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The Proof is Trivial!

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Here is a dinner table problem. :tongue:

Problem 250 *

Paint every point in the plane one of three colours.

i) Are there two points of the same colour exactly

1

(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

1

(cm) apart; b)

\sqrt{3}

(cm) apart; c)

2

cm apart;

(edited 10 years ago)

Reply 1741

james22

Original post by jack.hadamard

Here is a dinner table problem. :tongue:

Problem 248 *

Paint every point in the plane one of three colours.

i) Are there two points of the same colour exactly

1

(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

1

(cm) apart; b)

\sqrt{3}

(cm) apart; c)

2

cm apart;

Don't really understand this, doesn't this depend on how you paint the plane (what would be wrong with painting just 2 points 2 different colours then everything else the same so the answer is trivial(.

Reply 1742

jack.hadamard

Original post by james22

Don't really understand this, doesn't this depend on how you paint the plane.

The first part asks whether there are two such points, regardless of how you decide to paint it.

Reply 1743

james22

Original post by jack.hadamard

The first part asks whether there are two such points, regardless of how you decide to paint it.

Oh ok, this sounds like a good problem.

Reply 1744

henpen

Original post by Mladenov

Problem 249**

\left(\frac{n}{p} \right) = \left( \frac{n+k}{p} \right)

Is that the binomial coefficient or some other object?

Reply 1745

Mladenov

Original post by henpen

Is that the binomial coefficient or some other object?

Nope, it is the Legendre symbol.

Original post by jack.hadamard

Spoiler

Nice, and easy, combinatorial geometry. :tongue:

Solution 250

The first part generalizes to any positive number

d

.

In the case

d=1

, we use a circle of radius

\sqrt{3}

. Let its centre be the point

O

. 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

1

; in the second case, we consider the two circles with centres at

O

and

O'

(

O'

is the point on the circle of radius

\sqrt{3}

, which is of different colour) and radius

1

. The distance between the points of intersection of these two circles is

1

; these two points lie on circles of radius

1

. Hence, the result.

We construct a triangle with sides

1,\sqrt{3},2

and then complete it to a rectangle. Now, let the sides of length

\sqrt{3}

be bases of isosceles triangles with sides

\sqrt{3},1,1

.
The conclusion follows.

Reply 1746

jack.hadamard

Original post by Mladenov

Nice, and easy, combinatorial geometry. :tongue:

Yep.

I did the second part using a regular hexagon with unit side length (which amounts to the same thing).

(edited 10 years ago)

Reply 1747

Blutooth

Here is a dinner table problem. :tongue:

Problem 250 *

Paint every point in the plane one of three colours.

i) Are there two points of the same colour exactly

1

(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

1

(cm) apart; b)

\sqrt{3}

(cm) apart; c)

2

cm apart;

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.

(edited 10 years ago)

Reply 1748

Mladenov

Problem 251**

Let

A

and

B

be two disjoint finite non-empty 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

A \cup B

lie on a single line.

Reply 1749

jack.hadamard

Original post by Mladenov

Problem 251**

Why two stars?

Solution 251

Clearly, if

|A \cup B| = 2

, the result is true. Let

|A \cup B| \geq 3

and, without loss of generality,

|A| \geq 2

. We argue that if all the points of

A

lie on a single line, then all the points of

B

lie on the same line. Suppose, for a contradiction, that at least one point of

B

is not contained in the line which all the points of

A

lie on. Hence, from the following diagram,

we can construct a point,

a_3

, of

A

that is not on the line. This is because there is a point,

b_1

, between

a_1

and

a_2

(points on the line), and, hence, a point

a_3

between

b_1

and

b_2

. This point is not on the line, since

A

and

B

are disjoint. Now, consider the possibility

|A| \geq 3

with not all the points of

A

being on a line. We argue that for any three given points of

A

, not all on a line, there exist distinct three points of

A

not on a line. In fact, consider the diagram

and notice that

\bar{a}_1, \bar{a}_2

and

\bar{a}_3

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

A

is a finite set, all the points of

A

lie on a single line.

(edited 10 years ago)

Reply 1750

Mladenov

Original post by jack.hadamard

Spoiler

This.
Your argument is quite beautiful.

Reply 1751

Jkn

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

A = \{ (x,y) \in \mathbb{R}^2 : 1 \leq x^2+y^2 \leq 4 \}

is homeomorphic to the cylinder

C = \{ (x,y,z) \in \mathbb{R}^3 : x^2+y^2=1, 0 \leq z \leq 1 \}

Spoiler

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?

I observed that letting

z=\frac{4-x^2-y^2}{3}

would seem to map the interval

1 \le x^2+y^2 \le 4

on to

0 \le z \le 1

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 three-fold 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

[a,b]

can be transformed to

[c,d]

by using

f(x)=\frac{(x-a)(d-c)}{b-a}+c

, 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

r^2=x^2+y^2

?

Topology seems incredibly fun... if only I knew the language it was written in :')

Reply 1752

MW24595

Original post by jack.hadamard

Wow. That really is neat.

Reply 1753

Mladenov

Original post by Jkn

Spoiler

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

(x,y) \in A

to a point

(x,y,z) \in C

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..).
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

G

be a group and a smooth manifold. Suppose that the map

(a,b) \mapsto ab

is smooth. Show that

a \mapsto a^{-1}

is smooth and, hence, that

G

is a Lie group. Is the corresponding statement about topological groups true?

Reply 1754

FireGarden

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?

They're nothing particularly strange; for instance, a function

f:\mathbb{R}^2 \mapsto \mathbb{R}^3

could be defined

f: (x,y) \mapsto (x^2, y^2, xy)

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.

Reply 1755

FireGarden

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

[a,b]

can be transformed to

[c,d]

by using

f(x)=\frac{(x-a)(d-c)}{b-a}+c

, 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

r^2=x^2+y^2

Topology cares little about cardinality for equivalences. The two ubiquitous equivalences are homeomorphism (topologically identical) and homotopy equivalence (can be continuously deformed into each other), where the former is stricter, and encompasses the latter. It might surprise you to find out that

\mathbb{R}

is in fact 'equivalent' (homotopy equivalent) to

\{0\}

(yeah, just a single point.. such spaces are called contractible, and all euclidean spaces turn out to be so).

Reply 1756

jack.hadamard

Original post by Mladenov

This. Your argument is quite beautiful.

Thanks!

I should have thought of that. What happens if we drop the condition on the sets being finite? I tried to adapt the argument and construct such sets, whose union is not a line, but didn't succeed. In a more systematic approach, I tried to use subsets of rational, irrational and mixed points (various combinations) in the plane, but again it didn't work. Can you construct such sets?

Reply 1757

Jkn

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

(x,y) \in A

to a point

(x,y,z) \in C

such that this map is an isomorphism in Top.

I rem in you that I am pre-university :lol:

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..).

Why do you not find it very interesting? Btw, how much of an undergraduate syllabus have you already finished?

Original post by FireGarden

They're nothing particularly strange; for instance, a function

f:\mathbb{R}^2 \mapsto \mathbb{R}^3

could be defined

f: (x,y) \mapsto (x^2, y^2, xy)

Ah, okay!

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.

Hmm, I need to somehow get these subjects out of my head until wednesday. Literally just want STEP to be over so I can get on with more interesting maths :frown:

Anyway, I had not think that that notation was acceptable (I was thinking in terms of f(x)=... types of functions :lol:

) 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

f: (x,y,z) \mapsto (\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}, \sqrt{x^2+y^2}-1)

such that

f^{-1}: (x,y,z) \mapsto (x(z+1), y(z+1), 0)

\therefore

The given annulus

A

is homeomorphic the cylinder

C

.

We can generalise the result such that any annulus

A = \{ (x,y) \in \mathbb{R}^2 : a \leq x^2+y^2 \leq b \}

is homeomorphic to any cylinder

C = \{ (x,y,z) \in \mathbb{R}^3 : x^2+y^2=c, d \leq z \leq e \}

, where

a,b,c,d,e \in \mathbb{R}

.

Let

f: (x,y,z) \mapsto (\frac{x \sqrt{c}}{\sqrt{x^2+y^2}},\frac{y \sqrt{c}}{\sqrt{x^2+y^2}},\frac{(\sqrt{x^2+y^2}-\sqrt{a})(e-d)}{\sqrt{b}-\sqrt{a}}+d)

such that

f^{-1} : (x,y,z) \mapsto (\frac{x}{\sqrt{c}} \left( \frac{(z-d)(\sqrt{b}-\sqrt{a})}{e-d}+\sqrt{a} \right), \frac{y}{\sqrt{c}} \left( \frac{(z-d)(\sqrt{b}-\sqrt{a})}{e-d}+\sqrt{a} \right), 0)

.

Note that there are, in fact, two (or more?) ways to map an annulus on to a cylinder. We can let

f: (x,y,z) \mapsto (\frac{x \sqrt{c}}{\sqrt{x^2+y^2}},\frac{y \sqrt{c}}{\sqrt{x^2+y^2}},\frac{(\sqrt{b}-\sqrt{x^2+y^2})(e-d)}{\sqrt{b}-\sqrt{a}}+d)

such that

f^{-1} : (x,y,z) \mapsto (\frac{x}{\sqrt{c}} \left( \sqrt{b}-\frac{(z-d)(\sqrt{b}-\sqrt{a})}{e-d}\right), \frac{y}{\sqrt{c}} \left( \sqrt{b}- \frac{(z-d)(\sqrt{b}-\sqrt{a})}{e-d}\right), 0)

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.

\square

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 :smile:

Reply 1758

FireGarden

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 :smile:

Kind of. That is a pretty standard "first question on homeomorphism" problem.

Something harder would be:

Define

C = S^1 \times [0,1]

and let

((x,y),t) \sim ((x',y'),t') \iff t=t'=0

t=t'=1

. Show that

C/\sim

is homeomorphic to the sphere,

S^2

.

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 sphere-like!)

Reply 1759

Jkn

Original post by FireGarden

Kind of. That is a pretty standard "first question on homeomorphism" problem.

Something harder would be:

Define

C = S^1 \times [0,1]

and let

((x,y),t) \sim ((x',y'),t') \iff t=t'=0

t=t'=1

. Show that

C/\sim

is homeomorphic to the sphere,

S^2

*notation barrier* :lol:

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)