving it, yet it exists. It's a bit like "Between every two points there is only one line." Can we prove it? No. Why not? Because we cannot demonstrate using points or lines because in the real world they do not exist. Same thing with 1+2=3 or 1+1=2. If 1+2=3 is a theorem, then I actually don't know because I'm not out of high school yet.