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Construct a continuous 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
asked 1 hour ago
StammeringMathematician
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up vote
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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
asked 1 hour ago
StammeringMathematician
1,654219
I think "continuous" should be in the title.
– mr_e_man
16 mins ago
@mr_e_man Thanks for pointing it out
– StammeringMathematician
10 mins ago
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
asked 1 hour ago
StammeringMathematician
1,654219
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
asked 1 hour ago
StammeringMathematician
1,654219
asked 1 hour ago
StammeringMathematician
1,654219
edited 10 mins ago
asked 1 hour ago
StammeringMathematician
1,654219
asked 1 hour ago
StammeringMathematician
1,654219
asked 1 hour ago
StammeringMathematician
1,654219
1,654219
I think "continuous" should be in the title.
– mr_e_man
16 mins ago
@mr_e_man Thanks for pointing it out
– StammeringMathematician
10 mins ago
add a comment |Â
I think "continuous" should be in the title.
– mr_e_man
16 mins ago
@mr_e_man Thanks for pointing it out
– StammeringMathematician
10 mins ago
I think "continuous" should be in the title.
– mr_e_man
16 mins ago
I think "continuous" should be in the title.
– mr_e_man
16 mins ago
@mr_e_man Thanks for pointing it out
– StammeringMathematician
10 mins ago
@mr_e_man Thanks for pointing it out
– StammeringMathematician
10 mins ago
add a comment |Â
5 Answers
5
active
oldest
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up vote
4
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 1 hour ago
Guido A.
5,5861728
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up vote
4
down vote
Using piecewise linear functions (instead of $sin (pi x)$) makes this simpler. 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 (piecewise linear function) whose range contains $1-frac 1 n$ for each $n$ but does not contain $1$].
answered 1 hour ago
Kavi Rama Murthy
31.6k31543
add a comment |Â
up vote
4
down vote
The image of $sin(pi x)$ is closed because the peaks all reach 1 and -1. To make it open, we need the peaks to get arbitrarily close to some value, but never reach them. The easiest way is to use an amplitude modifier that asymptotes to a constant nonzero value at infinity, such as $tanh(x)$. Thus, we can use the function
$$
f(x) = sin(pi x)tanh(x),
$$
which is obviously continuous, zero at each integer, and can be easily shown to have image $(-1,1)$
answered 39 mins ago
eyeballfrog
4,902527
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up vote
2
down vote
I think you could just use ANY continuous real valued function $g:mathbbRrightarrowmathbbR$, such that $g(mathbbR)$ is not closed, and define
$$ f(x)=g(x)sin(pi x)$$
For example $f(x) = arctan(x)sin(pi x)$, defined and continuous on $mathbbR$, with $f(mathbbR)=]-fracpi2,fracpi2[$
answered 23 mins ago
Thomas Lesgourgues
635
I don't think it works with any such function, e.g. $g(x) = tfrac13 + tfrac23cdottfraccdotcos(2cdotpicdot x)$ has an open image but would give a closed image for $f$.
– leftaroundabout
10 mins ago
If you take for example $g(x) = e^x$ that has image $(0,infty)$, $f(x) := g(x)sin(pi x)$ is unbounded and vanishes at zero. Thus for each natural $N$ there exists $t > 0$ with $f(t) > N$. Now since $f(0) = 0$, by the intermediate value theorem $f([0,t])$ contains $[0,N]$ and by parity ($f$ is odd), we have that $[-N,N]$ is on the image of $f$. Thus, the image is actually the whole line, which is closed. Any other unbounded strictly positive function $g$ should fail for the same reason.
– Guido A.
3 mins ago
add a comment |Â
up vote
2
down vote
The simplest solution that would come to mind is to take that sine function and multiply it with an amplitude envolope that only approaches $1$ in the $pm$infinite limit:
$$
f(x) = frac1+2+cdotsin(picdot x)
$$
Incidentally, since you just said “not closed†but not whether it should be bounded, we could also just choose
$$
f!!!!/(!!!!/x!!!!/)!!!!/ =!!!!/ x!!!!/cdot!!!!/sin!!!!/!!!!!!!!/(pi!!!!/cdot!!!!/ x!!!!/)
$$
(The image of this is all of $mathbbR$ which is actually closed, as the commenters remarked.)
answered 28 mins ago
leftaroundabout
3,3661527
1
Isn't the image $mathbb R$ closed?
– mr_e_man
10 mins ago
Regarding your last example, isn't the image all the real line? The function is odd and unbounded, so by the intermediate value theorem the image should contain any interval $[0,n]$ (and $[-n,0]$, by parity).
– Guido A.
9 mins ago
You're right, the image of $xcdotsin(picdot x)$ is open but also closed.
– leftaroundabout
8 mins ago
Can you please tell me how make these graphs?
– StammeringMathematician
8 mins ago
1
@StammeringMathematicianplotWindow [fnPlot $ x -> (1+abs x)/(2+abs x)*sin(pi*x)]with dynamic-plot (a Haskell library).
– leftaroundabout
3 mins ago
 |Â
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
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 1 hour ago
Guido A.
5,5861728
add a comment |Â
up vote
4
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 1 hour ago
Guido A.
5,5861728
add a comment |Â
up vote
4
down vote
up vote
4
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 1 hour ago
Guido A.
5,5861728
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 1 hour ago
Guido A.
5,5861728
edited 1 hour ago
answered 1 hour ago
Guido A.
5,5861728
answered 1 hour ago
Guido A.
5,5861728
answered 1 hour ago
Guido A.
5,5861728
5,5861728
add a comment |Â
add a comment |Â
up vote
4
down vote
Using piecewise linear functions (instead of $sin (pi x)$) makes this simpler. 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 (piecewise linear function) whose range contains $1-frac 1 n$ for each $n$ but does not contain $1$].
answered 1 hour ago
Kavi Rama Murthy
31.6k31543
add a comment |Â
up vote
4
down vote
Using piecewise linear functions (instead of $sin (pi x)$) makes this simpler. 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 (piecewise linear function) whose range contains $1-frac 1 n$ for each $n$ but does not contain $1$].
answered 1 hour ago
Kavi Rama Murthy
31.6k31543
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Using piecewise linear functions (instead of $sin (pi x)$) makes this simpler. 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 (piecewise linear function) whose range contains $1-frac 1 n$ for each $n$ but does not contain $1$].
answered 1 hour ago
Kavi Rama Murthy
31.6k31543
Using piecewise linear functions (instead of $sin (pi x)$) makes this simpler. 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 (piecewise linear function) whose range contains $1-frac 1 n$ for each $n$ but does not contain $1$].
answered 1 hour ago
Kavi Rama Murthy
31.6k31543
edited 40 mins ago
answered 1 hour ago
Kavi Rama Murthy
31.6k31543
answered 1 hour ago
Kavi Rama Murthy
31.6k31543
answered 1 hour ago
Kavi Rama Murthy
31.6k31543
31.6k31543
add a comment |Â
add a comment |Â
up vote
4
down vote
The image of $sin(pi x)$ is closed because the peaks all reach 1 and -1. To make it open, we need the peaks to get arbitrarily close to some value, but never reach them. The easiest way is to use an amplitude modifier that asymptotes to a constant nonzero value at infinity, such as $tanh(x)$. Thus, we can use the function
$$
f(x) = sin(pi x)tanh(x),
$$
which is obviously continuous, zero at each integer, and can be easily shown to have image $(-1,1)$
answered 39 mins ago
eyeballfrog
4,902527
add a comment |Â
up vote
4
down vote
The image of $sin(pi x)$ is closed because the peaks all reach 1 and -1. To make it open, we need the peaks to get arbitrarily close to some value, but never reach them. The easiest way is to use an amplitude modifier that asymptotes to a constant nonzero value at infinity, such as $tanh(x)$. Thus, we can use the function
$$
f(x) = sin(pi x)tanh(x),
$$
which is obviously continuous, zero at each integer, and can be easily shown to have image $(-1,1)$
answered 39 mins ago
eyeballfrog
4,902527
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The image of $sin(pi x)$ is closed because the peaks all reach 1 and -1. To make it open, we need the peaks to get arbitrarily close to some value, but never reach them. The easiest way is to use an amplitude modifier that asymptotes to a constant nonzero value at infinity, such as $tanh(x)$. Thus, we can use the function
$$
f(x) = sin(pi x)tanh(x),
$$
which is obviously continuous, zero at each integer, and can be easily shown to have image $(-1,1)$
answered 39 mins ago
eyeballfrog
4,902527
The image of $sin(pi x)$ is closed because the peaks all reach 1 and -1. To make it open, we need the peaks to get arbitrarily close to some value, but never reach them. The easiest way is to use an amplitude modifier that asymptotes to a constant nonzero value at infinity, such as $tanh(x)$. Thus, we can use the function
$$
f(x) = sin(pi x)tanh(x),
$$
which is obviously continuous, zero at each integer, and can be easily shown to have image $(-1,1)$
answered 39 mins ago
eyeballfrog
4,902527
answered 39 mins ago
eyeballfrog
4,902527
answered 39 mins ago
eyeballfrog
4,902527
answered 39 mins ago
eyeballfrog
4,902527
4,902527
add a comment |Â
add a comment |Â
up vote
2
down vote
I think you could just use ANY continuous real valued function $g:mathbbRrightarrowmathbbR$, such that $g(mathbbR)$ is not closed, and define
$$ f(x)=g(x)sin(pi x)$$
For example $f(x) = arctan(x)sin(pi x)$, defined and continuous on $mathbbR$, with $f(mathbbR)=]-fracpi2,fracpi2[$
answered 23 mins ago
Thomas Lesgourgues
635
I don't think it works with any such function, e.g. $g(x) = tfrac13 + tfrac23cdottfraccdotcos(2cdotpicdot x)$ has an open image but would give a closed image for $f$.
– leftaroundabout
10 mins ago
If you take for example $g(x) = e^x$ that has image $(0,infty)$, $f(x) := g(x)sin(pi x)$ is unbounded and vanishes at zero. Thus for each natural $N$ there exists $t > 0$ with $f(t) > N$. Now since $f(0) = 0$, by the intermediate value theorem $f([0,t])$ contains $[0,N]$ and by parity ($f$ is odd), we have that $[-N,N]$ is on the image of $f$. Thus, the image is actually the whole line, which is closed. Any other unbounded strictly positive function $g$ should fail for the same reason.
– Guido A.
3 mins ago
add a comment |Â
up vote
2
down vote
I think you could just use ANY continuous real valued function $g:mathbbRrightarrowmathbbR$, such that $g(mathbbR)$ is not closed, and define
$$ f(x)=g(x)sin(pi x)$$
For example $f(x) = arctan(x)sin(pi x)$, defined and continuous on $mathbbR$, with $f(mathbbR)=]-fracpi2,fracpi2[$
answered 23 mins ago
Thomas Lesgourgues
635
I don't think it works with any such function, e.g. $g(x) = tfrac13 + tfrac23cdottfraccdotcos(2cdotpicdot x)$ has an open image but would give a closed image for $f$.
– leftaroundabout
10 mins ago
If you take for example $g(x) = e^x$ that has image $(0,infty)$, $f(x) := g(x)sin(pi x)$ is unbounded and vanishes at zero. Thus for each natural $N$ there exists $t > 0$ with $f(t) > N$. Now since $f(0) = 0$, by the intermediate value theorem $f([0,t])$ contains $[0,N]$ and by parity ($f$ is odd), we have that $[-N,N]$ is on the image of $f$. Thus, the image is actually the whole line, which is closed. Any other unbounded strictly positive function $g$ should fail for the same reason.
– Guido A.
3 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I think you could just use ANY continuous real valued function $g:mathbbRrightarrowmathbbR$, such that $g(mathbbR)$ is not closed, and define
$$ f(x)=g(x)sin(pi x)$$
For example $f(x) = arctan(x)sin(pi x)$, defined and continuous on $mathbbR$, with $f(mathbbR)=]-fracpi2,fracpi2[$
answered 23 mins ago
Thomas Lesgourgues
635
I think you could just use ANY continuous real valued function $g:mathbbRrightarrowmathbbR$, such that $g(mathbbR)$ is not closed, and define
$$ f(x)=g(x)sin(pi x)$$
For example $f(x) = arctan(x)sin(pi x)$, defined and continuous on $mathbbR$, with $f(mathbbR)=]-fracpi2,fracpi2[$
answered 23 mins ago
Thomas Lesgourgues
635
answered 23 mins ago
Thomas Lesgourgues
635
answered 23 mins ago
Thomas Lesgourgues
635
answered 23 mins ago
Thomas Lesgourgues
635
635
I don't think it works with any such function, e.g. $g(x) = tfrac13 + tfrac23cdottfraccdotcos(2cdotpicdot x)$ has an open image but would give a closed image for $f$.
– leftaroundabout
10 mins ago
If you take for example $g(x) = e^x$ that has image $(0,infty)$, $f(x) := g(x)sin(pi x)$ is unbounded and vanishes at zero. Thus for each natural $N$ there exists $t > 0$ with $f(t) > N$. Now since $f(0) = 0$, by the intermediate value theorem $f([0,t])$ contains $[0,N]$ and by parity ($f$ is odd), we have that $[-N,N]$ is on the image of $f$. Thus, the image is actually the whole line, which is closed. Any other unbounded strictly positive function $g$ should fail for the same reason.
– Guido A.
3 mins ago
add a comment |Â
I don't think it works with any such function, e.g. $g(x) = tfrac13 + tfrac23cdottfraccdotcos(2cdotpicdot x)$ has an open image but would give a closed image for $f$.
– leftaroundabout
10 mins ago
If you take for example $g(x) = e^x$ that has image $(0,infty)$, $f(x) := g(x)sin(pi x)$ is unbounded and vanishes at zero. Thus for each natural $N$ there exists $t > 0$ with $f(t) > N$. Now since $f(0) = 0$, by the intermediate value theorem $f([0,t])$ contains $[0,N]$ and by parity ($f$ is odd), we have that $[-N,N]$ is on the image of $f$. Thus, the image is actually the whole line, which is closed. Any other unbounded strictly positive function $g$ should fail for the same reason.
– Guido A.
3 mins ago
I don't think it works with any such function, e.g. $g(x) = tfrac13 + tfrac23cdottfraccdotcos(2cdotpicdot x)$ has an open image but would give a closed image for $f$.
– leftaroundabout
10 mins ago
I don't think it works with any such function, e.g. $g(x) = tfrac13 + tfrac23cdottfraccdotcos(2cdotpicdot x)$ has an open image but would give a closed image for $f$.
– leftaroundabout
10 mins ago
If you take for example $g(x) = e^x$ that has image $(0,infty)$, $f(x) := g(x)sin(pi x)$ is unbounded and vanishes at zero. Thus for each natural $N$ there exists $t > 0$ with $f(t) > N$. Now since $f(0) = 0$, by the intermediate value theorem $f([0,t])$ contains $[0,N]$ and by parity ($f$ is odd), we have that $[-N,N]$ is on the image of $f$. Thus, the image is actually the whole line, which is closed. Any other unbounded strictly positive function $g$ should fail for the same reason.
– Guido A.
3 mins ago
If you take for example $g(x) = e^x$ that has image $(0,infty)$, $f(x) := g(x)sin(pi x)$ is unbounded and vanishes at zero. Thus for each natural $N$ there exists $t > 0$ with $f(t) > N$. Now since $f(0) = 0$, by the intermediate value theorem $f([0,t])$ contains $[0,N]$ and by parity ($f$ is odd), we have that $[-N,N]$ is on the image of $f$. Thus, the image is actually the whole line, which is closed. Any other unbounded strictly positive function $g$ should fail for the same reason.
– Guido A.
3 mins ago
add a comment |Â
up vote
2
down vote
The simplest solution that would come to mind is to take that sine function and multiply it with an amplitude envolope that only approaches $1$ in the $pm$infinite limit:
$$
f(x) = frac1+2+cdotsin(picdot x)
$$
Incidentally, since you just said “not closed†but not whether it should be bounded, we could also just choose
$$
f!!!!/(!!!!/x!!!!/)!!!!/ =!!!!/ x!!!!/cdot!!!!/sin!!!!/!!!!!!!!/(pi!!!!/cdot!!!!/ x!!!!/)
$$
(The image of this is all of $mathbbR$ which is actually closed, as the commenters remarked.)
answered 28 mins ago
leftaroundabout
3,3661527
1
Isn't the image $mathbb R$ closed?
– mr_e_man
10 mins ago
Regarding your last example, isn't the image all the real line? The function is odd and unbounded, so by the intermediate value theorem the image should contain any interval $[0,n]$ (and $[-n,0]$, by parity).
– Guido A.
9 mins ago
You're right, the image of $xcdotsin(picdot x)$ is open but also closed.
– leftaroundabout
8 mins ago
Can you please tell me how make these graphs?
– StammeringMathematician
8 mins ago
1
@StammeringMathematicianplotWindow [fnPlot $ x -> (1+abs x)/(2+abs x)*sin(pi*x)]with dynamic-plot (a Haskell library).
– leftaroundabout
3 mins ago
 |Â
show 1 more comment
up vote
2
down vote
The simplest solution that would come to mind is to take that sine function and multiply it with an amplitude envolope that only approaches $1$ in the $pm$infinite limit:
$$
f(x) = frac1+2+cdotsin(picdot x)
$$
Incidentally, since you just said “not closed†but not whether it should be bounded, we could also just choose
$$
f!!!!/(!!!!/x!!!!/)!!!!/ =!!!!/ x!!!!/cdot!!!!/sin!!!!/!!!!!!!!/(pi!!!!/cdot!!!!/ x!!!!/)
$$
(The image of this is all of $mathbbR$ which is actually closed, as the commenters remarked.)
answered 28 mins ago
leftaroundabout
3,3661527
1
Isn't the image $mathbb R$ closed?
– mr_e_man
10 mins ago
Regarding your last example, isn't the image all the real line? The function is odd and unbounded, so by the intermediate value theorem the image should contain any interval $[0,n]$ (and $[-n,0]$, by parity).
– Guido A.
9 mins ago
You're right, the image of $xcdotsin(picdot x)$ is open but also closed.
– leftaroundabout
8 mins ago
Can you please tell me how make these graphs?
– StammeringMathematician
8 mins ago
1
@StammeringMathematicianplotWindow [fnPlot $ x -> (1+abs x)/(2+abs x)*sin(pi*x)]with dynamic-plot (a Haskell library).
– leftaroundabout
3 mins ago
 |Â
show 1 more comment
up vote
2
down vote
up vote
2
down vote
The simplest solution that would come to mind is to take that sine function and multiply it with an amplitude envolope that only approaches $1$ in the $pm$infinite limit:
$$
f(x) = frac1+2+cdotsin(picdot x)
$$
Incidentally, since you just said “not closed†but not whether it should be bounded, we could also just choose
$$
f!!!!/(!!!!/x!!!!/)!!!!/ =!!!!/ x!!!!/cdot!!!!/sin!!!!/!!!!!!!!/(pi!!!!/cdot!!!!/ x!!!!/)
$$
(The image of this is all of $mathbbR$ which is actually closed, as the commenters remarked.)
answered 28 mins ago
leftaroundabout
3,3661527
The simplest solution that would come to mind is to take that sine function and multiply it with an amplitude envolope that only approaches $1$ in the $pm$infinite limit:
$$
f(x) = frac1+2+cdotsin(picdot x)
$$
Incidentally, since you just said “not closed†but not whether it should be bounded, we could also just choose
$$
f!!!!/(!!!!/x!!!!/)!!!!/ =!!!!/ x!!!!/cdot!!!!/sin!!!!/!!!!!!!!/(pi!!!!/cdot!!!!/ x!!!!/)
$$
(The image of this is all of $mathbbR$ which is actually closed, as the commenters remarked.)
answered 28 mins ago
leftaroundabout
3,3661527
edited 4 mins ago
answered 28 mins ago
leftaroundabout
3,3661527
answered 28 mins ago
leftaroundabout
3,3661527
answered 28 mins ago
leftaroundabout
3,3661527
3,3661527
1
Isn't the image $mathbb R$ closed?
– mr_e_man
10 mins ago
Regarding your last example, isn't the image all the real line? The function is odd and unbounded, so by the intermediate value theorem the image should contain any interval $[0,n]$ (and $[-n,0]$, by parity).
– Guido A.
9 mins ago
You're right, the image of $xcdotsin(picdot x)$ is open but also closed.
– leftaroundabout
8 mins ago
Can you please tell me how make these graphs?
– StammeringMathematician
8 mins ago
1
@StammeringMathematicianplotWindow [fnPlot $ x -> (1+abs x)/(2+abs x)*sin(pi*x)]with dynamic-plot (a Haskell library).
– leftaroundabout
3 mins ago
 |Â
show 1 more comment
1
Isn't the image $mathbb R$ closed?
– mr_e_man
10 mins ago
Regarding your last example, isn't the image all the real line? The function is odd and unbounded, so by the intermediate value theorem the image should contain any interval $[0,n]$ (and $[-n,0]$, by parity).
– Guido A.
9 mins ago
You're right, the image of $xcdotsin(picdot x)$ is open but also closed.
– leftaroundabout
8 mins ago
Can you please tell me how make these graphs?
– StammeringMathematician
8 mins ago
1
@StammeringMathematicianplotWindow [fnPlot $ x -> (1+abs x)/(2+abs x)*sin(pi*x)]with dynamic-plot (a Haskell library).
– leftaroundabout
3 mins ago
1
1
Isn't the image $mathbb R$ closed?
– mr_e_man
10 mins ago
Isn't the image $mathbb R$ closed?
– mr_e_man
10 mins ago
Regarding your last example, isn't the image all the real line? The function is odd and unbounded, so by the intermediate value theorem the image should contain any interval $[0,n]$ (and $[-n,0]$, by parity).
– Guido A.
9 mins ago
Regarding your last example, isn't the image all the real line? The function is odd and unbounded, so by the intermediate value theorem the image should contain any interval $[0,n]$ (and $[-n,0]$, by parity).
– Guido A.
9 mins ago
You're right, the image of $xcdotsin(picdot x)$ is open but also closed.
– leftaroundabout
8 mins ago
You're right, the image of $xcdotsin(picdot x)$ is open but also closed.
– leftaroundabout
8 mins ago
Can you please tell me how make these graphs?
– StammeringMathematician
8 mins ago
Can you please tell me how make these graphs?
– StammeringMathematician
8 mins ago
1
1
@StammeringMathematician
plotWindow [fnPlot $ x -> (1+abs x)/(2+abs x)*sin(pi*x)] with dynamic-plot (a Haskell library).– leftaroundabout
3 mins ago
@StammeringMathematician
plotWindow [fnPlot $ x -> (1+abs x)/(2+abs x)*sin(pi*x)] with dynamic-plot (a Haskell library).– leftaroundabout
3 mins ago
 |Â
show 1 more comment
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I think "continuous" should be in the title.
– mr_e_man
16 mins ago
@mr_e_man Thanks for pointing it out
– StammeringMathematician
10 mins ago