Here is a dinner table problem.
Problem 250 *
Paint every point in the plane one of three colours.
i) Are there two points of the same colour exactly (cm) apart?
This time, paint every point one of six colours.
ii) Prove that there exist two points of the same colour with one of three possibilities:
a) they are (cm) apart; b) (cm) apart; c) cm apart;

jack.hadamard
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 21062013 21:41
Last edited by jack.hadamard; 22062013 at 01:27. Reason: problem number 
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 21062013 22:44
(Original post by jack.hadamard)
Here is a dinner table problem.
Problem 248 *
Paint every point in the plane one of three colours.
i) Are there two points of the same colour exactly (cm) apart?
This time, paint every point one of six colours.
ii) Prove that there exist two points of the same colour with one of three possibilities:
a) they are (cm) apart; b) (cm) apart; c) cm apart; 
jack.hadamard
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 1743
 21062013 23:16
(Original post by james22)
Don't really understand this, doesn't this depend on how you paint the plane. 
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 21062013 23:21
(Original post by jack.hadamard)
The first part asks whether there are two such points, regardless of how you decide to paint it. 
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 1745
 22062013 00:22

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 1746
 22062013 01:05
(Original post by henpen)
Is that the binomial coefficient or some other object?
(Original post by jack.hadamard)
Spoiler:ShowHere is a dinner table problem.
Problem 248 *
Paint every point in the plane one of three colours.
i) Are there two points of the same colour exactly (cm) apart?
This time, paint every point one of six colours.
ii) Prove that there exist two points of the same colour with one of three possibilities:
a) they are (cm) apart; b) (cm) apart; c) cm apart;
Solution 250
The first part generalizes to any positive number .
In the case , we use a circle of radius . Let its centre be the point . We have two possibilities: either all the points on the circle have the same colour as the centre of the circle, or there is a point of different colour. In the first case, we pick chord of length ; in the second case, we consider the two circles with centres at and ( is the point on the circle of radius , which is of different colour) and radius . The distance between the points of intersection of these two circles is ; these two points lie on circles of radius . Hence, the result.
We construct a triangle with sides and then complete it to a rectangle. Now, let the sides of length be bases of isosceles triangles with sides .
The conclusion follows. 
jack.hadamard
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 22062013 01:31
(Original post by Mladenov)
Nice, and easy, combinatorial geometry.Last edited by jack.hadamard; 22062013 at 01:33. 
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 1748
 22062013 02:06
Here is a dinner table problem.
Problem 250 *
Paint every point in the plane one of three colours.
i) Are there two points of the same colour exactly (cm) apart?
This time, paint every point one of six colours.
ii) Prove that there exist two points of the same colour with one of three possibilities:
a) they are (cm) apart; b) (cm) apart; c) cm apart;[/QUOTE]
Diagram:
Consider the following diagram of intersecting unit curves. Prove statement above by reductio. WLG suppose the origin is black, then the black circle must be a locus of points not black. Consider the point at (1,0). WLG assume it is red. Then THe red circle is the locus of points not red. Where the red and black intersect, those two points must be blue. Thus the two blue circles represent a locus of points not blue. Thus at the intersection we have two points that are not red blue or black. Contradiction. There must be points of the same colour 1 cm apart.
For the second part constructing a hexagon is the way to go.Last edited by Blutooth; 22062013 at 02:26. 
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 1749
 22062013 11:09
Problem 251**
Let and be two disjoint finite nonempty sets in the plane such that every segment joining two points in the same set contains a point from the other set. Show that all the points of the set lie on a single line. 
jack.hadamard
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 22062013 21:34
(Original post by Mladenov)
Problem 251**
Solution 251
Clearly, if , the result is true. Let and, without loss of generality, . We argue that if all the points of lie on a single line, then all the points of lie on the same line. Suppose, for a contradiction, that at least one point of is not contained in the line which all the points of lie on. Hence, from the following diagram,
we can construct a point, , of that is not on the line. This is because there is a point, , between and (points on the line), and, hence, a point between and . This point is not on the line, since and are disjoint. Now, consider the possibility with not all the points of being on a line. We argue that for any three given points of , not all on a line, there exist distinct three points of not on a line. In fact, consider the diagram
and notice that and are distinct and none of them is one of the previous points, since the sets are disjoint, and they are not on a line. Therefore, since this contradicts the assumption that is a finite set, all the points of lie on a single line.Last edited by jack.hadamard; 03072013 at 23:38. Reason: added solution number 
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 1751
 22062013 22:04
(Original post by jack.hadamard)
Spoiler:ShowWhy two stars?
Clearly, if , the result is true. Let and, without loss of generality, . We argue that if all the points of lie on a single line, then all the points of lie on the same line. Suppose, for a contradiction, that at least one point of is not contained in the line which all the points of lie on. Hence, from the following diagram,
we can construct a point, , of that is not on the line. This is because there is a point, , between and (points on the line), and, hence, a point between and . This point is not on the line, since and are disjoint. Now, consider the possibility with not all the points of being on a line. We argue that for any three given points of , not all on a line, there exist distinct three points of not on a line. In fact, consider the diagram
and notice that and are distinct and none of them is one of the previous points, since the sets are disjoint, and they are not on a line. Therefore, since this contradicts the assumption that is a finite set, all the points of lie on a single line.
Your argument is quite beautiful. 
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 1752
 23062013 08:48
(Original post by FireGarden)
Let's stick something different into the thread..
Problem 231* (*** if you don't look at the spoiler)
Prove the annulus is homeomorphic to the cylinder
Spoiler:ShowA glimpse into the geometric side of topology. A homeomorphism between two topological spaces says they are topologically identical, and consists of a pair of continuous functions , such that f and g are inverses of each other (more precisely, f(g(x)) = identity of C, g(f(x)) = identity of A). The question then is now simply to find such functions.
I observed that letting would seem to map the interval on to though, again, no idea what the convention is here for notation (though I'm probably being stupid).
I have a vague picture in my mind of reducing the distance of the points on the annulus by 1, squashing it threefold and then pulling the middle out to stretch it to form a cylinder and have loads of horribly scribbly equations that barely make sense (the above being an example of such).
Is this right? At this point I wouldn't mind you simply telling me the solution since I could likely use it as a template for a similar problem.
I was also wondering if this would require a proof that the cardinality of the sets containing the interval sizes are equal or whether or not this is automatically implied. I am aware that I am probably talking complete and b*ll**** but I think that you could say an interval can be transformed to by using , which seems rather cool (if it's correct that is?). Though again, not sure how to generalise this to more variables. Am I supposed to introduce a new variable r such that ?
Topology seems incredibly fun... if only I knew the language it was written in :') 
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 1753
 23062013 09:50
(Original post by jack.hadamard)
x 
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 1754
 23062013 15:54
(Original post by Jkn)
Spoiler:ShowNo idea what a function that maps more than one variable onto more than one variable would look like (in terms of notation) :/ Could you help me out?
I observed that letting would seem to map the interval on to though, again, no idea what the convention is here for notation (though I'm probably being stupid).
I have a vague picture in my mind of reducing the distance of the points on the annulus by 1, squashing it threefold and then pulling the middle out to stretch it to form a cylinder and have loads of horribly scribbly equations that barely make sense (the above being an example of such).
Is this right? At this point I wouldn't mind you simply telling me the solution since I could likely use it as a template for a similar problem.
I was also wondering if this would require a proof that the cardinality of the sets containing the interval sizes are equal or whether or not this is automatically implied. I am aware that I am probably talking complete and b*ll**** but I think that you could say an interval can be transformed to by using , which seems rather cool (if it's correct that is?). Though again, not sure how to generalise this to more variables. Am I supposed to introduce a new variable r such that ?
Topology seems incredibly fun... if only I knew the language it was written in :')
If you pick a book on General topology (I mean some serious book, Bourbaki, Kelley, etc.), you will see that it goes like: definition, definition, definition, lemma, theorem; definition, definition, definition, lemma, theorem.
Bearing this in mind, I am inclined to say that General topology is just a necessary tool. However, once you know it, you can go on to study much more interesting subjects, i.e. algebraic topology, differential geometry ( I want to make it clear that, when I say differential geometry, I do not mean something like Spivak's book  this is for engineers..).
I personally find general topology not so intuitive; on the other hand, topological groups, especially Lie groups, are quite natural. Just look at the definition of Lie group  is it not the only natural way to define something that is both a group and a smooth manifold?
Here is an exercise.
Problem 252***
Let be a group and a smooth manifold. Suppose that the map is smooth. Show that is smooth and, hence, that is a Lie group. Is the corresponding statement about topological groups true? 
FireGarden
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 23062013 18:14
(Original post by Jkn)
No idea what a function that maps more than one variable onto more than one variable would look like (in terms of notation) :/ Could you help me out?
The most help I could give without giving the game away, is to think of the annulus' outer circle "standing up" to make the cylinder; and to think of a function that would take the point on the annulus to the resulting point on the cylinder after this morphing.
As Mlad said above, many books on topology are very terse and formal: However I first read about Topology from a much more discursive book, targeted towards those who are in/completing their second year, called Essential Topology by Crossley, published by Springer. It is a very good introductory book that goes surprisingly far (well into algebraic topology), the only quirk it has is no mention of metric spaces. The only essential prerequisites are about algebra; particularly equivalence relations and (when getting into the algebraic side) comfort with groups. I highly recommend it. 
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 23062013 18:26
(Original post by Jkn)
I was also wondering if this would require a proof that the cardinality of the sets containing the interval sizes are equal or whether or not this is automatically implied. I am aware that I am probably talking complete and b*ll**** but I think that you could say an interval can be transformed to by using , which seems rather cool (if it's correct that is?). Though again, not sure how to generalise this to more variables. Am I supposed to introduce a new variable r such that ? 
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 23062013 19:02

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 23062013 20:42
(Original post by Mladenov)
You want to think of these objects as sets of points and, hence, all you need is to find a mapping that maps each point to a point such that this map is an isomorphism in Top.
If you pick a book on General topology (I mean some serious book, Bourbaki, Kelley, etc.), you will see that it goes like: definition, definition, definition, lemma, theorem; definition, definition, definition, lemma, theorem.
Bearing this in mind, I am inclined to say that General topology is just a necessary tool. However, once you know it, you can go on to study much more interesting subjects, i.e. algebraic topology, differential geometry ( I want to make it clear that, when I say differential geometry, I do not mean something like Spivak's book  this is for engineers..).
(Original post by FireGarden)
They're nothing particularly strange; for instance, a function could be defined
As Mlad said above, many books on topology are very terse and formal: However I first read about Topology from a much more discursive book, targeted towards those who are in/completing their second year, called Essential Topology by Crossley, published by Springer. It is a very good introductory book that goes surprisingly far (well into algebraic topology), the only quirk it has is no mention of metric spaces. The only essential prerequisites are about algebra; particularly equivalence relations and (when getting into the algebraic side) comfort with groups. I highly recommend it.
Anyway, I had not think that that notation was acceptable (I was thinking in terms of f(x)=... types of functions ) Just had a bit of a eureka moment by realising that a region can be mapped on to a circle by normalising it since this will conserve that direction.
The major issue I had was in choosing z such that it can be used in the map back to x and y.
Solution 231
Let such that .
The given annulus is homeomorphic the cylinder .
We can generalise the result such that any annulus is homeomorphic to any cylinder , where .
Let
such that .
Note that there are, in fact, two (or more?) ways to map an annulus on to a cylinder. We can let
such that which is analogous to the method I initially suggested of it being the centre being pulled up to the height of the cylinder and the rest of the annulus squashing in to form the lower parts of the cylinder. i.e. there is no point in the mapping that maps on to itself whereas, in the first solution, there is. This second formula can then be applied to give an alternate solution to the first problem.
So is this insanely easy compared to what you would learn on a topology course then? I would be interested in attempted more problems of this type 
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 23062013 22:49
(Original post by Jkn)
So is this insanely easy compared to what you would learn on a topology course then? I would be interested in attempted more problems of this type
Something harder would be:
Define and let or . Show that is homeomorphic to the sphere, .
This is a question about a cylinder, formed as the product space of the circle (s^1) and the closed interval, with an equivalence relation on it, which effectively is pulling a drawstring around the open ends of the cylinder (and geometrical intuition should confirm the result is spherelike!) 
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 1760
 23062013 23:13
(Original post by FireGarden)
Kind of. That is a pretty standard "first question on homeomorphism" problem.
Something harder would be:
Define and let or . Show that is homeomorphic to the sphere, .
This is a question about a cylinder, formed as the product space of the circle (s^1) and the closed interval, with an equivalence relation on it, which effectively is pulling a drawstring around the open ends of the cylinder (and geometrical intuition should confirm the result is spherelike!)
Where can topology be applied btw? Does it link well with number theory, calculus and/or theoretical physics? (would be apprehensive about taking the module if it did not link well with these specific areas)
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