Problem: proving that function is constant [on hold]

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Let $f colon mathbb{R} to mathbb{R} $ be a continuous function so that $ f(x) in mathbb I = mathbb Rsetminus mathbb Q, forall x in mathbb{R}$. Prove that $f$ is constant.

I tried assuming it's not a constant but I can't get to a contradiction with continuity.

functions continuity

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edited 11 hours ago

Surb

37k94375

asked 23 hours ago

user560461

525

put on hold as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R 9 hours ago

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." – user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What's $mathbb{I}$?
  – Dante Grevino
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanteGrevino it's set of irrational numbers
  – user560461
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
  – John Douma
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
  – Eric Duminil
  20 hours ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

1

Let $f colon mathbb{R} to mathbb{R} $ be a continuous function so that $ f(x) in mathbb I = mathbb Rsetminus mathbb Q, forall x in mathbb{R}$. Prove that $f$ is constant.

I tried assuming it's not a constant but I can't get to a contradiction with continuity.

functions continuity

share|cite|improve this question

edited 11 hours ago

Surb

37k94375

asked 23 hours ago

user560461

525

put on hold as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R 9 hours ago

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." – user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What's $mathbb{I}$?
  – Dante Grevino
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanteGrevino it's set of irrational numbers
  – user560461
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
  – John Douma
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
  – Eric Duminil
  20 hours ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

Let $f colon mathbb{R} to mathbb{R} $ be a continuous function so that $ f(x) in mathbb I = mathbb Rsetminus mathbb Q, forall x in mathbb{R}$. Prove that $f$ is constant.

I tried assuming it's not a constant but I can't get to a contradiction with continuity.

functions continuity

share|cite|improve this question

edited 11 hours ago

Surb

37k94375

asked 23 hours ago

user560461

525

Let $f colon mathbb{R} to mathbb{R} $ be a continuous function so that $ f(x) in mathbb I = mathbb Rsetminus mathbb Q, forall x in mathbb{R}$. Prove that $f$ is constant.

I tried assuming it's not a constant but I can't get to a contradiction with continuity.

functions continuity

functions continuity

share|cite|improve this question

edited 11 hours ago

Surb

37k94375

asked 23 hours ago

user560461

525

share|cite|improve this question

edited 11 hours ago

Surb

37k94375

asked 23 hours ago

user560461

525

share|cite|improve this question

share|cite|improve this question

edited 11 hours ago

Surb

37k94375

edited 11 hours ago

Surb

37k94375

edited 11 hours ago

Surb

37k94375

37k94375

asked 23 hours ago

user560461

525

asked 23 hours ago

user560461

525

asked 23 hours ago

user560461

525

525

put on hold as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R 9 hours ago

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." – user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R

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

put on hold as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R 9 hours ago

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." – user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What's $mathbb{I}$?
  – Dante Grevino
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanteGrevino it's set of irrational numbers
  – user560461
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
  – John Douma
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
  – Eric Duminil
  20 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What's $mathbb{I}$?
  – Dante Grevino
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanteGrevino it's set of irrational numbers
  – user560461
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
  – John Douma
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
  – Eric Duminil
  20 hours ago
  

  
  
  
  

2

2

What's $mathbb{I}$?
– Dante Grevino
23 hours ago

What's $mathbb{I}$?
– Dante Grevino
23 hours ago

@DanteGrevino it's set of irrational numbers
– user560461
23 hours ago

@DanteGrevino it's set of irrational numbers
– user560461
23 hours ago

5

5

Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
– John Douma
23 hours ago

Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
– John Douma
23 hours ago

Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
– Eric Duminil
20 hours ago

Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
– Eric Duminil
20 hours ago

add a comment | 

4 Answers
4

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up vote
10
down vote

Assuming we know 1. The Intermediate Value Theorem and 2. that between any two irrationals there is a rational, then the proof by contradiction should work fine.

If $f(a)<f(b)$ then there has to be a rational number $r$ with $f(a)<r<f(b).$ By IVT, there is a $c$ such that $f(c)=r$, contradiction.

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answered 23 hours ago

B. Goddard

17.9k21340

add a comment | 

up vote
9
down vote

The image $f(mathbb{R})$ is path-connected and the only non-empty path-connected subspaces of $mathbb{I}$ are the points. So $f$ is constant.

share|cite|improve this answer

answered 23 hours ago

Dante Grevino

1463

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Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment | 

up vote
2
down vote

The continuous image of a connected set is connected.

The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.

Done.

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answered 17 hours ago

MPW

29.4k11856

add a comment | 

up vote
2
down vote

Suppose that $f$ is not constant. Then there are $i_1, i_2 in mathbb I$ such that $i_1 ne i_2$ and $i_1,i_2 in f(mathbb R)$. WLOG we can assume that $i_1 <i_2$. Now pick some rational number in $(i_1, i_2)$. The intermediate value theorem gives $r in f(mathbb R)$, a contradiction.

share|cite|improve this answer

edited 17 hours ago

MPW

29.4k11856

answered 23 hours ago

Fred

41.9k1642

add a comment | 

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
10
down vote

Assuming we know 1. The Intermediate Value Theorem and 2. that between any two irrationals there is a rational, then the proof by contradiction should work fine.

If $f(a)<f(b)$ then there has to be a rational number $r$ with $f(a)<r<f(b).$ By IVT, there is a $c$ such that $f(c)=r$, contradiction.

share|cite|improve this answer

answered 23 hours ago

B. Goddard

17.9k21340

add a comment | 

up vote
10
down vote

Assuming we know 1. The Intermediate Value Theorem and 2. that between any two irrationals there is a rational, then the proof by contradiction should work fine.

If $f(a)<f(b)$ then there has to be a rational number $r$ with $f(a)<r<f(b).$ By IVT, there is a $c$ such that $f(c)=r$, contradiction.

share|cite|improve this answer

answered 23 hours ago

B. Goddard

17.9k21340

add a comment | 

up vote
10
down vote

up vote
10
down vote

Assuming we know 1. The Intermediate Value Theorem and 2. that between any two irrationals there is a rational, then the proof by contradiction should work fine.

If $f(a)<f(b)$ then there has to be a rational number $r$ with $f(a)<r<f(b).$ By IVT, there is a $c$ such that $f(c)=r$, contradiction.

share|cite|improve this answer

answered 23 hours ago

B. Goddard

17.9k21340

Assuming we know 1. The Intermediate Value Theorem and 2. that between any two irrationals there is a rational, then the proof by contradiction should work fine.

If $f(a)<f(b)$ then there has to be a rational number $r$ with $f(a)<r<f(b).$ By IVT, there is a $c$ such that $f(c)=r$, contradiction.

share|cite|improve this answer

answered 23 hours ago

B. Goddard

17.9k21340

share|cite|improve this answer

share|cite|improve this answer

answered 23 hours ago

B. Goddard

17.9k21340

answered 23 hours ago

B. Goddard

17.9k21340

answered 23 hours ago

B. Goddard

17.9k21340

17.9k21340

add a comment | 

add a comment | 

up vote
9
down vote

The image $f(mathbb{R})$ is path-connected and the only non-empty path-connected subspaces of $mathbb{I}$ are the points. So $f$ is constant.

share|cite|improve this answer

answered 23 hours ago

Dante Grevino

1463

New contributor

Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
9
down vote

The image $f(mathbb{R})$ is path-connected and the only non-empty path-connected subspaces of $mathbb{I}$ are the points. So $f$ is constant.

share|cite|improve this answer

answered 23 hours ago

Dante Grevino

1463

New contributor

Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
9
down vote

up vote
9
down vote

The image $f(mathbb{R})$ is path-connected and the only non-empty path-connected subspaces of $mathbb{I}$ are the points. So $f$ is constant.

share|cite|improve this answer

answered 23 hours ago

Dante Grevino

1463

New contributor

Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

The image $f(mathbb{R})$ is path-connected and the only non-empty path-connected subspaces of $mathbb{I}$ are the points. So $f$ is constant.

share|cite|improve this answer

answered 23 hours ago

Dante Grevino

1463

New contributor

Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this answer

share|cite|improve this answer

answered 23 hours ago

Dante Grevino

1463

New contributor

Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 23 hours ago

Dante Grevino

1463

answered 23 hours ago

Dante Grevino

1463

1463

New contributor

Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

up vote
2
down vote

The continuous image of a connected set is connected.

The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.

Done.

share|cite|improve this answer

answered 17 hours ago

MPW

29.4k11856

add a comment | 

up vote
2
down vote

The continuous image of a connected set is connected.

The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.

Done.

share|cite|improve this answer

answered 17 hours ago

MPW

29.4k11856

add a comment | 

up vote
2
down vote

up vote
2
down vote

The continuous image of a connected set is connected.

The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.

Done.

share|cite|improve this answer

answered 17 hours ago

MPW

29.4k11856

The continuous image of a connected set is connected.

The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.

Done.

share|cite|improve this answer

answered 17 hours ago

MPW

29.4k11856

share|cite|improve this answer

share|cite|improve this answer

answered 17 hours ago

MPW

29.4k11856

answered 17 hours ago

MPW

29.4k11856

answered 17 hours ago

MPW

29.4k11856

29.4k11856

add a comment | 

add a comment | 

up vote
2
down vote

Suppose that $f$ is not constant. Then there are $i_1, i_2 in mathbb I$ such that $i_1 ne i_2$ and $i_1,i_2 in f(mathbb R)$. WLOG we can assume that $i_1 <i_2$. Now pick some rational number in $(i_1, i_2)$. The intermediate value theorem gives $r in f(mathbb R)$, a contradiction.

share|cite|improve this answer

edited 17 hours ago

MPW

29.4k11856

answered 23 hours ago

Fred

41.9k1642

add a comment | 

up vote
2
down vote

Suppose that $f$ is not constant. Then there are $i_1, i_2 in mathbb I$ such that $i_1 ne i_2$ and $i_1,i_2 in f(mathbb R)$. WLOG we can assume that $i_1 <i_2$. Now pick some rational number in $(i_1, i_2)$. The intermediate value theorem gives $r in f(mathbb R)$, a contradiction.

share|cite|improve this answer

edited 17 hours ago

MPW

29.4k11856

answered 23 hours ago

Fred

41.9k1642

add a comment | 

up vote
2
down vote

up vote
2
down vote

Suppose that $f$ is not constant. Then there are $i_1, i_2 in mathbb I$ such that $i_1 ne i_2$ and $i_1,i_2 in f(mathbb R)$. WLOG we can assume that $i_1 <i_2$. Now pick some rational number in $(i_1, i_2)$. The intermediate value theorem gives $r in f(mathbb R)$, a contradiction.

share|cite|improve this answer

edited 17 hours ago

MPW

29.4k11856

answered 23 hours ago

Fred

41.9k1642

Suppose that $f$ is not constant. Then there are $i_1, i_2 in mathbb I$ such that $i_1 ne i_2$ and $i_1,i_2 in f(mathbb R)$. WLOG we can assume that $i_1 <i_2$. Now pick some rational number in $(i_1, i_2)$. The intermediate value theorem gives $r in f(mathbb R)$, a contradiction.

share|cite|improve this answer

edited 17 hours ago

MPW

29.4k11856

answered 23 hours ago

Fred

41.9k1642

share|cite|improve this answer

share|cite|improve this answer

edited 17 hours ago

MPW

29.4k11856

edited 17 hours ago

MPW

29.4k11856

edited 17 hours ago

MPW

29.4k11856

29.4k11856

answered 23 hours ago

Fred

41.9k1642

answered 23 hours ago

Fred

41.9k1642

answered 23 hours ago

Fred

41.9k1642

41.9k1642

add a comment | 

add a comment | 

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