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Construct a real valued function which takes zero on integers and such that image of function is not closed.
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I am trying to construct a continuous real valued function $f:mathbbRto mathbbR$ which takes zero on all integer points(that is $f(k)=0$ for all $kin mathbbZ$) and Image(f) is not closed in $mathbbR$
I had $f(x)=sin(pi x) $ in mind. But image of $f(x)$ is closed. 
I have a feeling that we can use some clever idea to modify this function such that it satisfy our given condition.
real-analysis general-topology functions continuity
edited 5 mins ago
Parcly Taxel
36.3k136994
asked 57 mins ago
StammeringMathematician
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up vote
2
down vote
favorite
I am trying to construct a continuous real valued function $f:mathbbRto mathbbR$ which takes zero on all integer points(that is $f(k)=0$ for all $kin mathbbZ$) and Image(f) is not closed in $mathbbR$
I had $f(x)=sin(pi x) $ in mind. But image of $f(x)$ is closed.
I have a feeling that we can use some clever idea to modify this function such that it satisfy our given condition.
real-analysis general-topology functions continuity
edited 5 mins ago
Parcly Taxel
36.3k136994
asked 57 mins ago
StammeringMathematician
1,652219
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to construct a continuous real valued function $f:mathbbRto mathbbR$ which takes zero on all integer points(that is $f(k)=0$ for all $kin mathbbZ$) and Image(f) is not closed in $mathbbR$
I had $f(x)=sin(pi x) $ in mind. But image of $f(x)$ is closed.
I have a feeling that we can use some clever idea to modify this function such that it satisfy our given condition.
real-analysis general-topology functions continuity
edited 5 mins ago
Parcly Taxel
36.3k136994
asked 57 mins ago
StammeringMathematician
1,652219
I am trying to construct a continuous real valued function $f:mathbbRto mathbbR$ which takes zero on all integer points(that is $f(k)=0$ for all $kin mathbbZ$) and Image(f) is not closed in $mathbbR$
I had $f(x)=sin(pi x) $ in mind. But image of $f(x)$ is closed.
I have a feeling that we can use some clever idea to modify this function such that it satisfy our given condition.
real-analysis general-topology functions continuity
real-analysis general-topology functions continuity
edited 5 mins ago
Parcly Taxel
36.3k136994
asked 57 mins ago
StammeringMathematician
1,652219
edited 5 mins ago
Parcly Taxel
36.3k136994
asked 57 mins ago
StammeringMathematician
1,652219
edited 5 mins ago
Parcly Taxel
36.3k136994
edited 5 mins ago
Parcly Taxel
36.3k136994
edited 5 mins ago
Parcly Taxel
36.3k136994
36.3k136994
asked 57 mins ago
StammeringMathematician
1,652219
asked 57 mins ago
StammeringMathematician
1,652219
asked 57 mins ago
StammeringMathematician
1,652219
1,652219
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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up vote
3
down vote
For each $n neq 0$ draw the triangle with vertices $(n,0),(n+1,0)$ and $(n+frac 1 2, 1-frac 1 n)$. You will immediately see how to construct an example. [You will get a function whose range contains $1-frac 1 n$ for each $n$ but does not contain $1$].
answered 46 mins ago
Kavi Rama Murthy
31.5k31543
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up vote
3
down vote
If $f$ verifies the desired propery, its restriction $f|_[n,n+1]$ gives a continuous function on $[n,n+1]$ that is zero on the edges of the interval, for any $n in mathbbZ$. Reciprocally, if we have $f_n : [n,n+1] to mathbbR$ continuous with $f_n(n) = f_n(n+1) = 0$ for each integer $n$, by the gluing lemma this gives a continuous function $f: mathbbR to mathbbR$ with $f(n) = f_n(n) = 0$. This means that we can approach the problem somewhat 'locally', i.e. we can fix an interval $[n,n+1]$. Now, the function
$$
f_n (t) = mu_nsin(pi(t-n))
$$
takes values on $mu_n[-1,1] = [-mu_n,mu_n]$ and $f_n(n) = f_n(n+1) = 0$. Thus, the family $(f_n)_n$ induces a continuous function $f$ that vanishes at $mathbbZ$ and
$$
f(mathbbR) = bigcup_n in mathbbZ f_n([n,n+1]) = bigcup_n in mathbbZ[-mu_n,mu_n]
$$
so the problem reduces to choosing a sequence $(mu_n)_n$ so that the former union is open. One possible choice is $mu_n = 1-frac1n$ so that
$$
f(mathbbR) = bigcup_nin mathbbZ[-mu_n,mu_n] = bigcup_nin mathbbN[-1+frac1n,1-frac1n] = (-1,1).
$$
answered 33 mins ago
Guido A.
5,5761728
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
For each $n neq 0$ draw the triangle with vertices $(n,0),(n+1,0)$ and $(n+frac 1 2, 1-frac 1 n)$. You will immediately see how to construct an example. [You will get a function whose range contains $1-frac 1 n$ for each $n$ but does not contain $1$].
answered 46 mins ago
Kavi Rama Murthy
31.5k31543
add a comment |Â
up vote
3
down vote
For each $n neq 0$ draw the triangle with vertices $(n,0),(n+1,0)$ and $(n+frac 1 2, 1-frac 1 n)$. You will immediately see how to construct an example. [You will get a function whose range contains $1-frac 1 n$ for each $n$ but does not contain $1$].
answered 46 mins ago
Kavi Rama Murthy
31.5k31543
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For each $n neq 0$ draw the triangle with vertices $(n,0),(n+1,0)$ and $(n+frac 1 2, 1-frac 1 n)$. You will immediately see how to construct an example. [You will get a function whose range contains $1-frac 1 n$ for each $n$ but does not contain $1$].
answered 46 mins ago
Kavi Rama Murthy
31.5k31543
For each $n neq 0$ draw the triangle with vertices $(n,0),(n+1,0)$ and $(n+frac 1 2, 1-frac 1 n)$. You will immediately see how to construct an example. [You will get a function whose range contains $1-frac 1 n$ for each $n$ but does not contain $1$].
answered 46 mins ago
Kavi Rama Murthy
31.5k31543
answered 46 mins ago
Kavi Rama Murthy
31.5k31543
answered 46 mins ago
Kavi Rama Murthy
31.5k31543
answered 46 mins ago
Kavi Rama Murthy
31.5k31543
31.5k31543
add a comment |Â
add a comment |Â
up vote
3
down vote
If $f$ verifies the desired propery, its restriction $f|_[n,n+1]$ gives a continuous function on $[n,n+1]$ that is zero on the edges of the interval, for any $n in mathbbZ$. Reciprocally, if we have $f_n : [n,n+1] to mathbbR$ continuous with $f_n(n) = f_n(n+1) = 0$ for each integer $n$, by the gluing lemma this gives a continuous function $f: mathbbR to mathbbR$ with $f(n) = f_n(n) = 0$. This means that we can approach the problem somewhat 'locally', i.e. we can fix an interval $[n,n+1]$. Now, the function
$$
f_n (t) = mu_nsin(pi(t-n))
$$
takes values on $mu_n[-1,1] = [-mu_n,mu_n]$ and $f_n(n) = f_n(n+1) = 0$. Thus, the family $(f_n)_n$ induces a continuous function $f$ that vanishes at $mathbbZ$ and
$$
f(mathbbR) = bigcup_n in mathbbZ f_n([n,n+1]) = bigcup_n in mathbbZ[-mu_n,mu_n]
$$
so the problem reduces to choosing a sequence $(mu_n)_n$ so that the former union is open. One possible choice is $mu_n = 1-frac1n$ so that
$$
f(mathbbR) = bigcup_nin mathbbZ[-mu_n,mu_n] = bigcup_nin mathbbN[-1+frac1n,1-frac1n] = (-1,1).
$$
answered 33 mins ago
Guido A.
5,5761728
add a comment |Â
up vote
3
down vote
If $f$ verifies the desired propery, its restriction $f|_[n,n+1]$ gives a continuous function on $[n,n+1]$ that is zero on the edges of the interval, for any $n in mathbbZ$. Reciprocally, if we have $f_n : [n,n+1] to mathbbR$ continuous with $f_n(n) = f_n(n+1) = 0$ for each integer $n$, by the gluing lemma this gives a continuous function $f: mathbbR to mathbbR$ with $f(n) = f_n(n) = 0$. This means that we can approach the problem somewhat 'locally', i.e. we can fix an interval $[n,n+1]$. Now, the function
$$
f_n (t) = mu_nsin(pi(t-n))
$$
takes values on $mu_n[-1,1] = [-mu_n,mu_n]$ and $f_n(n) = f_n(n+1) = 0$. Thus, the family $(f_n)_n$ induces a continuous function $f$ that vanishes at $mathbbZ$ and
$$
f(mathbbR) = bigcup_n in mathbbZ f_n([n,n+1]) = bigcup_n in mathbbZ[-mu_n,mu_n]
$$
so the problem reduces to choosing a sequence $(mu_n)_n$ so that the former union is open. One possible choice is $mu_n = 1-frac1n$ so that
$$
f(mathbbR) = bigcup_nin mathbbZ[-mu_n,mu_n] = bigcup_nin mathbbN[-1+frac1n,1-frac1n] = (-1,1).
$$
answered 33 mins ago
Guido A.
5,5761728
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If $f$ verifies the desired propery, its restriction $f|_[n,n+1]$ gives a continuous function on $[n,n+1]$ that is zero on the edges of the interval, for any $n in mathbbZ$. Reciprocally, if we have $f_n : [n,n+1] to mathbbR$ continuous with $f_n(n) = f_n(n+1) = 0$ for each integer $n$, by the gluing lemma this gives a continuous function $f: mathbbR to mathbbR$ with $f(n) = f_n(n) = 0$. This means that we can approach the problem somewhat 'locally', i.e. we can fix an interval $[n,n+1]$. Now, the function
$$
f_n (t) = mu_nsin(pi(t-n))
$$
takes values on $mu_n[-1,1] = [-mu_n,mu_n]$ and $f_n(n) = f_n(n+1) = 0$. Thus, the family $(f_n)_n$ induces a continuous function $f$ that vanishes at $mathbbZ$ and
$$
f(mathbbR) = bigcup_n in mathbbZ f_n([n,n+1]) = bigcup_n in mathbbZ[-mu_n,mu_n]
$$
so the problem reduces to choosing a sequence $(mu_n)_n$ so that the former union is open. One possible choice is $mu_n = 1-frac1n$ so that
$$
f(mathbbR) = bigcup_nin mathbbZ[-mu_n,mu_n] = bigcup_nin mathbbN[-1+frac1n,1-frac1n] = (-1,1).
$$
answered 33 mins ago
Guido A.
5,5761728
If $f$ verifies the desired propery, its restriction $f|_[n,n+1]$ gives a continuous function on $[n,n+1]$ that is zero on the edges of the interval, for any $n in mathbbZ$. Reciprocally, if we have $f_n : [n,n+1] to mathbbR$ continuous with $f_n(n) = f_n(n+1) = 0$ for each integer $n$, by the gluing lemma this gives a continuous function $f: mathbbR to mathbbR$ with $f(n) = f_n(n) = 0$. This means that we can approach the problem somewhat 'locally', i.e. we can fix an interval $[n,n+1]$. Now, the function
$$
f_n (t) = mu_nsin(pi(t-n))
$$
takes values on $mu_n[-1,1] = [-mu_n,mu_n]$ and $f_n(n) = f_n(n+1) = 0$. Thus, the family $(f_n)_n$ induces a continuous function $f$ that vanishes at $mathbbZ$ and
$$
f(mathbbR) = bigcup_n in mathbbZ f_n([n,n+1]) = bigcup_n in mathbbZ[-mu_n,mu_n]
$$
so the problem reduces to choosing a sequence $(mu_n)_n$ so that the former union is open. One possible choice is $mu_n = 1-frac1n$ so that
$$
f(mathbbR) = bigcup_nin mathbbZ[-mu_n,mu_n] = bigcup_nin mathbbN[-1+frac1n,1-frac1n] = (-1,1).
$$
answered 33 mins ago
Guido A.
5,5761728
edited 26 mins ago
answered 33 mins ago
Guido A.
5,5761728
answered 33 mins ago
Guido A.
5,5761728
answered 33 mins ago
Guido A.
5,5761728
5,5761728
add a comment |Â
add a comment |Â
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