Problem: proving that function is constant [closed]
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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
edited Nov 20 at 0:14
Surb
37.1k94375
asked Nov 19 at 12:24
user560461
525
closed as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R Nov 20 at 2:58
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.
add a comment |
up vote
4
down vote
favorite
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
edited Nov 20 at 0:14
Surb
37.1k94375
asked Nov 19 at 12:24
user560461
525
closed as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R Nov 20 at 2:58
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
Nov 19 at 12:26
@DanteGrevino it's set of irrational numbers
– user560461
Nov 19 at 12:27
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
Nov 19 at 12:29
1
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
Nov 19 at 15:50
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
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
edited Nov 20 at 0:14
Surb
37.1k94375
asked Nov 19 at 12:24
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
edited Nov 20 at 0:14
Surb
37.1k94375
asked Nov 19 at 12:24
user560461
525
edited Nov 20 at 0:14
Surb
37.1k94375
asked Nov 19 at 12:24
user560461
525
edited Nov 20 at 0:14
Surb
37.1k94375
edited Nov 20 at 0:14
Surb
37.1k94375
edited Nov 20 at 0:14
Surb
37.1k94375
37.1k94375
asked Nov 19 at 12:24
user560461
525
asked Nov 19 at 12:24
user560461
525
asked Nov 19 at 12:24
user560461
525
525
closed as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R Nov 20 at 2:58
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.
closed as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R Nov 20 at 2:58
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
Nov 19 at 12:26
@DanteGrevino it's set of irrational numbers
– user560461
Nov 19 at 12:27
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
Nov 19 at 12:29
1
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
Nov 19 at 15:50
add a comment |
2
What's $mathbb{I}$?
– Dante Grevino
Nov 19 at 12:26
@DanteGrevino it's set of irrational numbers
– user560461
Nov 19 at 12:27
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
Nov 19 at 12:29
1
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
Nov 19 at 15:50
2
2
What's $mathbb{I}$?
– Dante Grevino
Nov 19 at 12:26
What's $mathbb{I}$?
– Dante Grevino
Nov 19 at 12:26
@DanteGrevino it's set of irrational numbers
– user560461
Nov 19 at 12:27
@DanteGrevino it's set of irrational numbers
– user560461
Nov 19 at 12:27
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
Nov 19 at 12:29
Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
– John Douma
Nov 19 at 12:29
1
1
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
Nov 19 at 15:50
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
Nov 19 at 15:50
add a comment |
4 Answers
4
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oldest
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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.
answered Nov 19 at 12:31
B. Goddard
18k21340
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up vote
9
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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.
answered Nov 19 at 12:34
Dante Grevino
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2
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The continuous image of a connected set is connected.
The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.
Done.
answered Nov 19 at 18:06
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.
edited Nov 19 at 18:19
MPW
29.4k11856
answered Nov 19 at 12:31
Fred
42.2k1642
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.
answered Nov 19 at 12:31
B. Goddard
18k21340
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.
answered Nov 19 at 12:31
B. Goddard
18k21340
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.
answered Nov 19 at 12:31
B. Goddard
18k21340
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.
answered Nov 19 at 12:31
B. Goddard
18k21340
answered Nov 19 at 12:31
B. Goddard
18k21340
answered Nov 19 at 12:31
B. Goddard
18k21340
answered Nov 19 at 12:31
B. Goddard
18k21340
18k21340
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.
answered Nov 19 at 12:34
Dante Grevino
3565
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.
answered Nov 19 at 12:34
Dante Grevino
3565
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.
answered Nov 19 at 12:34
Dante Grevino
3565
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.
answered Nov 19 at 12:34
Dante Grevino
3565
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 Nov 19 at 12:34
Dante Grevino
3565
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 Nov 19 at 12:34
Dante Grevino
3565
answered Nov 19 at 12:34
Dante Grevino
3565
3565
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.
answered Nov 19 at 18:06
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.
answered Nov 19 at 18:06
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.
answered Nov 19 at 18:06
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.
answered Nov 19 at 18:06
MPW
29.4k11856
answered Nov 19 at 18:06
MPW
29.4k11856
answered Nov 19 at 18:06
MPW
29.4k11856
answered Nov 19 at 18:06
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.
edited Nov 19 at 18:19
MPW
29.4k11856
answered Nov 19 at 12:31
Fred
42.2k1642
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.
edited Nov 19 at 18:19
MPW
29.4k11856
answered Nov 19 at 12:31
Fred
42.2k1642
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.
edited Nov 19 at 18:19
MPW
29.4k11856
answered Nov 19 at 12:31
Fred
42.2k1642
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.
edited Nov 19 at 18:19
MPW
29.4k11856
answered Nov 19 at 12:31
Fred
42.2k1642
edited Nov 19 at 18:19
MPW
29.4k11856
edited Nov 19 at 18:19
MPW
29.4k11856
edited Nov 19 at 18:19
MPW
29.4k11856
29.4k11856
answered Nov 19 at 12:31
Fred
42.2k1642
answered Nov 19 at 12:31
Fred
42.2k1642
answered Nov 19 at 12:31
Fred
42.2k1642
42.2k1642
add a comment |
add a comment |
2
What's $mathbb{I}$?
– Dante Grevino
Nov 19 at 12:26
@DanteGrevino it's set of irrational numbers
– user560461
Nov 19 at 12:27
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
Nov 19 at 12:29
1
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
Nov 19 at 15:50