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given a continuous function from $f:mathbb{Q}to mathbb{Q}$ [duplicate]
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This question already has an answer here:
Does every continuous map from $mathbb{Q}$ to $mathbb{Q}$ extends continuously as a map from $mathbb{R}$ to $mathbb{R}$?
3 answers
Given a continuous function from $f:mathbb{Q}to mathbb{Q}$ there exists a continuous function $g:mathbb{R}to mathbb{R}$ such that g restricts to f on $mathbb{Q}$
I could think of one example which cannot be extended namely $f(x)=1 if x> sqrt{2} $ and $f(x)=0 if x< sqrt{2} .$
I am interested in knowing more examples. I am looking for different kind of examples other than mine. The reason why my example works because the left limit and right limit of $f$ at $sqrt{2}$ do not match. Are there different kinds of examples which exploit some other property of discontinuous functions or of totally different nature?
This gives only one example I am interested in knowing other examples and is there any clasification of such examples?
real-analysis continuity examples-counterexamples
edited Feb 1 at 23:42
José Carlos Santos
174k23133242
asked Feb 1 at 21:26
user345777user345777
432312
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marked as duplicate by Leucippus, Cesareo, Lord Shark the Unknown, rtybase, José Carlos Santos real-analysis
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Feb 2 at 10:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Does every continuous map from $mathbb{Q}$ to $mathbb{Q}$ extends continuously as a map from $mathbb{R}$ to $mathbb{R}$?
3 answers
Given a continuous function from $f:mathbb{Q}to mathbb{Q}$ there exists a continuous function $g:mathbb{R}to mathbb{R}$ such that g restricts to f on $mathbb{Q}$
I could think of one example which cannot be extended namely $f(x)=1 if x> sqrt{2} $ and $f(x)=0 if x< sqrt{2} .$
I am interested in knowing more examples. I am looking for different kind of examples other than mine. The reason why my example works because the left limit and right limit of $f$ at $sqrt{2}$ do not match. Are there different kinds of examples which exploit some other property of discontinuous functions or of totally different nature?
This gives only one example I am interested in knowing other examples and is there any clasification of such examples?
real-analysis continuity examples-counterexamples
edited Feb 1 at 23:42
José Carlos Santos
174k23133242
asked Feb 1 at 21:26
user345777user345777
432312
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marked as duplicate by Leucippus, Cesareo, Lord Shark the Unknown, rtybase, José Carlos Santos real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Feb 2 at 10:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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I have edited the question ffffforall, do you still think it is duplicate?
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– user345777
Feb 1 at 21:50
add a comment |
$begingroup$
This question already has an answer here:
Does every continuous map from $mathbb{Q}$ to $mathbb{Q}$ extends continuously as a map from $mathbb{R}$ to $mathbb{R}$?
3 answers
Given a continuous function from $f:mathbb{Q}to mathbb{Q}$ there exists a continuous function $g:mathbb{R}to mathbb{R}$ such that g restricts to f on $mathbb{Q}$
I could think of one example which cannot be extended namely $f(x)=1 if x> sqrt{2} $ and $f(x)=0 if x< sqrt{2} .$
I am interested in knowing more examples. I am looking for different kind of examples other than mine. The reason why my example works because the left limit and right limit of $f$ at $sqrt{2}$ do not match. Are there different kinds of examples which exploit some other property of discontinuous functions or of totally different nature?
This gives only one example I am interested in knowing other examples and is there any clasification of such examples?
real-analysis continuity examples-counterexamples
edited Feb 1 at 23:42
José Carlos Santos
174k23133242
asked Feb 1 at 21:26
user345777user345777
432312
$endgroup$
This question already has an answer here:
Does every continuous map from $mathbb{Q}$ to $mathbb{Q}$ extends continuously as a map from $mathbb{R}$ to $mathbb{R}$?
3 answers
Given a continuous function from $f:mathbb{Q}to mathbb{Q}$ there exists a continuous function $g:mathbb{R}to mathbb{R}$ such that g restricts to f on $mathbb{Q}$
I could think of one example which cannot be extended namely $f(x)=1 if x> sqrt{2} $ and $f(x)=0 if x< sqrt{2} .$
I am interested in knowing more examples. I am looking for different kind of examples other than mine. The reason why my example works because the left limit and right limit of $f$ at $sqrt{2}$ do not match. Are there different kinds of examples which exploit some other property of discontinuous functions or of totally different nature?
This gives only one example I am interested in knowing other examples and is there any clasification of such examples?
This question already has an answer here:
Does every continuous map from $mathbb{Q}$ to $mathbb{Q}$ extends continuously as a map from $mathbb{R}$ to $mathbb{R}$?
3 answers
real-analysis continuity examples-counterexamples
real-analysis continuity examples-counterexamples
edited Feb 1 at 23:42
José Carlos Santos
174k23133242
asked Feb 1 at 21:26
user345777user345777
432312
edited Feb 1 at 23:42
José Carlos Santos
174k23133242
asked Feb 1 at 21:26
user345777user345777
432312
edited Feb 1 at 23:42
José Carlos Santos
174k23133242
edited Feb 1 at 23:42
José Carlos Santos
174k23133242
edited Feb 1 at 23:42
José Carlos Santos
174k23133242
174k23133242
asked Feb 1 at 21:26
user345777user345777
432312
asked Feb 1 at 21:26
user345777user345777
432312
asked Feb 1 at 21:26
user345777user345777
432312
432312
marked as duplicate by Leucippus, Cesareo, Lord Shark the Unknown, rtybase, José Carlos Santos real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Feb 2 at 10:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Leucippus, Cesareo, Lord Shark the Unknown, rtybase, José Carlos Santos real-analysis
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Feb 2 at 10:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
I have edited the question ffffforall, do you still think it is duplicate?
$endgroup$
– user345777
Feb 1 at 21:50
add a comment |
$begingroup$
I have edited the question ffffforall, do you still think it is duplicate?
$endgroup$
– user345777
Feb 1 at 21:50
$begingroup$
I have edited the question ffffforall, do you still think it is duplicate?
$endgroup$
– user345777
Feb 1 at 21:50
$begingroup$
I have edited the question ffffforall, do you still think it is duplicate?
$endgroup$
– user345777
Feb 1 at 21:50
add a comment |
1 Answer
1
active
oldest
votes
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Your example is fine. More generaly, no continuous function from $mathbb{Q}$ into itself whose range is finite and it has more than one element can be extended to a continuous function from $mathbb{R}$ to itself.
On the other hand, every uniformly continuous function from $mathbb{Q}$ into itself can be extended to a continuous function from $mathbb{R}$ to itself.
answered Feb 1 at 21:33
José Carlos SantosJosé Carlos Santos
174k23133242
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$begingroup$
Could you please explain why every uniformly continuous function can be extended?
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– user345777
Feb 1 at 21:45
$begingroup$
I could, but there is already an answer here.
$endgroup$
– José Carlos Santos
Feb 1 at 21:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your example is fine. More generaly, no continuous function from $mathbb{Q}$ into itself whose range is finite and it has more than one element can be extended to a continuous function from $mathbb{R}$ to itself.
On the other hand, every uniformly continuous function from $mathbb{Q}$ into itself can be extended to a continuous function from $mathbb{R}$ to itself.
answered Feb 1 at 21:33
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
$begingroup$
Could you please explain why every uniformly continuous function can be extended?
$endgroup$
– user345777
Feb 1 at 21:45
$begingroup$
I could, but there is already an answer here.
$endgroup$
– José Carlos Santos
Feb 1 at 21:56
add a comment |
$begingroup$
Your example is fine. More generaly, no continuous function from $mathbb{Q}$ into itself whose range is finite and it has more than one element can be extended to a continuous function from $mathbb{R}$ to itself.
On the other hand, every uniformly continuous function from $mathbb{Q}$ into itself can be extended to a continuous function from $mathbb{R}$ to itself.
answered Feb 1 at 21:33
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
$begingroup$
Could you please explain why every uniformly continuous function can be extended?
$endgroup$
– user345777
Feb 1 at 21:45
$begingroup$
I could, but there is already an answer here.
$endgroup$
– José Carlos Santos
Feb 1 at 21:56
add a comment |
$begingroup$
Your example is fine. More generaly, no continuous function from $mathbb{Q}$ into itself whose range is finite and it has more than one element can be extended to a continuous function from $mathbb{R}$ to itself.
On the other hand, every uniformly continuous function from $mathbb{Q}$ into itself can be extended to a continuous function from $mathbb{R}$ to itself.
answered Feb 1 at 21:33
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
Your example is fine. More generaly, no continuous function from $mathbb{Q}$ into itself whose range is finite and it has more than one element can be extended to a continuous function from $mathbb{R}$ to itself.
On the other hand, every uniformly continuous function from $mathbb{Q}$ into itself can be extended to a continuous function from $mathbb{R}$ to itself.
answered Feb 1 at 21:33
José Carlos SantosJosé Carlos Santos
174k23133242
answered Feb 1 at 21:33
José Carlos SantosJosé Carlos Santos
174k23133242
answered Feb 1 at 21:33
José Carlos SantosJosé Carlos Santos
174k23133242
answered Feb 1 at 21:33
José Carlos SantosJosé Carlos Santos
174k23133242
174k23133242
$begingroup$
Could you please explain why every uniformly continuous function can be extended?
$endgroup$
– user345777
Feb 1 at 21:45
$begingroup$
I could, but there is already an answer here.
$endgroup$
– José Carlos Santos
Feb 1 at 21:56
add a comment |
$begingroup$
Could you please explain why every uniformly continuous function can be extended?
$endgroup$
– user345777
Feb 1 at 21:45
$begingroup$
I could, but there is already an answer here.
$endgroup$
– José Carlos Santos
Feb 1 at 21:56
$begingroup$
Could you please explain why every uniformly continuous function can be extended?
$endgroup$
– user345777
Feb 1 at 21:45
$begingroup$
Could you please explain why every uniformly continuous function can be extended?
$endgroup$
– user345777
Feb 1 at 21:45
$begingroup$
I could, but there is already an answer here.
$endgroup$
– José Carlos Santos
Feb 1 at 21:56
$begingroup$
I could, but there is already an answer here.
$endgroup$
– José Carlos Santos
Feb 1 at 21:56
add a comment |
$begingroup$
I have edited the question ffffforall, do you still think it is duplicate?
$endgroup$
– user345777
Feb 1 at 21:50