Mixing
Heine theorem without subsequences
$begingroup$
I wanted to prove the following without the need of subsequences :
let $f$ be a continuous fonction on a closed interval of $mathbb{R}$ then it is uniformly continuous.
I found this proof but can't understand it
Let us assume every increasing sequence on [0;1] converges (if anyone can help me proving this) and the IVT
then let us create a sequence $u_n$
$begin{cases}u_0=0 \
u_{n+1}=1 text{ if } forall x>x_k, |f(x)-f(x_k)|<varepsilon text{ and we stop here} \
text{otherwise } u_{n+1}=min(x) text{ s.t } |f(x)-f(x_k)|=varepsilon\
end{cases}$
if this sequence is finite, let $eta=max(x_{n+1}-x_n)$ then $forall (x,y),forall varepsilon >0$
$$|x-y|leq eta implies |f(x)-f(y)|<2varepsilon $$
Otherwise we would have a contradiction.
Can anyone help me understanding
Thank you for your time.
proof-verification proof-writing proof-explanation
edited Dec 11 '18 at 15:01
Bernard
119k740113
asked Dec 11 '18 at 14:55
T.D.T.D.
128114
$endgroup$
add a comment |
$begingroup$
I wanted to prove the following without the need of subsequences :
let $f$ be a continuous fonction on a closed interval of $mathbb{R}$ then it is uniformly continuous.
I found this proof but can't understand it
Let us assume every increasing sequence on [0;1] converges (if anyone can help me proving this) and the IVT
then let us create a sequence $u_n$
$begin{cases}u_0=0 \
u_{n+1}=1 text{ if } forall x>x_k, |f(x)-f(x_k)|<varepsilon text{ and we stop here} \
text{otherwise } u_{n+1}=min(x) text{ s.t } |f(x)-f(x_k)|=varepsilon\
end{cases}$
if this sequence is finite, let $eta=max(x_{n+1}-x_n)$ then $forall (x,y),forall varepsilon >0$
$$|x-y|leq eta implies |f(x)-f(y)|<2varepsilon $$
Otherwise we would have a contradiction.
Can anyone help me understanding
Thank you for your time.
proof-verification proof-writing proof-explanation
edited Dec 11 '18 at 15:01
Bernard
119k740113
asked Dec 11 '18 at 14:55
T.D.T.D.
128114
$endgroup$
$begingroup$
It looks like $x_k$ is $u_k$...
$endgroup$
– CopyPasteIt
Jan 6 at 15:48
$begingroup$
Have you looked at en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem ? Same argument of your setup.
$endgroup$
– CopyPasteIt
Jan 6 at 15:50
$begingroup$
I cant understand it thus I ask here
$endgroup$
– T.D.
Jan 7 at 19:30
add a comment |
$begingroup$
I wanted to prove the following without the need of subsequences :
let $f$ be a continuous fonction on a closed interval of $mathbb{R}$ then it is uniformly continuous.
I found this proof but can't understand it
Let us assume every increasing sequence on [0;1] converges (if anyone can help me proving this) and the IVT
then let us create a sequence $u_n$
$begin{cases}u_0=0 \
u_{n+1}=1 text{ if } forall x>x_k, |f(x)-f(x_k)|<varepsilon text{ and we stop here} \
text{otherwise } u_{n+1}=min(x) text{ s.t } |f(x)-f(x_k)|=varepsilon\
end{cases}$
if this sequence is finite, let $eta=max(x_{n+1}-x_n)$ then $forall (x,y),forall varepsilon >0$
$$|x-y|leq eta implies |f(x)-f(y)|<2varepsilon $$
Otherwise we would have a contradiction.
Can anyone help me understanding
Thank you for your time.
proof-verification proof-writing proof-explanation
edited Dec 11 '18 at 15:01
Bernard
119k740113
asked Dec 11 '18 at 14:55
T.D.T.D.
128114
$endgroup$
I wanted to prove the following without the need of subsequences :
let $f$ be a continuous fonction on a closed interval of $mathbb{R}$ then it is uniformly continuous.
I found this proof but can't understand it
Let us assume every increasing sequence on [0;1] converges (if anyone can help me proving this) and the IVT
then let us create a sequence $u_n$
$begin{cases}u_0=0 \
u_{n+1}=1 text{ if } forall x>x_k, |f(x)-f(x_k)|<varepsilon text{ and we stop here} \
text{otherwise } u_{n+1}=min(x) text{ s.t } |f(x)-f(x_k)|=varepsilon\
end{cases}$
if this sequence is finite, let $eta=max(x_{n+1}-x_n)$ then $forall (x,y),forall varepsilon >0$
$$|x-y|leq eta implies |f(x)-f(y)|<2varepsilon $$
Otherwise we would have a contradiction.
Can anyone help me understanding
Thank you for your time.
proof-verification proof-writing proof-explanation
proof-verification proof-writing proof-explanation
edited Dec 11 '18 at 15:01
Bernard
119k740113
asked Dec 11 '18 at 14:55
T.D.T.D.
128114
edited Dec 11 '18 at 15:01
Bernard
119k740113
asked Dec 11 '18 at 14:55
T.D.T.D.
128114
edited Dec 11 '18 at 15:01
Bernard
119k740113
edited Dec 11 '18 at 15:01
Bernard
119k740113
edited Dec 11 '18 at 15:01
Bernard
119k740113
119k740113
asked Dec 11 '18 at 14:55
T.D.T.D.
128114
asked Dec 11 '18 at 14:55
T.D.T.D.
128114
asked Dec 11 '18 at 14:55
T.D.T.D.
128114
128114
$begingroup$
It looks like $x_k$ is $u_k$...
$endgroup$
– CopyPasteIt
Jan 6 at 15:48
$begingroup$
Have you looked at en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem ? Same argument of your setup.
$endgroup$
– CopyPasteIt
Jan 6 at 15:50
$begingroup$
I cant understand it thus I ask here
$endgroup$
– T.D.
Jan 7 at 19:30
add a comment |
$begingroup$
It looks like $x_k$ is $u_k$...
$endgroup$
– CopyPasteIt
Jan 6 at 15:48
$begingroup$
Have you looked at en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem ? Same argument of your setup.
$endgroup$
– CopyPasteIt
Jan 6 at 15:50
$begingroup$
I cant understand it thus I ask here
$endgroup$
– T.D.
Jan 7 at 19:30
$begingroup$
It looks like $x_k$ is $u_k$...
$endgroup$
– CopyPasteIt
Jan 6 at 15:48
$begingroup$
It looks like $x_k$ is $u_k$...
$endgroup$
– CopyPasteIt
Jan 6 at 15:48
$begingroup$
Have you looked at en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem ? Same argument of your setup.
$endgroup$
– CopyPasteIt
Jan 6 at 15:50
$begingroup$
Have you looked at en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem ? Same argument of your setup.
$endgroup$
– CopyPasteIt
Jan 6 at 15:50
$begingroup$
I cant understand it thus I ask here
$endgroup$
– T.D.
Jan 7 at 19:30
$begingroup$
I cant understand it thus I ask here
$endgroup$
– T.D.
Jan 7 at 19:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any closed interval of real numbers has the following property; For any infinite subset of the interval there exists a point in the interval for which we can find an infinite amount of elements of said subset which are as close as we wish. Since you are trying to prove something I assume you will want to prove this statement as well.
To prove this statement think about the following procedure. Let's say we have some infinite subset of a real interval. Now we may cut the interval in half and at least one of the halves must contain an infinite number of members of said subset. We may continue this procedure as often as we wish, or else the infiniteness of the subset would clearly be contradicted.
Now we may look at some sequence of intervals, whose length gets halved with each step, and every element of the sequence contains infinitely many points of the subset.
To finish the proof show that there exists a number which is a member of all the intervals of the sequence and that this numbers has the desired property.( use the least upper bound property of the reals)
To prove your original question assume, for a contradiction, that you have a continious but non uniformly continuous function on a closed real intervall. Using the nonuniformity construct a sequence of points so that the nth member of the sequence has a point which has a distance of 1/n or less and the distance between the projection of the points is larger than a fixed number. Now use the first statemet to produce a point which is infinetly close to infinetly many members of this sequence. And finally use this point to produce a contradiction to the assumed continuity of the function.
answered Jan 8 at 12:41
Jagol95Jagol95
1057
$endgroup$
add a comment |
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$begingroup$
Any closed interval of real numbers has the following property; For any infinite subset of the interval there exists a point in the interval for which we can find an infinite amount of elements of said subset which are as close as we wish. Since you are trying to prove something I assume you will want to prove this statement as well.
To prove this statement think about the following procedure. Let's say we have some infinite subset of a real interval. Now we may cut the interval in half and at least one of the halves must contain an infinite number of members of said subset. We may continue this procedure as often as we wish, or else the infiniteness of the subset would clearly be contradicted.
Now we may look at some sequence of intervals, whose length gets halved with each step, and every element of the sequence contains infinitely many points of the subset.
To finish the proof show that there exists a number which is a member of all the intervals of the sequence and that this numbers has the desired property.( use the least upper bound property of the reals)
To prove your original question assume, for a contradiction, that you have a continious but non uniformly continuous function on a closed real intervall. Using the nonuniformity construct a sequence of points so that the nth member of the sequence has a point which has a distance of 1/n or less and the distance between the projection of the points is larger than a fixed number. Now use the first statemet to produce a point which is infinetly close to infinetly many members of this sequence. And finally use this point to produce a contradiction to the assumed continuity of the function.
answered Jan 8 at 12:41
Jagol95Jagol95
1057
$endgroup$
add a comment |
$begingroup$
Any closed interval of real numbers has the following property; For any infinite subset of the interval there exists a point in the interval for which we can find an infinite amount of elements of said subset which are as close as we wish. Since you are trying to prove something I assume you will want to prove this statement as well.
To prove this statement think about the following procedure. Let's say we have some infinite subset of a real interval. Now we may cut the interval in half and at least one of the halves must contain an infinite number of members of said subset. We may continue this procedure as often as we wish, or else the infiniteness of the subset would clearly be contradicted.
Now we may look at some sequence of intervals, whose length gets halved with each step, and every element of the sequence contains infinitely many points of the subset.
To finish the proof show that there exists a number which is a member of all the intervals of the sequence and that this numbers has the desired property.( use the least upper bound property of the reals)
To prove your original question assume, for a contradiction, that you have a continious but non uniformly continuous function on a closed real intervall. Using the nonuniformity construct a sequence of points so that the nth member of the sequence has a point which has a distance of 1/n or less and the distance between the projection of the points is larger than a fixed number. Now use the first statemet to produce a point which is infinetly close to infinetly many members of this sequence. And finally use this point to produce a contradiction to the assumed continuity of the function.
answered Jan 8 at 12:41
Jagol95Jagol95
1057
$endgroup$
add a comment |
$begingroup$
Any closed interval of real numbers has the following property; For any infinite subset of the interval there exists a point in the interval for which we can find an infinite amount of elements of said subset which are as close as we wish. Since you are trying to prove something I assume you will want to prove this statement as well.
To prove this statement think about the following procedure. Let's say we have some infinite subset of a real interval. Now we may cut the interval in half and at least one of the halves must contain an infinite number of members of said subset. We may continue this procedure as often as we wish, or else the infiniteness of the subset would clearly be contradicted.
Now we may look at some sequence of intervals, whose length gets halved with each step, and every element of the sequence contains infinitely many points of the subset.
To finish the proof show that there exists a number which is a member of all the intervals of the sequence and that this numbers has the desired property.( use the least upper bound property of the reals)
To prove your original question assume, for a contradiction, that you have a continious but non uniformly continuous function on a closed real intervall. Using the nonuniformity construct a sequence of points so that the nth member of the sequence has a point which has a distance of 1/n or less and the distance between the projection of the points is larger than a fixed number. Now use the first statemet to produce a point which is infinetly close to infinetly many members of this sequence. And finally use this point to produce a contradiction to the assumed continuity of the function.
answered Jan 8 at 12:41
Jagol95Jagol95
1057
$endgroup$
Any closed interval of real numbers has the following property; For any infinite subset of the interval there exists a point in the interval for which we can find an infinite amount of elements of said subset which are as close as we wish. Since you are trying to prove something I assume you will want to prove this statement as well.
To prove this statement think about the following procedure. Let's say we have some infinite subset of a real interval. Now we may cut the interval in half and at least one of the halves must contain an infinite number of members of said subset. We may continue this procedure as often as we wish, or else the infiniteness of the subset would clearly be contradicted.
Now we may look at some sequence of intervals, whose length gets halved with each step, and every element of the sequence contains infinitely many points of the subset.
To finish the proof show that there exists a number which is a member of all the intervals of the sequence and that this numbers has the desired property.( use the least upper bound property of the reals)
To prove your original question assume, for a contradiction, that you have a continious but non uniformly continuous function on a closed real intervall. Using the nonuniformity construct a sequence of points so that the nth member of the sequence has a point which has a distance of 1/n or less and the distance between the projection of the points is larger than a fixed number. Now use the first statemet to produce a point which is infinetly close to infinetly many members of this sequence. And finally use this point to produce a contradiction to the assumed continuity of the function.
answered Jan 8 at 12:41
Jagol95Jagol95
1057
edited Jan 8 at 12:46
answered Jan 8 at 12:41
Jagol95Jagol95
1057
answered Jan 8 at 12:41
Jagol95Jagol95
1057
answered Jan 8 at 12:41
Jagol95Jagol95
1057
1057
add a comment |
add a comment |
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$begingroup$
It looks like $x_k$ is $u_k$...
$endgroup$
– CopyPasteIt
Jan 6 at 15:48
$begingroup$
Have you looked at en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem ? Same argument of your setup.
$endgroup$
– CopyPasteIt
Jan 6 at 15:50
$begingroup$
I cant understand it thus I ask here
$endgroup$
– T.D.
Jan 7 at 19:30