On continuous functions The Next CEO of Stack Overflowconvergence of $alpha$-Hölder-continuous functionsBasic calculus questionCheck my proof of the “Boundedness theorem”Show that $f$ is identically zero if $|f(x)|leqint_0^xf(t)dt$Proving set of bounded continuous functions is an open setFinding an appropriate continuous functionsIs there a bijective continuous function from closed to open intervalls?Difference between sup and inf of $|f(x)|$ less than difference between sup and inf of $f(x)$Giving examples for real functions with specific parametersSupremum metric on closed subset of continuous functions
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On continuous functions
The Next CEO of Stack Overflowconvergence of $alpha$-Hölder-continuous functionsBasic calculus questionCheck my proof of the “Boundedness theorem”Show that $f$ is identically zero if $|f(x)|leqint_0^xf(t)dt$Proving set of bounded continuous functions is an open setFinding an appropriate continuous functionsIs there a bijective continuous function from closed to open intervalls?Difference between sup and inf of $|f(x)|$ less than difference between sup and inf of $f(x)$Giving examples for real functions with specific parametersSupremum metric on closed subset of continuous functions
$begingroup$
Let $f:[0,1]to mathbb R$ be continuous and such that $f(0)<0$ and $f(1)=0$. Show that the set $x in (0,1] : f(x)=0 $ has a minimum.
So I though because the function is continuous, it is bounded. Since it is bounded it must have a inf and a sup. Since a inf exists in a bounded set I could achrimedian proof it is the minimum. But my professor told me that since the set he wants a minimum in is unbounded my logic wont work. So what am I doing wrong?
calculus functions
$endgroup$
add a comment |
$begingroup$
Let $f:[0,1]to mathbb R$ be continuous and such that $f(0)<0$ and $f(1)=0$. Show that the set $x in (0,1] : f(x)=0 $ has a minimum.
So I though because the function is continuous, it is bounded. Since it is bounded it must have a inf and a sup. Since a inf exists in a bounded set I could achrimedian proof it is the minimum. But my professor told me that since the set he wants a minimum in is unbounded my logic wont work. So what am I doing wrong?
calculus functions
$endgroup$
$begingroup$
The set is closed in $[0,1].$
$endgroup$
– Dbchatto67
Mar 27 at 19:34
$begingroup$
Not in the set im looking for a minimum in?
$endgroup$
– JJ Burt
Mar 27 at 19:38
add a comment |
$begingroup$
Let $f:[0,1]to mathbb R$ be continuous and such that $f(0)<0$ and $f(1)=0$. Show that the set $x in (0,1] : f(x)=0 $ has a minimum.
So I though because the function is continuous, it is bounded. Since it is bounded it must have a inf and a sup. Since a inf exists in a bounded set I could achrimedian proof it is the minimum. But my professor told me that since the set he wants a minimum in is unbounded my logic wont work. So what am I doing wrong?
calculus functions
$endgroup$
Let $f:[0,1]to mathbb R$ be continuous and such that $f(0)<0$ and $f(1)=0$. Show that the set $x in (0,1] : f(x)=0 $ has a minimum.
So I though because the function is continuous, it is bounded. Since it is bounded it must have a inf and a sup. Since a inf exists in a bounded set I could achrimedian proof it is the minimum. But my professor told me that since the set he wants a minimum in is unbounded my logic wont work. So what am I doing wrong?
calculus functions
calculus functions
edited Mar 27 at 19:34
Bernard
123k741117
123k741117
asked Mar 27 at 19:31
JJ BurtJJ Burt
215
215
$begingroup$
The set is closed in $[0,1].$
$endgroup$
– Dbchatto67
Mar 27 at 19:34
$begingroup$
Not in the set im looking for a minimum in?
$endgroup$
– JJ Burt
Mar 27 at 19:38
add a comment |
$begingroup$
The set is closed in $[0,1].$
$endgroup$
– Dbchatto67
Mar 27 at 19:34
$begingroup$
Not in the set im looking for a minimum in?
$endgroup$
– JJ Burt
Mar 27 at 19:38
$begingroup$
The set is closed in $[0,1].$
$endgroup$
– Dbchatto67
Mar 27 at 19:34
$begingroup$
The set is closed in $[0,1].$
$endgroup$
– Dbchatto67
Mar 27 at 19:34
$begingroup$
Not in the set im looking for a minimum in?
$endgroup$
– JJ Burt
Mar 27 at 19:38
$begingroup$
Not in the set im looking for a minimum in?
$endgroup$
– JJ Burt
Mar 27 at 19:38
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The point of the exercise is this:
The set $x in (0,1] : f(x)=0 $ obviously has an infimum, because it is non-empty and bounded below. But is this infimum necessarily a minimum? That is, does the infimum itself belong to the set? This is what you have to prove.
It seems to me that you missed this. It also seems that you must have misunderstood your professor -- what he allegedly said makes no sense to me..
(Also, I presume you mean "archimedean proof". But what do you mean by this?)
$endgroup$
add a comment |
$begingroup$
Continuity and $f(0)<0$ implies that there exists a small $epsilon>0$ such that if $0leq x<epsilon$, $f(x) < 0$. Then, the zeros of $f$ are contained in the compact set $[epsilon,1]$. Now let $x_0 = inf_[epsilon,1]left x : f(x) = 0 right$. If this is a finite set, then we are done. If not, consider a sequence from the set converging to $x_0$. Since $f$ is continuous, $f(x_0)=0$ and certainly $x_0in[epsilon,1]$ due to compactness. Hence $x_0$ is the desired minimum.
$endgroup$
add a comment |
$begingroup$
Hint $:$ $x in (0,1] : f(x) = 0 = f^-1 (0).$ Since $f$ is continuous so $f^-1 (0)$ is a closed subspace of $[0,1]$ (Since $0 $ is a closed subset of $Bbb R$). Since $[0,1]$ is a closed subspace of $Bbb R$ with usual topology so $f^-1 (0)$ is a closed subspace of $Bbb R$ with usual topology.
$endgroup$
$begingroup$
Inverse image of $0 $ under $f$ is a closed subspace of $[0,1]$ since $0 $ is a closed subspace of $Bbb R$ and $f$ is continuous.
$endgroup$
– Dbchatto67
Mar 27 at 19:41
$begingroup$
Very confused we havnt used much topology in this class so I am not very familiar with it.
$endgroup$
– JJ Burt
Mar 27 at 19:41
$begingroup$
Closed subspace of $[0,1]$ is of the form $A cap [0,1]$ where $A$ is a closed subset of $Bbb R.$ Hence the subsets which are closed in $[0,1]$ must also be closed in $Bbb R$ since intersection of two closed subsets of $Bbb R$ is also a closed subset of $Bbb R.$
$endgroup$
– Dbchatto67
Mar 27 at 19:43
$begingroup$
Do you know @JJ Burt that every closed subset of $Bbb R$ attains it's infimum?
$endgroup$
– Dbchatto67
Mar 27 at 19:55
$begingroup$
that does sound familiar so becuase its a subset?
$endgroup$
– JJ Burt
Mar 27 at 20:00
|
show 1 more comment
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The point of the exercise is this:
The set $x in (0,1] : f(x)=0 $ obviously has an infimum, because it is non-empty and bounded below. But is this infimum necessarily a minimum? That is, does the infimum itself belong to the set? This is what you have to prove.
It seems to me that you missed this. It also seems that you must have misunderstood your professor -- what he allegedly said makes no sense to me..
(Also, I presume you mean "archimedean proof". But what do you mean by this?)
$endgroup$
add a comment |
$begingroup$
The point of the exercise is this:
The set $x in (0,1] : f(x)=0 $ obviously has an infimum, because it is non-empty and bounded below. But is this infimum necessarily a minimum? That is, does the infimum itself belong to the set? This is what you have to prove.
It seems to me that you missed this. It also seems that you must have misunderstood your professor -- what he allegedly said makes no sense to me..
(Also, I presume you mean "archimedean proof". But what do you mean by this?)
$endgroup$
add a comment |
$begingroup$
The point of the exercise is this:
The set $x in (0,1] : f(x)=0 $ obviously has an infimum, because it is non-empty and bounded below. But is this infimum necessarily a minimum? That is, does the infimum itself belong to the set? This is what you have to prove.
It seems to me that you missed this. It also seems that you must have misunderstood your professor -- what he allegedly said makes no sense to me..
(Also, I presume you mean "archimedean proof". But what do you mean by this?)
$endgroup$
The point of the exercise is this:
The set $x in (0,1] : f(x)=0 $ obviously has an infimum, because it is non-empty and bounded below. But is this infimum necessarily a minimum? That is, does the infimum itself belong to the set? This is what you have to prove.
It seems to me that you missed this. It also seems that you must have misunderstood your professor -- what he allegedly said makes no sense to me..
(Also, I presume you mean "archimedean proof". But what do you mean by this?)
answered Mar 27 at 19:55
TonyKTonyK
43.7k358137
43.7k358137
add a comment |
add a comment |
$begingroup$
Continuity and $f(0)<0$ implies that there exists a small $epsilon>0$ such that if $0leq x<epsilon$, $f(x) < 0$. Then, the zeros of $f$ are contained in the compact set $[epsilon,1]$. Now let $x_0 = inf_[epsilon,1]left x : f(x) = 0 right$. If this is a finite set, then we are done. If not, consider a sequence from the set converging to $x_0$. Since $f$ is continuous, $f(x_0)=0$ and certainly $x_0in[epsilon,1]$ due to compactness. Hence $x_0$ is the desired minimum.
$endgroup$
add a comment |
$begingroup$
Continuity and $f(0)<0$ implies that there exists a small $epsilon>0$ such that if $0leq x<epsilon$, $f(x) < 0$. Then, the zeros of $f$ are contained in the compact set $[epsilon,1]$. Now let $x_0 = inf_[epsilon,1]left x : f(x) = 0 right$. If this is a finite set, then we are done. If not, consider a sequence from the set converging to $x_0$. Since $f$ is continuous, $f(x_0)=0$ and certainly $x_0in[epsilon,1]$ due to compactness. Hence $x_0$ is the desired minimum.
$endgroup$
add a comment |
$begingroup$
Continuity and $f(0)<0$ implies that there exists a small $epsilon>0$ such that if $0leq x<epsilon$, $f(x) < 0$. Then, the zeros of $f$ are contained in the compact set $[epsilon,1]$. Now let $x_0 = inf_[epsilon,1]left x : f(x) = 0 right$. If this is a finite set, then we are done. If not, consider a sequence from the set converging to $x_0$. Since $f$ is continuous, $f(x_0)=0$ and certainly $x_0in[epsilon,1]$ due to compactness. Hence $x_0$ is the desired minimum.
$endgroup$
Continuity and $f(0)<0$ implies that there exists a small $epsilon>0$ such that if $0leq x<epsilon$, $f(x) < 0$. Then, the zeros of $f$ are contained in the compact set $[epsilon,1]$. Now let $x_0 = inf_[epsilon,1]left x : f(x) = 0 right$. If this is a finite set, then we are done. If not, consider a sequence from the set converging to $x_0$. Since $f$ is continuous, $f(x_0)=0$ and certainly $x_0in[epsilon,1]$ due to compactness. Hence $x_0$ is the desired minimum.
answered Mar 27 at 20:02
MatthewPeterMatthewPeter
191
191
add a comment |
add a comment |
$begingroup$
Hint $:$ $x in (0,1] : f(x) = 0 = f^-1 (0).$ Since $f$ is continuous so $f^-1 (0)$ is a closed subspace of $[0,1]$ (Since $0 $ is a closed subset of $Bbb R$). Since $[0,1]$ is a closed subspace of $Bbb R$ with usual topology so $f^-1 (0)$ is a closed subspace of $Bbb R$ with usual topology.
$endgroup$
$begingroup$
Inverse image of $0 $ under $f$ is a closed subspace of $[0,1]$ since $0 $ is a closed subspace of $Bbb R$ and $f$ is continuous.
$endgroup$
– Dbchatto67
Mar 27 at 19:41
$begingroup$
Very confused we havnt used much topology in this class so I am not very familiar with it.
$endgroup$
– JJ Burt
Mar 27 at 19:41
$begingroup$
Closed subspace of $[0,1]$ is of the form $A cap [0,1]$ where $A$ is a closed subset of $Bbb R.$ Hence the subsets which are closed in $[0,1]$ must also be closed in $Bbb R$ since intersection of two closed subsets of $Bbb R$ is also a closed subset of $Bbb R.$
$endgroup$
– Dbchatto67
Mar 27 at 19:43
$begingroup$
Do you know @JJ Burt that every closed subset of $Bbb R$ attains it's infimum?
$endgroup$
– Dbchatto67
Mar 27 at 19:55
$begingroup$
that does sound familiar so becuase its a subset?
$endgroup$
– JJ Burt
Mar 27 at 20:00
|
show 1 more comment
$begingroup$
Hint $:$ $x in (0,1] : f(x) = 0 = f^-1 (0).$ Since $f$ is continuous so $f^-1 (0)$ is a closed subspace of $[0,1]$ (Since $0 $ is a closed subset of $Bbb R$). Since $[0,1]$ is a closed subspace of $Bbb R$ with usual topology so $f^-1 (0)$ is a closed subspace of $Bbb R$ with usual topology.
$endgroup$
$begingroup$
Inverse image of $0 $ under $f$ is a closed subspace of $[0,1]$ since $0 $ is a closed subspace of $Bbb R$ and $f$ is continuous.
$endgroup$
– Dbchatto67
Mar 27 at 19:41
$begingroup$
Very confused we havnt used much topology in this class so I am not very familiar with it.
$endgroup$
– JJ Burt
Mar 27 at 19:41
$begingroup$
Closed subspace of $[0,1]$ is of the form $A cap [0,1]$ where $A$ is a closed subset of $Bbb R.$ Hence the subsets which are closed in $[0,1]$ must also be closed in $Bbb R$ since intersection of two closed subsets of $Bbb R$ is also a closed subset of $Bbb R.$
$endgroup$
– Dbchatto67
Mar 27 at 19:43
$begingroup$
Do you know @JJ Burt that every closed subset of $Bbb R$ attains it's infimum?
$endgroup$
– Dbchatto67
Mar 27 at 19:55
$begingroup$
that does sound familiar so becuase its a subset?
$endgroup$
– JJ Burt
Mar 27 at 20:00
|
show 1 more comment
$begingroup$
Hint $:$ $x in (0,1] : f(x) = 0 = f^-1 (0).$ Since $f$ is continuous so $f^-1 (0)$ is a closed subspace of $[0,1]$ (Since $0 $ is a closed subset of $Bbb R$). Since $[0,1]$ is a closed subspace of $Bbb R$ with usual topology so $f^-1 (0)$ is a closed subspace of $Bbb R$ with usual topology.
$endgroup$
Hint $:$ $x in (0,1] : f(x) = 0 = f^-1 (0).$ Since $f$ is continuous so $f^-1 (0)$ is a closed subspace of $[0,1]$ (Since $0 $ is a closed subset of $Bbb R$). Since $[0,1]$ is a closed subspace of $Bbb R$ with usual topology so $f^-1 (0)$ is a closed subspace of $Bbb R$ with usual topology.
edited Mar 27 at 19:52
answered Mar 27 at 19:35
Dbchatto67Dbchatto67
2,419422
2,419422
$begingroup$
Inverse image of $0 $ under $f$ is a closed subspace of $[0,1]$ since $0 $ is a closed subspace of $Bbb R$ and $f$ is continuous.
$endgroup$
– Dbchatto67
Mar 27 at 19:41
$begingroup$
Very confused we havnt used much topology in this class so I am not very familiar with it.
$endgroup$
– JJ Burt
Mar 27 at 19:41
$begingroup$
Closed subspace of $[0,1]$ is of the form $A cap [0,1]$ where $A$ is a closed subset of $Bbb R.$ Hence the subsets which are closed in $[0,1]$ must also be closed in $Bbb R$ since intersection of two closed subsets of $Bbb R$ is also a closed subset of $Bbb R.$
$endgroup$
– Dbchatto67
Mar 27 at 19:43
$begingroup$
Do you know @JJ Burt that every closed subset of $Bbb R$ attains it's infimum?
$endgroup$
– Dbchatto67
Mar 27 at 19:55
$begingroup$
that does sound familiar so becuase its a subset?
$endgroup$
– JJ Burt
Mar 27 at 20:00
|
show 1 more comment
$begingroup$
Inverse image of $0 $ under $f$ is a closed subspace of $[0,1]$ since $0 $ is a closed subspace of $Bbb R$ and $f$ is continuous.
$endgroup$
– Dbchatto67
Mar 27 at 19:41
$begingroup$
Very confused we havnt used much topology in this class so I am not very familiar with it.
$endgroup$
– JJ Burt
Mar 27 at 19:41
$begingroup$
Closed subspace of $[0,1]$ is of the form $A cap [0,1]$ where $A$ is a closed subset of $Bbb R.$ Hence the subsets which are closed in $[0,1]$ must also be closed in $Bbb R$ since intersection of two closed subsets of $Bbb R$ is also a closed subset of $Bbb R.$
$endgroup$
– Dbchatto67
Mar 27 at 19:43
$begingroup$
Do you know @JJ Burt that every closed subset of $Bbb R$ attains it's infimum?
$endgroup$
– Dbchatto67
Mar 27 at 19:55
$begingroup$
that does sound familiar so becuase its a subset?
$endgroup$
– JJ Burt
Mar 27 at 20:00
$begingroup$
Inverse image of $0 $ under $f$ is a closed subspace of $[0,1]$ since $0 $ is a closed subspace of $Bbb R$ and $f$ is continuous.
$endgroup$
– Dbchatto67
Mar 27 at 19:41
$begingroup$
Inverse image of $0 $ under $f$ is a closed subspace of $[0,1]$ since $0 $ is a closed subspace of $Bbb R$ and $f$ is continuous.
$endgroup$
– Dbchatto67
Mar 27 at 19:41
$begingroup$
Very confused we havnt used much topology in this class so I am not very familiar with it.
$endgroup$
– JJ Burt
Mar 27 at 19:41
$begingroup$
Very confused we havnt used much topology in this class so I am not very familiar with it.
$endgroup$
– JJ Burt
Mar 27 at 19:41
$begingroup$
Closed subspace of $[0,1]$ is of the form $A cap [0,1]$ where $A$ is a closed subset of $Bbb R.$ Hence the subsets which are closed in $[0,1]$ must also be closed in $Bbb R$ since intersection of two closed subsets of $Bbb R$ is also a closed subset of $Bbb R.$
$endgroup$
– Dbchatto67
Mar 27 at 19:43
$begingroup$
Closed subspace of $[0,1]$ is of the form $A cap [0,1]$ where $A$ is a closed subset of $Bbb R.$ Hence the subsets which are closed in $[0,1]$ must also be closed in $Bbb R$ since intersection of two closed subsets of $Bbb R$ is also a closed subset of $Bbb R.$
$endgroup$
– Dbchatto67
Mar 27 at 19:43
$begingroup$
Do you know @JJ Burt that every closed subset of $Bbb R$ attains it's infimum?
$endgroup$
– Dbchatto67
Mar 27 at 19:55
$begingroup$
Do you know @JJ Burt that every closed subset of $Bbb R$ attains it's infimum?
$endgroup$
– Dbchatto67
Mar 27 at 19:55
$begingroup$
that does sound familiar so becuase its a subset?
$endgroup$
– JJ Burt
Mar 27 at 20:00
$begingroup$
that does sound familiar so becuase its a subset?
$endgroup$
– JJ Burt
Mar 27 at 20:00
|
show 1 more comment
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$begingroup$
The set is closed in $[0,1].$
$endgroup$
– Dbchatto67
Mar 27 at 19:34
$begingroup$
Not in the set im looking for a minimum in?
$endgroup$
– JJ Burt
Mar 27 at 19:38