If $f:mathbb{R}tomathbb{R}$ is continuous, and $f(mathbb{Q})subsetmathbb{N}$, then is $f$ unbounded,...
$begingroup$
Suppose $f:mathbb{R}tomathbb{R}$ is a continuous function such that $ f(mathbb{Q})subsetmathbb{N}$. What can be said about $f$? Is it (1) unbounded, (2) constant, or (3) non-constant bounded?
I think it should be constant, as $mathbb{Q}$ is dense in $mathbb{R}$.
Kindly help! Thanks & Regards
real-analysis functions
edited Jan 22 at 19:17
Blue
48.6k870156
asked Jan 22 at 18:51
Devendra Singh RanaDevendra Singh Rana
7751416
$endgroup$
add a comment |
$begingroup$
Suppose $f:mathbb{R}tomathbb{R}$ is a continuous function such that $ f(mathbb{Q})subsetmathbb{N}$. What can be said about $f$? Is it (1) unbounded, (2) constant, or (3) non-constant bounded?
I think it should be constant, as $mathbb{Q}$ is dense in $mathbb{R}$.
Kindly help! Thanks & Regards
real-analysis functions
edited Jan 22 at 19:17
Blue
48.6k870156
asked Jan 22 at 18:51
Devendra Singh RanaDevendra Singh Rana
7751416
$endgroup$
3
$begingroup$
You are right .
$endgroup$
– Hagen von Eitzen
Jan 22 at 18:54
1
$begingroup$
Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
$endgroup$
– Sangchul Lee
Jan 22 at 19:22
add a comment |
$begingroup$
Suppose $f:mathbb{R}tomathbb{R}$ is a continuous function such that $ f(mathbb{Q})subsetmathbb{N}$. What can be said about $f$? Is it (1) unbounded, (2) constant, or (3) non-constant bounded?
I think it should be constant, as $mathbb{Q}$ is dense in $mathbb{R}$.
Kindly help! Thanks & Regards
real-analysis functions
edited Jan 22 at 19:17
Blue
48.6k870156
asked Jan 22 at 18:51
Devendra Singh RanaDevendra Singh Rana
7751416
$endgroup$
Suppose $f:mathbb{R}tomathbb{R}$ is a continuous function such that $ f(mathbb{Q})subsetmathbb{N}$. What can be said about $f$? Is it (1) unbounded, (2) constant, or (3) non-constant bounded?
I think it should be constant, as $mathbb{Q}$ is dense in $mathbb{R}$.
Kindly help! Thanks & Regards
real-analysis functions
real-analysis functions
edited Jan 22 at 19:17
Blue
48.6k870156
asked Jan 22 at 18:51
Devendra Singh RanaDevendra Singh Rana
7751416
edited Jan 22 at 19:17
Blue
48.6k870156
asked Jan 22 at 18:51
Devendra Singh RanaDevendra Singh Rana
7751416
edited Jan 22 at 19:17
Blue
48.6k870156
edited Jan 22 at 19:17
Blue
48.6k870156
edited Jan 22 at 19:17
Blue
48.6k870156
48.6k870156
asked Jan 22 at 18:51
Devendra Singh RanaDevendra Singh Rana
7751416
asked Jan 22 at 18:51
Devendra Singh RanaDevendra Singh Rana
7751416
asked Jan 22 at 18:51
Devendra Singh RanaDevendra Singh Rana
7751416
7751416
3
$begingroup$
You are right .
$endgroup$
– Hagen von Eitzen
Jan 22 at 18:54
1
$begingroup$
Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
$endgroup$
– Sangchul Lee
Jan 22 at 19:22
add a comment |
3
$begingroup$
You are right .
$endgroup$
– Hagen von Eitzen
Jan 22 at 18:54
1
$begingroup$
Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
$endgroup$
– Sangchul Lee
Jan 22 at 19:22
3
3
$begingroup$
You are right .
$endgroup$
– Hagen von Eitzen
Jan 22 at 18:54
$begingroup$
You are right .
$endgroup$
– Hagen von Eitzen
Jan 22 at 18:54
1
1
$begingroup$
Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
$endgroup$
– Sangchul Lee
Jan 22 at 19:22
$begingroup$
Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
$endgroup$
– Sangchul Lee
Jan 22 at 19:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, $f$ must be a constant. Suppose not. Pick $q_1, q_2 in mathbb{Q}$ such that $f(q_1) > f(q_2)$. Without loss of generality, assume $q_1 < q_2$. Pick any noninteger $y$ between $f(q_1)$ and $f(q_2)$. Then by intermediate value theorem, there exists $x in (q_1, q_2)$ such that $f(x) = y$. Any rational number sufficiently close to $x$ should not take integer value, a contradiction.
answered Jan 22 at 19:08
user295959user295959
664310
$endgroup$
add a comment |
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$begingroup$
Yes, $f$ must be a constant. Suppose not. Pick $q_1, q_2 in mathbb{Q}$ such that $f(q_1) > f(q_2)$. Without loss of generality, assume $q_1 < q_2$. Pick any noninteger $y$ between $f(q_1)$ and $f(q_2)$. Then by intermediate value theorem, there exists $x in (q_1, q_2)$ such that $f(x) = y$. Any rational number sufficiently close to $x$ should not take integer value, a contradiction.
answered Jan 22 at 19:08
user295959user295959
664310
$endgroup$
add a comment |
$begingroup$
Yes, $f$ must be a constant. Suppose not. Pick $q_1, q_2 in mathbb{Q}$ such that $f(q_1) > f(q_2)$. Without loss of generality, assume $q_1 < q_2$. Pick any noninteger $y$ between $f(q_1)$ and $f(q_2)$. Then by intermediate value theorem, there exists $x in (q_1, q_2)$ such that $f(x) = y$. Any rational number sufficiently close to $x$ should not take integer value, a contradiction.
answered Jan 22 at 19:08
user295959user295959
664310
$endgroup$
add a comment |
$begingroup$
Yes, $f$ must be a constant. Suppose not. Pick $q_1, q_2 in mathbb{Q}$ such that $f(q_1) > f(q_2)$. Without loss of generality, assume $q_1 < q_2$. Pick any noninteger $y$ between $f(q_1)$ and $f(q_2)$. Then by intermediate value theorem, there exists $x in (q_1, q_2)$ such that $f(x) = y$. Any rational number sufficiently close to $x$ should not take integer value, a contradiction.
answered Jan 22 at 19:08
user295959user295959
664310
$endgroup$
Yes, $f$ must be a constant. Suppose not. Pick $q_1, q_2 in mathbb{Q}$ such that $f(q_1) > f(q_2)$. Without loss of generality, assume $q_1 < q_2$. Pick any noninteger $y$ between $f(q_1)$ and $f(q_2)$. Then by intermediate value theorem, there exists $x in (q_1, q_2)$ such that $f(x) = y$. Any rational number sufficiently close to $x$ should not take integer value, a contradiction.
answered Jan 22 at 19:08
user295959user295959
664310
answered Jan 22 at 19:08
user295959user295959
664310
answered Jan 22 at 19:08
user295959user295959
664310
answered Jan 22 at 19:08
user295959user295959
664310
664310
add a comment |
add a comment |
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3
$begingroup$
You are right .
$endgroup$
– Hagen von Eitzen
Jan 22 at 18:54
1
$begingroup$
Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
$endgroup$
– Sangchul Lee
Jan 22 at 19:22