Composition of Darboux functions [closed]
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If $f$ and $g$ are two Darboux functions, is their composition also a Darboux function? I believe it is, but I can't prove it.
real-analysis
asked Dec 17 '18 at 18:48
JustAnAmateurJustAnAmateur
825
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closed as off-topic by Saad, mrtaurho, RRL, Ben, José Carlos Santos Dec 18 '18 at 8:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
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$begingroup$
If $f$ and $g$ are two Darboux functions, is their composition also a Darboux function? I believe it is, but I can't prove it.
real-analysis
asked Dec 17 '18 at 18:48
JustAnAmateurJustAnAmateur
825
$endgroup$
closed as off-topic by Saad, mrtaurho, RRL, Ben, José Carlos Santos Dec 18 '18 at 8:26
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, mrtaurho, RRL, Ben, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
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Could you elaborate on why you believe it is and what you tried to prove it?
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– eyeballfrog
Dec 18 '18 at 5:28
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$begingroup$
If $f$ and $g$ are two Darboux functions, is their composition also a Darboux function? I believe it is, but I can't prove it.
real-analysis
asked Dec 17 '18 at 18:48
JustAnAmateurJustAnAmateur
825
$endgroup$
If $f$ and $g$ are two Darboux functions, is their composition also a Darboux function? I believe it is, but I can't prove it.
real-analysis
real-analysis
asked Dec 17 '18 at 18:48
JustAnAmateurJustAnAmateur
825
asked Dec 17 '18 at 18:48
JustAnAmateurJustAnAmateur
825
asked Dec 17 '18 at 18:48
JustAnAmateurJustAnAmateur
825
asked Dec 17 '18 at 18:48
JustAnAmateurJustAnAmateur
825
asked Dec 17 '18 at 18:48
JustAnAmateurJustAnAmateur
825
825
closed as off-topic by Saad, mrtaurho, RRL, Ben, José Carlos Santos Dec 18 '18 at 8:26
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, mrtaurho, RRL, Ben, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, mrtaurho, RRL, Ben, José Carlos Santos Dec 18 '18 at 8:26
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, mrtaurho, RRL, Ben, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
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Could you elaborate on why you believe it is and what you tried to prove it?
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– eyeballfrog
Dec 18 '18 at 5:28
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Could you elaborate on why you believe it is and what you tried to prove it?
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– eyeballfrog
Dec 18 '18 at 5:28
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Could you elaborate on why you believe it is and what you tried to prove it?
$endgroup$
– eyeballfrog
Dec 18 '18 at 5:28
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Could you elaborate on why you believe it is and what you tried to prove it?
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– eyeballfrog
Dec 18 '18 at 5:28
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2 Answers
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Let $a<b$ and $f(g(a))<t<f(g(b))$.
$f$ is Darboux, so there is $g(a)<s<g(b)$ such that $f(s)=t$. (Assuming $g(a)<g(b)$, otherwise swap)
$g$ is Darboux, so there is $a<r<b$ such that $g(r)=s$.
But then we have $a<r<b$ and $f(g(r))=t$.
answered Dec 17 '18 at 19:09
FedericoFederico
4,919514
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Hint: By definition, for a Darboux function the image of an interval is an interval.
answered Dec 17 '18 at 19:52
ir7ir7
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $a<b$ and $f(g(a))<t<f(g(b))$.
$f$ is Darboux, so there is $g(a)<s<g(b)$ such that $f(s)=t$. (Assuming $g(a)<g(b)$, otherwise swap)
$g$ is Darboux, so there is $a<r<b$ such that $g(r)=s$.
But then we have $a<r<b$ and $f(g(r))=t$.
answered Dec 17 '18 at 19:09
FedericoFederico
4,919514
$endgroup$
add a comment |
$begingroup$
Let $a<b$ and $f(g(a))<t<f(g(b))$.
$f$ is Darboux, so there is $g(a)<s<g(b)$ such that $f(s)=t$. (Assuming $g(a)<g(b)$, otherwise swap)
$g$ is Darboux, so there is $a<r<b$ such that $g(r)=s$.
But then we have $a<r<b$ and $f(g(r))=t$.
answered Dec 17 '18 at 19:09
FedericoFederico
4,919514
$endgroup$
add a comment |
$begingroup$
Let $a<b$ and $f(g(a))<t<f(g(b))$.
$f$ is Darboux, so there is $g(a)<s<g(b)$ such that $f(s)=t$. (Assuming $g(a)<g(b)$, otherwise swap)
$g$ is Darboux, so there is $a<r<b$ such that $g(r)=s$.
But then we have $a<r<b$ and $f(g(r))=t$.
answered Dec 17 '18 at 19:09
FedericoFederico
4,919514
$endgroup$
Let $a<b$ and $f(g(a))<t<f(g(b))$.
$f$ is Darboux, so there is $g(a)<s<g(b)$ such that $f(s)=t$. (Assuming $g(a)<g(b)$, otherwise swap)
$g$ is Darboux, so there is $a<r<b$ such that $g(r)=s$.
But then we have $a<r<b$ and $f(g(r))=t$.
answered Dec 17 '18 at 19:09
FedericoFederico
4,919514
answered Dec 17 '18 at 19:09
FedericoFederico
4,919514
answered Dec 17 '18 at 19:09
FedericoFederico
4,919514
answered Dec 17 '18 at 19:09
FedericoFederico
4,919514
4,919514
add a comment |
add a comment |
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Hint: By definition, for a Darboux function the image of an interval is an interval.
answered Dec 17 '18 at 19:52
ir7ir7
4,13311015
$endgroup$
add a comment |
$begingroup$
Hint: By definition, for a Darboux function the image of an interval is an interval.
answered Dec 17 '18 at 19:52
ir7ir7
4,13311015
$endgroup$
add a comment |
$begingroup$
Hint: By definition, for a Darboux function the image of an interval is an interval.
answered Dec 17 '18 at 19:52
ir7ir7
4,13311015
$endgroup$
Hint: By definition, for a Darboux function the image of an interval is an interval.
answered Dec 17 '18 at 19:52
ir7ir7
4,13311015
answered Dec 17 '18 at 19:52
ir7ir7
4,13311015
answered Dec 17 '18 at 19:52
ir7ir7
4,13311015
answered Dec 17 '18 at 19:52
ir7ir7
4,13311015
4,13311015
add a comment |
add a comment |
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Could you elaborate on why you believe it is and what you tried to prove it?
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– eyeballfrog
Dec 18 '18 at 5:28