Is there any easy proof of Brouwer's fixed point theorem? [closed]

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I am taking a course in game theory. For proving the Nash equilibrium we require Brouwer's fixed point theorem. But I have not taken a topology course so I am finding the proof difficult to understand. You may explain the same Brouwer's in little easy way.

general-topology game-theory nash-equilibrium

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edited Aug 25 at 13:57

Jendrik Stelzner

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asked Aug 25 at 13:44

RockDock

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closed as off-topic by amWhy, Xander Henderson, Jendrik Stelzner, user91500, Shailesh Aug 26 at 9:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Jendrik Stelzner, user91500, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  There's an easy proof based on Sperner's lemma
  – saulspatz
  Aug 25 at 13:48
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Which proof do you find difficult to understand?
  – Michael Greinecker♦
  Aug 25 at 15:14
  

  
  
  
  

add a comment | 

up vote
5
down vote

favorite

1

I am taking a course in game theory. For proving the Nash equilibrium we require Brouwer's fixed point theorem. But I have not taken a topology course so I am finding the proof difficult to understand. You may explain the same Brouwer's in little easy way.

general-topology game-theory nash-equilibrium

share|cite|improve this question

edited Aug 25 at 13:57

Jendrik Stelzner

7,63121037

asked Aug 25 at 13:44

RockDock

262

closed as off-topic by amWhy, Xander Henderson, Jendrik Stelzner, user91500, Shailesh Aug 26 at 9:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Jendrik Stelzner, user91500, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  There's an easy proof based on Sperner's lemma
  – saulspatz
  Aug 25 at 13:48
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Which proof do you find difficult to understand?
  – Michael Greinecker♦
  Aug 25 at 15:14
  

  
  
  
  

add a comment | 

up vote
5
down vote

favorite

1

up vote
5
down vote

favorite

1

1

I am taking a course in game theory. For proving the Nash equilibrium we require Brouwer's fixed point theorem. But I have not taken a topology course so I am finding the proof difficult to understand. You may explain the same Brouwer's in little easy way.

general-topology game-theory nash-equilibrium

share|cite|improve this question

edited Aug 25 at 13:57

Jendrik Stelzner

7,63121037

asked Aug 25 at 13:44

RockDock

262

I am taking a course in game theory. For proving the Nash equilibrium we require Brouwer's fixed point theorem. But I have not taken a topology course so I am finding the proof difficult to understand. You may explain the same Brouwer's in little easy way.

general-topology game-theory nash-equilibrium

share|cite|improve this question

edited Aug 25 at 13:57

Jendrik Stelzner

7,63121037

asked Aug 25 at 13:44

RockDock

262

share|cite|improve this question

share|cite|improve this question

edited Aug 25 at 13:57

Jendrik Stelzner

7,63121037

edited Aug 25 at 13:57

Jendrik Stelzner

7,63121037

edited Aug 25 at 13:57

Jendrik Stelzner

7,63121037

7,63121037

asked Aug 25 at 13:44

RockDock

262

asked Aug 25 at 13:44

RockDock

262

asked Aug 25 at 13:44

RockDock

262

262

closed as off-topic by amWhy, Xander Henderson, Jendrik Stelzner, user91500, Shailesh Aug 26 at 9:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Jendrik Stelzner, user91500, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Xander Henderson, Jendrik Stelzner, user91500, Shailesh Aug 26 at 9:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Jendrik Stelzner, user91500, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  There's an easy proof based on Sperner's lemma
  – saulspatz
  Aug 25 at 13:48
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Which proof do you find difficult to understand?
  – Michael Greinecker♦
  Aug 25 at 15:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  There's an easy proof based on Sperner's lemma
  – saulspatz
  Aug 25 at 13:48
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Which proof do you find difficult to understand?
  – Michael Greinecker♦
  Aug 25 at 15:14
  

  
  
  
  

5

5

There's an easy proof based on Sperner's lemma
– saulspatz
Aug 25 at 13:48

There's an easy proof based on Sperner's lemma
– saulspatz
Aug 25 at 13:48

2

2

Which proof do you find difficult to understand?
– Michael Greinecker♦
Aug 25 at 15:14

Which proof do you find difficult to understand?
– Michael Greinecker♦
Aug 25 at 15:14

add a comment | 

5 Answers
5

active

oldest

votes

up vote
4
down vote

Of course "easy" has a large degree of subjectivity to it. But there is a famous proof due to Emanuel Sperner that was (and still is) striking for minimizing the amount of topology needed.

These course notes of Jacob Fox give an exposition using Sperner's Lemma. The document starts from scratch and proves Brouwer's Theorem (for all $n$) in just under 2.5 pages.

share|cite|improve this answer

edited Aug 25 at 21:25

answered Aug 25 at 20:26

Pete L. Clark

78.9k9157307

add a comment | 

up vote
3
down vote

I think so. Hirsch's proof, which is in Guillemin and Pollack.

It goes (something like) this:

Suppose $p:D^2to D^2$ doesn't have a fixed point. Define $r:D^2to S^1$ by $r(x)$ is the point on the segment from $x$ to $p(x)$ on $S^1$. $r$ would be a retraction of the disk onto the circle. This is impossible.

share|cite|improve this answer

answered Aug 25 at 14:00

Chris Custer

6,2752622

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Though this proof might be geometrically intuitive, regular point theorem, Sard's theorem and classification of $1$-manifold are not easy at all.
  – Cave Johnson
  Aug 25 at 14:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. I'm assuming you know they have different homology, say.
  – Chris Custer
  Aug 25 at 14:08
  

  
  
  
  

add a comment | 

up vote
3
down vote

Aigner gives an elementary proof for the case $n = 2$ in Proofs from the BOOK.

But if I remember correctly you need Brouwer's fixed point theorem for arbitrary $n$ in the proof of Nash.

saulspatz has a source for this (and the general case) in his comment to the question.

share|cite|improve this answer

edited Aug 25 at 15:41

answered Aug 25 at 15:09

red_trumpet

523216

add a comment | 

up vote
2
down vote

You could have a look at John Milnor's take. His methods are all quite elementary (the most difficult prerequisite needed here is the well-known-but-rarely-proved change of variables theorem). However, in my opinion this proof is indeed quite mysterious and John himself seems to agree with that.

share|cite|improve this answer

answered Aug 25 at 16:12

Thibaut Demaerel

476311

add a comment | 

up vote
1
down vote

Topological Spaces From Distance To Nighborhood by Gerard Buskes and Arnoud van Rooij provide a nice 'intuitive proof' (Section~4.21). It is not exact, but it has plenty of diagrams and explanations.

share|cite|improve this answer

answered Aug 25 at 13:49

Bill Wallis

2,2361826

add a comment | 

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

Of course "easy" has a large degree of subjectivity to it. But there is a famous proof due to Emanuel Sperner that was (and still is) striking for minimizing the amount of topology needed.

These course notes of Jacob Fox give an exposition using Sperner's Lemma. The document starts from scratch and proves Brouwer's Theorem (for all $n$) in just under 2.5 pages.

share|cite|improve this answer

edited Aug 25 at 21:25

answered Aug 25 at 20:26

Pete L. Clark

78.9k9157307

add a comment | 

up vote
4
down vote

Of course "easy" has a large degree of subjectivity to it. But there is a famous proof due to Emanuel Sperner that was (and still is) striking for minimizing the amount of topology needed.

These course notes of Jacob Fox give an exposition using Sperner's Lemma. The document starts from scratch and proves Brouwer's Theorem (for all $n$) in just under 2.5 pages.

share|cite|improve this answer

edited Aug 25 at 21:25

answered Aug 25 at 20:26

Pete L. Clark

78.9k9157307

add a comment | 

up vote
4
down vote

up vote
4
down vote

Of course "easy" has a large degree of subjectivity to it. But there is a famous proof due to Emanuel Sperner that was (and still is) striking for minimizing the amount of topology needed.

These course notes of Jacob Fox give an exposition using Sperner's Lemma. The document starts from scratch and proves Brouwer's Theorem (for all $n$) in just under 2.5 pages.

share|cite|improve this answer

edited Aug 25 at 21:25

answered Aug 25 at 20:26

Pete L. Clark

78.9k9157307

Of course "easy" has a large degree of subjectivity to it. But there is a famous proof due to Emanuel Sperner that was (and still is) striking for minimizing the amount of topology needed.

These course notes of Jacob Fox give an exposition using Sperner's Lemma. The document starts from scratch and proves Brouwer's Theorem (for all $n$) in just under 2.5 pages.

share|cite|improve this answer

edited Aug 25 at 21:25

answered Aug 25 at 20:26

Pete L. Clark

78.9k9157307

share|cite|improve this answer

share|cite|improve this answer

edited Aug 25 at 21:25

edited Aug 25 at 21:25

edited Aug 25 at 21:25

answered Aug 25 at 20:26

Pete L. Clark

78.9k9157307

answered Aug 25 at 20:26

Pete L. Clark

78.9k9157307

answered Aug 25 at 20:26

Pete L. Clark

78.9k9157307

78.9k9157307

add a comment | 

add a comment | 

up vote
3
down vote

I think so. Hirsch's proof, which is in Guillemin and Pollack.

It goes (something like) this:

Suppose $p:D^2to D^2$ doesn't have a fixed point. Define $r:D^2to S^1$ by $r(x)$ is the point on the segment from $x$ to $p(x)$ on $S^1$. $r$ would be a retraction of the disk onto the circle. This is impossible.

share|cite|improve this answer

answered Aug 25 at 14:00

Chris Custer

6,2752622

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Though this proof might be geometrically intuitive, regular point theorem, Sard's theorem and classification of $1$-manifold are not easy at all.
  – Cave Johnson
  Aug 25 at 14:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. I'm assuming you know they have different homology, say.
  – Chris Custer
  Aug 25 at 14:08
  

  
  
  
  

add a comment | 

up vote
3
down vote

I think so. Hirsch's proof, which is in Guillemin and Pollack.

It goes (something like) this:

Suppose $p:D^2to D^2$ doesn't have a fixed point. Define $r:D^2to S^1$ by $r(x)$ is the point on the segment from $x$ to $p(x)$ on $S^1$. $r$ would be a retraction of the disk onto the circle. This is impossible.

share|cite|improve this answer

answered Aug 25 at 14:00

Chris Custer

6,2752622

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Though this proof might be geometrically intuitive, regular point theorem, Sard's theorem and classification of $1$-manifold are not easy at all.
  – Cave Johnson
  Aug 25 at 14:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. I'm assuming you know they have different homology, say.
  – Chris Custer
  Aug 25 at 14:08
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

I think so. Hirsch's proof, which is in Guillemin and Pollack.

It goes (something like) this:

Suppose $p:D^2to D^2$ doesn't have a fixed point. Define $r:D^2to S^1$ by $r(x)$ is the point on the segment from $x$ to $p(x)$ on $S^1$. $r$ would be a retraction of the disk onto the circle. This is impossible.

share|cite|improve this answer

answered Aug 25 at 14:00

Chris Custer

6,2752622

I think so. Hirsch's proof, which is in Guillemin and Pollack.

It goes (something like) this:

Suppose $p:D^2to D^2$ doesn't have a fixed point. Define $r:D^2to S^1$ by $r(x)$ is the point on the segment from $x$ to $p(x)$ on $S^1$. $r$ would be a retraction of the disk onto the circle. This is impossible.

share|cite|improve this answer

answered Aug 25 at 14:00

Chris Custer

6,2752622

share|cite|improve this answer

share|cite|improve this answer

answered Aug 25 at 14:00

Chris Custer

6,2752622

answered Aug 25 at 14:00

Chris Custer

6,2752622

answered Aug 25 at 14:00

Chris Custer

6,2752622

6,2752622

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Though this proof might be geometrically intuitive, regular point theorem, Sard's theorem and classification of $1$-manifold are not easy at all.
  – Cave Johnson
  Aug 25 at 14:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. I'm assuming you know they have different homology, say.
  – Chris Custer
  Aug 25 at 14:08
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Though this proof might be geometrically intuitive, regular point theorem, Sard's theorem and classification of $1$-manifold are not easy at all.
  – Cave Johnson
  Aug 25 at 14:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. I'm assuming you know they have different homology, say.
  – Chris Custer
  Aug 25 at 14:08
  

  
  
  
  

3

3

Though this proof might be geometrically intuitive, regular point theorem, Sard's theorem and classification of $1$-manifold are not easy at all.
– Cave Johnson
Aug 25 at 14:06

Though this proof might be geometrically intuitive, regular point theorem, Sard's theorem and classification of $1$-manifold are not easy at all.
– Cave Johnson
Aug 25 at 14:06

Yes. I'm assuming you know they have different homology, say.
– Chris Custer
Aug 25 at 14:08

Yes. I'm assuming you know they have different homology, say.
– Chris Custer
Aug 25 at 14:08

add a comment | 

up vote
3
down vote

Aigner gives an elementary proof for the case $n = 2$ in Proofs from the BOOK.

But if I remember correctly you need Brouwer's fixed point theorem for arbitrary $n$ in the proof of Nash.

saulspatz has a source for this (and the general case) in his comment to the question.

share|cite|improve this answer

edited Aug 25 at 15:41

answered Aug 25 at 15:09

red_trumpet

523216

add a comment | 

up vote
3
down vote

Aigner gives an elementary proof for the case $n = 2$ in Proofs from the BOOK.

But if I remember correctly you need Brouwer's fixed point theorem for arbitrary $n$ in the proof of Nash.

saulspatz has a source for this (and the general case) in his comment to the question.

share|cite|improve this answer

edited Aug 25 at 15:41

answered Aug 25 at 15:09

red_trumpet

523216

add a comment | 

up vote
3
down vote

up vote
3
down vote

Aigner gives an elementary proof for the case $n = 2$ in Proofs from the BOOK.

But if I remember correctly you need Brouwer's fixed point theorem for arbitrary $n$ in the proof of Nash.

saulspatz has a source for this (and the general case) in his comment to the question.

share|cite|improve this answer

edited Aug 25 at 15:41

answered Aug 25 at 15:09

red_trumpet

523216

Aigner gives an elementary proof for the case $n = 2$ in Proofs from the BOOK.

But if I remember correctly you need Brouwer's fixed point theorem for arbitrary $n$ in the proof of Nash.

saulspatz has a source for this (and the general case) in his comment to the question.

share|cite|improve this answer

edited Aug 25 at 15:41

answered Aug 25 at 15:09

red_trumpet

523216

share|cite|improve this answer

share|cite|improve this answer

edited Aug 25 at 15:41

edited Aug 25 at 15:41

edited Aug 25 at 15:41

answered Aug 25 at 15:09

red_trumpet

523216

answered Aug 25 at 15:09

red_trumpet

523216

answered Aug 25 at 15:09

red_trumpet

523216

523216

add a comment | 

add a comment | 

up vote
2
down vote

You could have a look at John Milnor's take. His methods are all quite elementary (the most difficult prerequisite needed here is the well-known-but-rarely-proved change of variables theorem). However, in my opinion this proof is indeed quite mysterious and John himself seems to agree with that.

share|cite|improve this answer

answered Aug 25 at 16:12

Thibaut Demaerel

476311

add a comment | 

up vote
2
down vote

You could have a look at John Milnor's take. His methods are all quite elementary (the most difficult prerequisite needed here is the well-known-but-rarely-proved change of variables theorem). However, in my opinion this proof is indeed quite mysterious and John himself seems to agree with that.

share|cite|improve this answer

answered Aug 25 at 16:12

Thibaut Demaerel

476311

add a comment | 

up vote
2
down vote

up vote
2
down vote

You could have a look at John Milnor's take. His methods are all quite elementary (the most difficult prerequisite needed here is the well-known-but-rarely-proved change of variables theorem). However, in my opinion this proof is indeed quite mysterious and John himself seems to agree with that.

share|cite|improve this answer

answered Aug 25 at 16:12

Thibaut Demaerel

476311

You could have a look at John Milnor's take. His methods are all quite elementary (the most difficult prerequisite needed here is the well-known-but-rarely-proved change of variables theorem). However, in my opinion this proof is indeed quite mysterious and John himself seems to agree with that.

share|cite|improve this answer

answered Aug 25 at 16:12

Thibaut Demaerel

476311

share|cite|improve this answer

share|cite|improve this answer

answered Aug 25 at 16:12

Thibaut Demaerel

476311

answered Aug 25 at 16:12

Thibaut Demaerel

476311

answered Aug 25 at 16:12

Thibaut Demaerel

476311

476311

add a comment | 

add a comment | 

up vote
1
down vote

Topological Spaces From Distance To Nighborhood by Gerard Buskes and Arnoud van Rooij provide a nice 'intuitive proof' (Section~4.21). It is not exact, but it has plenty of diagrams and explanations.

share|cite|improve this answer

answered Aug 25 at 13:49

Bill Wallis

2,2361826

add a comment | 

up vote
1
down vote

Topological Spaces From Distance To Nighborhood by Gerard Buskes and Arnoud van Rooij provide a nice 'intuitive proof' (Section~4.21). It is not exact, but it has plenty of diagrams and explanations.

share|cite|improve this answer

answered Aug 25 at 13:49

Bill Wallis

2,2361826

add a comment | 

up vote
1
down vote

up vote
1
down vote

Topological Spaces From Distance To Nighborhood by Gerard Buskes and Arnoud van Rooij provide a nice 'intuitive proof' (Section~4.21). It is not exact, but it has plenty of diagrams and explanations.

share|cite|improve this answer

answered Aug 25 at 13:49

Bill Wallis

2,2361826

Topological Spaces From Distance To Nighborhood by Gerard Buskes and Arnoud van Rooij provide a nice 'intuitive proof' (Section~4.21). It is not exact, but it has plenty of diagrams and explanations.

share|cite|improve this answer

answered Aug 25 at 13:49

Bill Wallis

2,2361826

share|cite|improve this answer

share|cite|improve this answer

answered Aug 25 at 13:49

Bill Wallis

2,2361826

answered Aug 25 at 13:49

Bill Wallis

2,2361826

answered Aug 25 at 13:49

Bill Wallis

2,2361826

2,2361826

add a comment | 

add a comment |