SUPER-FUN-HAPPY-MATHS-TIME

[kamil9876]

The question means "for a constant r" show that there are infinitely many z. You are right that there are infinitely many r's but ultimately you want to show that for each r, there are infinitely many z. That is, if we select an r and keep it constant, then we find infinitely many z, and that this condition holds for any r>0.5d. For each r the set of z is a circle, however the union of all these circles is a plane, yes. Evaporade's disk just has the unions of some circles and so he has shown this to by true for any r>0.5d in general for k=3(however to make this more explicit it would probably be good to add in the circle that the points z lies on but imagination can easily fill that in heh)

[evaporade]

r > 0  means any r > 0, not a constant r > 0. I used a dotted circle (look at the picture) to mean the plane is infinite, not because I was thinking about a constant r.

[kamil9876]

Yes, any r>0. But specifically, if we select say r=3 (provided d<1.5), we find an infinite set of z that satisfy |z-x|=|z-y|=3. If we select r=4, we find infinite set of z that satisfy |z-x|=|z-y|=4 etc.

It's possible, generally speaking, for a set to have finite members, but a union of infinite such sets to have infinite members, hence it's important for this problem to show that each subset(each selected r value) has infinite members. But I think your diagram shows that (the disk is a set of circles(subsets) and each circle contains infinite points)  so all good

[evaporade]

There is only one set - the 'plane' that is 'perpendicular' to and passes through the 'line' xy at its midpoint. z belongs to this set for (a) and (b), and there is no z in this set or outside this set satisfying requirement (c). There is no subset of this 'plane' to be considered in this problem. 

[evaporade]

By the way could you please name the set with finite number of members, which you referred to?

[zzdfa]

r > 0  means any r > 0, not a constant r > 0. I used a dotted circle (look at the picture) to mean the plane is infinite, not because I was thinking about a constant r.

So for a question such as:

the answer would be infinity?

By the way could you please name the set with finite number of members, which you referred to?

i think he was just using it as an example,

that it is possible to form a set with an infinite number of elements, by taking the union of an infinite number of finite sets, thus if one wanted to prove that each individual set was infinite, it does not suffice to show that the union of them is infinite.
e.g.
The natural numbers = {1} U {2,3,4.....}
the union is infinite but one of the sets is finite.

[evaporade]

So what is the point in considering the union of sets when you can simply name the infinite set? Besides, this question is not about set theory. Part (a) is a simple question about 'the sum of two sides of a triangle > third side' in higher dimensions.

[kamil9876]

yep zzdfa answered my question exactly. I did not want to reiterate myself.

evaporade: I never said "finite sets" in this particular problem, but in general terms(even said "generally speaking"). However a proof of such a statement would have to rule out the possibility of an infinite set of finite set. I was going to ask you what is your answer to "how would this statement(a) need to be modified for k=2?" because with your logic the answer would be "this statement is also true for k=2" but according to mine and zzdfa's understanding it is not as it a case of an infinite set of finite sets. (a line being a set of points).

[kamil9876]

So what is the point in considering the union of sets when you can simply name the infinite set?

thus if one wanted to prove that each individual set was infinite, it does not suffice to show that the union of them is infinite.
e.g.
The natural numbers = {1} U {2,3,4.....}
the union is infinite but one of the sets is finite.

[zzdfa]

Your diagram indicates the set of all |z-x|=|z-y|=r for any r
The question asks that,
given any r, is there an infinite number of points such that |z-x|=|z-y|=r

there's a difference

[evaporade]

I don't see any difference. The question did not even use the word given, it said suppose r > 0, i.e. |z - x| = |z - y| > 0 .

[evaporade]

"how would this statement(a) need to be modified for k=2?"

No modification required for (a), (b) and (c). The infinite set of z in this case is the perpendicular bisector of xy.

[kamil9876]

There is a a modification required if you take mine and zzdfa's understanding:

"if k=2 then there are only 2 such z's for any r".

Maybe my english is bad, but I think the question does mean 'for any given r' since it is introduced in the opening sentence just like x,y and d are and we do consider x,y and d as constant hence same for r. This is just a matter of understanding the english.

But as to whether there is a difference between the two interpretations, there certainly is for reasons reiterated many times above.

In fact you could just as easily say that x and y are not constant(introduced in the exact same sentence as r) and then the answer would be so trivial.

Btw: Interpreting it as "for any given r" is a much more fun problem(hence the thread name) and the statements then do indeed require modifications for k=2.

[evaporade]

Quote "Maybe my english is bad, but I think the question does mean 'for any given r' since it is introduced in the opening sentence just like x,y and d are and we do consider x,y and d as constant hence same for r".



Let us make things simple, consider the following:

Suppose x,y E R, |x - y| = d > 0. Are you suggesting that the value of x and the value of y are given constants?

[kamil9876]

Forget the word 'constant' then. All I mean is that the question can be stated as "prove that for every given x, y and r (such that 2r>|x-y|) there are infinitely many z such that...". This is what I meant by constant, just like in your diagram x and y are constant and fixed points because you are proving that 'for every given x and y...'