$f$ is continuous on $mathbb R -{0}$, but cannot be made into a continuous function by change on a set of...
$begingroup$
Consider $fcolon mathbb R to mathbb R$ given by $$ f(x) = begin{cases} 1 & x ge 0\ 0 & x < 0end{cases} $$
$f$ is continuous on $mathbb R -{0}$, but cannot be made into a continuous function by change on a set of measure zero.
since there exists dense set of irrational numbers after removing a set of measure zero ,i can show there exists $(x_n),(y_n)$ of irrational number converges to 0, but $(f(x_n)) $ converges to 0 and $(f(y_n))$ converges to 1.Therefore, this function is discontinous at x=0(I didn't add explicit sequence but is this idea of proof correct)
real-analysis proof-verification continuity lebesgue-measure alternative-proof
asked Nov 27 '18 at 15:19
Cloud JRCloud JR
878517
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closed as off-topic by Andrés E. Caicedo, user90369, user10354138, Lord_Farin, Xander Henderson Dec 3 '18 at 20:14
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." – Andrés E. Caicedo, user90369, user10354138, Lord_Farin, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider $fcolon mathbb R to mathbb R$ given by $$ f(x) = begin{cases} 1 & x ge 0\ 0 & x < 0end{cases} $$
$f$ is continuous on $mathbb R -{0}$, but cannot be made into a continuous function by change on a set of measure zero.
since there exists dense set of irrational numbers after removing a set of measure zero ,i can show there exists $(x_n),(y_n)$ of irrational number converges to 0, but $(f(x_n)) $ converges to 0 and $(f(y_n))$ converges to 1.Therefore, this function is discontinous at x=0(I didn't add explicit sequence but is this idea of proof correct)
real-analysis proof-verification continuity lebesgue-measure alternative-proof
asked Nov 27 '18 at 15:19
Cloud JRCloud JR
878517
$endgroup$
closed as off-topic by Andrés E. Caicedo, user90369, user10354138, Lord_Farin, Xander Henderson Dec 3 '18 at 20:14
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." – Andrés E. Caicedo, user90369, user10354138, Lord_Farin, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
It is not even remotely true that all negative irrationals are zero. For your actual question: zero has nothing to do with it, in fact (and neither do irrationals): this is an entirely topological fact: the function is just two lines which don't join up.
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– user3482749
Nov 27 '18 at 15:27
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You talk of "this idea of proof" but I don't see an idea. All you said is "I think the problem occurs at 0". This is not an idea of proof, since nothing you have said sheds any light on how to argue.
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– Andrés E. Caicedo
Nov 27 '18 at 15:33
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Again, there is no argument. Obviously the function is discontinuous. So what? You have not yet begun to address the question. By the way, you accepted the answer you were given, but there is some serious work still required to turn it into a full answer, and I am not sure you know yet how, given the level of confusion your suggested attempt indicates.
$endgroup$
– Andrés E. Caicedo
Nov 27 '18 at 15:57
add a comment |
$begingroup$
Consider $fcolon mathbb R to mathbb R$ given by $$ f(x) = begin{cases} 1 & x ge 0\ 0 & x < 0end{cases} $$
$f$ is continuous on $mathbb R -{0}$, but cannot be made into a continuous function by change on a set of measure zero.
since there exists dense set of irrational numbers after removing a set of measure zero ,i can show there exists $(x_n),(y_n)$ of irrational number converges to 0, but $(f(x_n)) $ converges to 0 and $(f(y_n))$ converges to 1.Therefore, this function is discontinous at x=0(I didn't add explicit sequence but is this idea of proof correct)
real-analysis proof-verification continuity lebesgue-measure alternative-proof
asked Nov 27 '18 at 15:19
Cloud JRCloud JR
878517
$endgroup$
Consider $fcolon mathbb R to mathbb R$ given by $$ f(x) = begin{cases} 1 & x ge 0\ 0 & x < 0end{cases} $$
$f$ is continuous on $mathbb R -{0}$, but cannot be made into a continuous function by change on a set of measure zero.
since there exists dense set of irrational numbers after removing a set of measure zero ,i can show there exists $(x_n),(y_n)$ of irrational number converges to 0, but $(f(x_n)) $ converges to 0 and $(f(y_n))$ converges to 1.Therefore, this function is discontinous at x=0(I didn't add explicit sequence but is this idea of proof correct)
real-analysis proof-verification continuity lebesgue-measure alternative-proof
real-analysis proof-verification continuity lebesgue-measure alternative-proof
asked Nov 27 '18 at 15:19
Cloud JRCloud JR
878517
asked Nov 27 '18 at 15:19
Cloud JRCloud JR
878517
edited Nov 27 '18 at 15:37
Cloud JR
asked Nov 27 '18 at 15:19
Cloud JRCloud JR
878517
asked Nov 27 '18 at 15:19
Cloud JRCloud JR
878517
asked Nov 27 '18 at 15:19
Cloud JRCloud JR
878517
878517
closed as off-topic by Andrés E. Caicedo, user90369, user10354138, Lord_Farin, Xander Henderson Dec 3 '18 at 20:14
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." – Andrés E. Caicedo, user90369, user10354138, Lord_Farin, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Andrés E. Caicedo, user90369, user10354138, Lord_Farin, Xander Henderson Dec 3 '18 at 20:14
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." – Andrés E. Caicedo, user90369, user10354138, Lord_Farin, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
It is not even remotely true that all negative irrationals are zero. For your actual question: zero has nothing to do with it, in fact (and neither do irrationals): this is an entirely topological fact: the function is just two lines which don't join up.
$endgroup$
– user3482749
Nov 27 '18 at 15:27
$begingroup$
You talk of "this idea of proof" but I don't see an idea. All you said is "I think the problem occurs at 0". This is not an idea of proof, since nothing you have said sheds any light on how to argue.
$endgroup$
– Andrés E. Caicedo
Nov 27 '18 at 15:33
$begingroup$
Again, there is no argument. Obviously the function is discontinuous. So what? You have not yet begun to address the question. By the way, you accepted the answer you were given, but there is some serious work still required to turn it into a full answer, and I am not sure you know yet how, given the level of confusion your suggested attempt indicates.
$endgroup$
– Andrés E. Caicedo
Nov 27 '18 at 15:57
add a comment |
1
$begingroup$
It is not even remotely true that all negative irrationals are zero. For your actual question: zero has nothing to do with it, in fact (and neither do irrationals): this is an entirely topological fact: the function is just two lines which don't join up.
$endgroup$
– user3482749
Nov 27 '18 at 15:27
$begingroup$
You talk of "this idea of proof" but I don't see an idea. All you said is "I think the problem occurs at 0". This is not an idea of proof, since nothing you have said sheds any light on how to argue.
$endgroup$
– Andrés E. Caicedo
Nov 27 '18 at 15:33
$begingroup$
Again, there is no argument. Obviously the function is discontinuous. So what? You have not yet begun to address the question. By the way, you accepted the answer you were given, but there is some serious work still required to turn it into a full answer, and I am not sure you know yet how, given the level of confusion your suggested attempt indicates.
$endgroup$
– Andrés E. Caicedo
Nov 27 '18 at 15:57
1
1
$begingroup$
It is not even remotely true that all negative irrationals are zero. For your actual question: zero has nothing to do with it, in fact (and neither do irrationals): this is an entirely topological fact: the function is just two lines which don't join up.
$endgroup$
– user3482749
Nov 27 '18 at 15:27
$begingroup$
It is not even remotely true that all negative irrationals are zero. For your actual question: zero has nothing to do with it, in fact (and neither do irrationals): this is an entirely topological fact: the function is just two lines which don't join up.
$endgroup$
– user3482749
Nov 27 '18 at 15:27
$begingroup$
You talk of "this idea of proof" but I don't see an idea. All you said is "I think the problem occurs at 0". This is not an idea of proof, since nothing you have said sheds any light on how to argue.
$endgroup$
– Andrés E. Caicedo
Nov 27 '18 at 15:33
$begingroup$
You talk of "this idea of proof" but I don't see an idea. All you said is "I think the problem occurs at 0". This is not an idea of proof, since nothing you have said sheds any light on how to argue.
$endgroup$
– Andrés E. Caicedo
Nov 27 '18 at 15:33
$begingroup$
Again, there is no argument. Obviously the function is discontinuous. So what? You have not yet begun to address the question. By the way, you accepted the answer you were given, but there is some serious work still required to turn it into a full answer, and I am not sure you know yet how, given the level of confusion your suggested attempt indicates.
$endgroup$
– Andrés E. Caicedo
Nov 27 '18 at 15:57
$begingroup$
Again, there is no argument. Obviously the function is discontinuous. So what? You have not yet begun to address the question. By the way, you accepted the answer you were given, but there is some serious work still required to turn it into a full answer, and I am not sure you know yet how, given the level of confusion your suggested attempt indicates.
$endgroup$
– Andrés E. Caicedo
Nov 27 '18 at 15:57
add a comment |
1 Answer
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oldest
votes
$begingroup$
Your proof is, at least, incomplete. You don't explain why, after removing a set with measure $0$, you can always find such a sequence of irrational numbers.
I think that the introduction of irrational numbers in this problem is not the best approach. All you need is to use the fact if $Asubsetmathbb R$ has measure $0$, then $A^complement$ is dense.
answered Nov 27 '18 at 15:26
José Carlos SantosJosé Carlos Santos
157k22126227
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your proof is, at least, incomplete. You don't explain why, after removing a set with measure $0$, you can always find such a sequence of irrational numbers.
I think that the introduction of irrational numbers in this problem is not the best approach. All you need is to use the fact if $Asubsetmathbb R$ has measure $0$, then $A^complement$ is dense.
answered Nov 27 '18 at 15:26
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
add a comment |
$begingroup$
Your proof is, at least, incomplete. You don't explain why, after removing a set with measure $0$, you can always find such a sequence of irrational numbers.
I think that the introduction of irrational numbers in this problem is not the best approach. All you need is to use the fact if $Asubsetmathbb R$ has measure $0$, then $A^complement$ is dense.
answered Nov 27 '18 at 15:26
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
add a comment |
$begingroup$
Your proof is, at least, incomplete. You don't explain why, after removing a set with measure $0$, you can always find such a sequence of irrational numbers.
I think that the introduction of irrational numbers in this problem is not the best approach. All you need is to use the fact if $Asubsetmathbb R$ has measure $0$, then $A^complement$ is dense.
answered Nov 27 '18 at 15:26
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
Your proof is, at least, incomplete. You don't explain why, after removing a set with measure $0$, you can always find such a sequence of irrational numbers.
I think that the introduction of irrational numbers in this problem is not the best approach. All you need is to use the fact if $Asubsetmathbb R$ has measure $0$, then $A^complement$ is dense.
answered Nov 27 '18 at 15:26
José Carlos SantosJosé Carlos Santos
157k22126227
answered Nov 27 '18 at 15:26
José Carlos SantosJosé Carlos Santos
157k22126227
answered Nov 27 '18 at 15:26
José Carlos SantosJosé Carlos Santos
157k22126227
answered Nov 27 '18 at 15:26
José Carlos SantosJosé Carlos Santos
157k22126227
157k22126227
add a comment |
add a comment |
1
$begingroup$
It is not even remotely true that all negative irrationals are zero. For your actual question: zero has nothing to do with it, in fact (and neither do irrationals): this is an entirely topological fact: the function is just two lines which don't join up.
$endgroup$
– user3482749
Nov 27 '18 at 15:27
$begingroup$
You talk of "this idea of proof" but I don't see an idea. All you said is "I think the problem occurs at 0". This is not an idea of proof, since nothing you have said sheds any light on how to argue.
$endgroup$
– Andrés E. Caicedo
Nov 27 '18 at 15:33
$begingroup$
Again, there is no argument. Obviously the function is discontinuous. So what? You have not yet begun to address the question. By the way, you accepted the answer you were given, but there is some serious work still required to turn it into a full answer, and I am not sure you know yet how, given the level of confusion your suggested attempt indicates.
$endgroup$
– Andrés E. Caicedo
Nov 27 '18 at 15:57