Mixing
Understanding proof of monotonic increasing function and continuity
Clash Royale CLAN TAG#URR8PPP
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I found this statement with the proof:
But I don't understand the proof. Where is the contradiction? We have a nonempty interval $J$ contained in the nonempty interval $(f(a),f(b))$. Where is the problem?
real-analysis continuity monotone-functions
asked Jul 24 at 12:41
mathstackuser
40410
add a comment |Â
up vote
0
down vote
favorite
I found this statement with the proof:
But I don't understand the proof. Where is the contradiction? We have a nonempty interval $J$ contained in the nonempty interval $(f(a),f(b))$. Where is the problem?
real-analysis continuity monotone-functions
asked Jul 24 at 12:41
mathstackuser
40410
The contradiction comes in that for $yin J$, the equation $y = f(x)$ has no solution.
– BindersFull
Jul 24 at 12:58
Why is there no solution?
– mathstackuser
Jul 24 at 13:16
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I found this statement with the proof:
But I don't understand the proof. Where is the contradiction? We have a nonempty interval $J$ contained in the nonempty interval $(f(a),f(b))$. Where is the problem?
real-analysis continuity monotone-functions
asked Jul 24 at 12:41
mathstackuser
40410
I found this statement with the proof:
But I don't understand the proof. Where is the contradiction? We have a nonempty interval $J$ contained in the nonempty interval $(f(a),f(b))$. Where is the problem?
real-analysis continuity monotone-functions
asked Jul 24 at 12:41
mathstackuser
40410
asked Jul 24 at 12:41
mathstackuser
40410
asked Jul 24 at 12:41
mathstackuser
40410
asked Jul 24 at 12:41
mathstackuser
40410
40410
The contradiction comes in that for $yin J$, the equation $y = f(x)$ has no solution.
– BindersFull
Jul 24 at 12:58
Why is there no solution?
– mathstackuser
Jul 24 at 13:16
add a comment |Â
The contradiction comes in that for $yin J$, the equation $y = f(x)$ has no solution.
– BindersFull
Jul 24 at 12:58
Why is there no solution?
– mathstackuser
Jul 24 at 13:16
The contradiction comes in that for $yin J$, the equation $y = f(x)$ has no solution.
– BindersFull
Jul 24 at 12:58
The contradiction comes in that for $yin J$, the equation $y = f(x)$ has no solution.
– BindersFull
Jul 24 at 12:58
Why is there no solution?
– mathstackuser
Jul 24 at 13:16
Why is there no solution?
– mathstackuser
Jul 24 at 13:16
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
In the above form the proof isn't done well. You know that $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Hence at least one of the intervals $J^- = (y_0^-, f(x_0))$ and $J^+ = (f(x_0),y_0^+)$ is non-empty and does not contain any point of the form $f(x)$. This comes from the fact that $f(x_0)$ is contained in none of these intervals and $f(x) le y_0^-$ for $x < x_0$ and $f(x) ge y_0^+$ for $x > x_0$. W.l.o.g. assume $J^- ne emptyset$ and $J^- cap f(I) = emptyset$.
This is a contradiction: By assumption $f(I)$ is an interval which must contain $J^-$ because $f(a) le y_0^-$ and $y_0^+ le f(b)$.
answered Jul 24 at 13:03
Paul Frost
3,623420
Of course one of the intervals is non empty. But why does it not contain any point of the form $f(x)$? And isn't $(f(a),f(b))$ always an interval, just by the definition of the notation?
– mathstackuser
Jul 24 at 13:18
I shall edit my answer.
– Paul Frost
Jul 24 at 13:21
add a comment |Â
up vote
1
down vote
There have been two answers, but the comments have shown that the arguments at first glance appeared a little obscure or cumbersome. I believe that everything is clarified now, either by enhancing the answers or by additional comments. But let us give another proof which is hopefully more transparent from the beginning.
Observe that in the theorem it is not specified whether $I$ is an open, closed or half-open interval and whether it is bounded or not. We shall prove the stronger result that $f$ is continuous on $I$.
Assume that $f$ is not continuous in a point $x_0 in I$. Then $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Note that only one of $y_0^pm$ is defined if $x_0$ is a boundary point of $I$. We only consider the case $y_0^- < f(x_0)$, the other case is similar. The case under consideration can only occur when $I' = I cap (-infty,x_0) ne emptyset$. For $x < x_0$ we have $f(x) le y_0^-$, for $x > x_0$ we have $f(x) ge f(x_0)$. Therefore $(y_0^-,f(x_0) cap f(I) = emptyset$. For $xi in I'$ we have $f(xi) le y_0^-$ so that $(y_0^-,f(x_0) subset [f(xi),f(x_0)] subset f(I)$ because $f(I)$ is an interval. This is the desired contradiction.
answered Jul 24 at 16:17
Paul Frost
3,623420
add a comment |Â
up vote
0
down vote
This is essentially Paul Frost's answer, in a slighly less verbose form.
The range $f(I)$ contains the interval $(f(a),f(b))$, so every point in the interval $(f(a),f(b))$ is of the form $f(x)$ for some $xin I$. On the other hand, only one endpoint of the non-empty interval $J$ is a value of $f$ - because $f$ jumps from $y_0^-$ to $y_0^+$. That is the contradiction.
answered Jul 24 at 12:53
uniquesolution
7,630721
Don't we have $f((a,b))subseteq (f(a),f(b))$ because $f$ is monotone increasing? It loks lik you claim it the other way round? And what do you mean by endpoints of the interval? They are open...
– mathstackuser
Jul 24 at 13:38
I am assuming that $I$ is an interval containing the points $a$ and $b$. Therefore $f(a)$ and $f(b)$ are values of $f$ that belong to the range, $f(I)$. Since the range is an interval, it must contain the interval $(f(a),f(b))$. In order to prove the inclusion that you wrote down we need continuity, but we don't have it yet. Yes, the interval is open, but surely you understand what are its end points, no?
– uniquesolution
Jul 24 at 13:46
... after all, you are using the notation $(a,b)$ to designate an open interval, by means of the "end points" $a,b$.
– uniquesolution
Jul 24 at 13:56
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
In the above form the proof isn't done well. You know that $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Hence at least one of the intervals $J^- = (y_0^-, f(x_0))$ and $J^+ = (f(x_0),y_0^+)$ is non-empty and does not contain any point of the form $f(x)$. This comes from the fact that $f(x_0)$ is contained in none of these intervals and $f(x) le y_0^-$ for $x < x_0$ and $f(x) ge y_0^+$ for $x > x_0$. W.l.o.g. assume $J^- ne emptyset$ and $J^- cap f(I) = emptyset$.
This is a contradiction: By assumption $f(I)$ is an interval which must contain $J^-$ because $f(a) le y_0^-$ and $y_0^+ le f(b)$.
answered Jul 24 at 13:03
Paul Frost
3,623420
Of course one of the intervals is non empty. But why does it not contain any point of the form $f(x)$? And isn't $(f(a),f(b))$ always an interval, just by the definition of the notation?
– mathstackuser
Jul 24 at 13:18
I shall edit my answer.
– Paul Frost
Jul 24 at 13:21
add a comment |Â
up vote
2
down vote
accepted
In the above form the proof isn't done well. You know that $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Hence at least one of the intervals $J^- = (y_0^-, f(x_0))$ and $J^+ = (f(x_0),y_0^+)$ is non-empty and does not contain any point of the form $f(x)$. This comes from the fact that $f(x_0)$ is contained in none of these intervals and $f(x) le y_0^-$ for $x < x_0$ and $f(x) ge y_0^+$ for $x > x_0$. W.l.o.g. assume $J^- ne emptyset$ and $J^- cap f(I) = emptyset$.
This is a contradiction: By assumption $f(I)$ is an interval which must contain $J^-$ because $f(a) le y_0^-$ and $y_0^+ le f(b)$.
answered Jul 24 at 13:03
Paul Frost
3,623420
Of course one of the intervals is non empty. But why does it not contain any point of the form $f(x)$? And isn't $(f(a),f(b))$ always an interval, just by the definition of the notation?
– mathstackuser
Jul 24 at 13:18
I shall edit my answer.
– Paul Frost
Jul 24 at 13:21
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
In the above form the proof isn't done well. You know that $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Hence at least one of the intervals $J^- = (y_0^-, f(x_0))$ and $J^+ = (f(x_0),y_0^+)$ is non-empty and does not contain any point of the form $f(x)$. This comes from the fact that $f(x_0)$ is contained in none of these intervals and $f(x) le y_0^-$ for $x < x_0$ and $f(x) ge y_0^+$ for $x > x_0$. W.l.o.g. assume $J^- ne emptyset$ and $J^- cap f(I) = emptyset$.
This is a contradiction: By assumption $f(I)$ is an interval which must contain $J^-$ because $f(a) le y_0^-$ and $y_0^+ le f(b)$.
answered Jul 24 at 13:03
Paul Frost
3,623420
In the above form the proof isn't done well. You know that $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Hence at least one of the intervals $J^- = (y_0^-, f(x_0))$ and $J^+ = (f(x_0),y_0^+)$ is non-empty and does not contain any point of the form $f(x)$. This comes from the fact that $f(x_0)$ is contained in none of these intervals and $f(x) le y_0^-$ for $x < x_0$ and $f(x) ge y_0^+$ for $x > x_0$. W.l.o.g. assume $J^- ne emptyset$ and $J^- cap f(I) = emptyset$.
This is a contradiction: By assumption $f(I)$ is an interval which must contain $J^-$ because $f(a) le y_0^-$ and $y_0^+ le f(b)$.
answered Jul 24 at 13:03
Paul Frost
3,623420
edited Jul 24 at 13:46
answered Jul 24 at 13:03
Paul Frost
3,623420
answered Jul 24 at 13:03
Paul Frost
3,623420
answered Jul 24 at 13:03
Paul Frost
3,623420
3,623420
Of course one of the intervals is non empty. But why does it not contain any point of the form $f(x)$? And isn't $(f(a),f(b))$ always an interval, just by the definition of the notation?
– mathstackuser
Jul 24 at 13:18
I shall edit my answer.
– Paul Frost
Jul 24 at 13:21
add a comment |Â
Of course one of the intervals is non empty. But why does it not contain any point of the form $f(x)$? And isn't $(f(a),f(b))$ always an interval, just by the definition of the notation?
– mathstackuser
Jul 24 at 13:18
I shall edit my answer.
– Paul Frost
Jul 24 at 13:21
Of course one of the intervals is non empty. But why does it not contain any point of the form $f(x)$? And isn't $(f(a),f(b))$ always an interval, just by the definition of the notation?
– mathstackuser
Jul 24 at 13:18
Of course one of the intervals is non empty. But why does it not contain any point of the form $f(x)$? And isn't $(f(a),f(b))$ always an interval, just by the definition of the notation?
– mathstackuser
Jul 24 at 13:18
I shall edit my answer.
– Paul Frost
Jul 24 at 13:21
I shall edit my answer.
– Paul Frost
Jul 24 at 13:21
add a comment |Â
up vote
1
down vote
There have been two answers, but the comments have shown that the arguments at first glance appeared a little obscure or cumbersome. I believe that everything is clarified now, either by enhancing the answers or by additional comments. But let us give another proof which is hopefully more transparent from the beginning.
Observe that in the theorem it is not specified whether $I$ is an open, closed or half-open interval and whether it is bounded or not. We shall prove the stronger result that $f$ is continuous on $I$.
Assume that $f$ is not continuous in a point $x_0 in I$. Then $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Note that only one of $y_0^pm$ is defined if $x_0$ is a boundary point of $I$. We only consider the case $y_0^- < f(x_0)$, the other case is similar. The case under consideration can only occur when $I' = I cap (-infty,x_0) ne emptyset$. For $x < x_0$ we have $f(x) le y_0^-$, for $x > x_0$ we have $f(x) ge f(x_0)$. Therefore $(y_0^-,f(x_0) cap f(I) = emptyset$. For $xi in I'$ we have $f(xi) le y_0^-$ so that $(y_0^-,f(x_0) subset [f(xi),f(x_0)] subset f(I)$ because $f(I)$ is an interval. This is the desired contradiction.
answered Jul 24 at 16:17
Paul Frost
3,623420
add a comment |Â
up vote
1
down vote
There have been two answers, but the comments have shown that the arguments at first glance appeared a little obscure or cumbersome. I believe that everything is clarified now, either by enhancing the answers or by additional comments. But let us give another proof which is hopefully more transparent from the beginning.
Observe that in the theorem it is not specified whether $I$ is an open, closed or half-open interval and whether it is bounded or not. We shall prove the stronger result that $f$ is continuous on $I$.
Assume that $f$ is not continuous in a point $x_0 in I$. Then $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Note that only one of $y_0^pm$ is defined if $x_0$ is a boundary point of $I$. We only consider the case $y_0^- < f(x_0)$, the other case is similar. The case under consideration can only occur when $I' = I cap (-infty,x_0) ne emptyset$. For $x < x_0$ we have $f(x) le y_0^-$, for $x > x_0$ we have $f(x) ge f(x_0)$. Therefore $(y_0^-,f(x_0) cap f(I) = emptyset$. For $xi in I'$ we have $f(xi) le y_0^-$ so that $(y_0^-,f(x_0) subset [f(xi),f(x_0)] subset f(I)$ because $f(I)$ is an interval. This is the desired contradiction.
answered Jul 24 at 16:17
Paul Frost
3,623420
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There have been two answers, but the comments have shown that the arguments at first glance appeared a little obscure or cumbersome. I believe that everything is clarified now, either by enhancing the answers or by additional comments. But let us give another proof which is hopefully more transparent from the beginning.
Observe that in the theorem it is not specified whether $I$ is an open, closed or half-open interval and whether it is bounded or not. We shall prove the stronger result that $f$ is continuous on $I$.
Assume that $f$ is not continuous in a point $x_0 in I$. Then $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Note that only one of $y_0^pm$ is defined if $x_0$ is a boundary point of $I$. We only consider the case $y_0^- < f(x_0)$, the other case is similar. The case under consideration can only occur when $I' = I cap (-infty,x_0) ne emptyset$. For $x < x_0$ we have $f(x) le y_0^-$, for $x > x_0$ we have $f(x) ge f(x_0)$. Therefore $(y_0^-,f(x_0) cap f(I) = emptyset$. For $xi in I'$ we have $f(xi) le y_0^-$ so that $(y_0^-,f(x_0) subset [f(xi),f(x_0)] subset f(I)$ because $f(I)$ is an interval. This is the desired contradiction.
answered Jul 24 at 16:17
Paul Frost
3,623420
There have been two answers, but the comments have shown that the arguments at first glance appeared a little obscure or cumbersome. I believe that everything is clarified now, either by enhancing the answers or by additional comments. But let us give another proof which is hopefully more transparent from the beginning.
Observe that in the theorem it is not specified whether $I$ is an open, closed or half-open interval and whether it is bounded or not. We shall prove the stronger result that $f$ is continuous on $I$.
Assume that $f$ is not continuous in a point $x_0 in I$. Then $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Note that only one of $y_0^pm$ is defined if $x_0$ is a boundary point of $I$. We only consider the case $y_0^- < f(x_0)$, the other case is similar. The case under consideration can only occur when $I' = I cap (-infty,x_0) ne emptyset$. For $x < x_0$ we have $f(x) le y_0^-$, for $x > x_0$ we have $f(x) ge f(x_0)$. Therefore $(y_0^-,f(x_0) cap f(I) = emptyset$. For $xi in I'$ we have $f(xi) le y_0^-$ so that $(y_0^-,f(x_0) subset [f(xi),f(x_0)] subset f(I)$ because $f(I)$ is an interval. This is the desired contradiction.
answered Jul 24 at 16:17
Paul Frost
3,623420
answered Jul 24 at 16:17
Paul Frost
3,623420
answered Jul 24 at 16:17
Paul Frost
3,623420
answered Jul 24 at 16:17
Paul Frost
3,623420
3,623420
add a comment |Â
add a comment |Â
up vote
0
down vote
This is essentially Paul Frost's answer, in a slighly less verbose form.
The range $f(I)$ contains the interval $(f(a),f(b))$, so every point in the interval $(f(a),f(b))$ is of the form $f(x)$ for some $xin I$. On the other hand, only one endpoint of the non-empty interval $J$ is a value of $f$ - because $f$ jumps from $y_0^-$ to $y_0^+$. That is the contradiction.
answered Jul 24 at 12:53
uniquesolution
7,630721
Don't we have $f((a,b))subseteq (f(a),f(b))$ because $f$ is monotone increasing? It loks lik you claim it the other way round? And what do you mean by endpoints of the interval? They are open...
– mathstackuser
Jul 24 at 13:38
I am assuming that $I$ is an interval containing the points $a$ and $b$. Therefore $f(a)$ and $f(b)$ are values of $f$ that belong to the range, $f(I)$. Since the range is an interval, it must contain the interval $(f(a),f(b))$. In order to prove the inclusion that you wrote down we need continuity, but we don't have it yet. Yes, the interval is open, but surely you understand what are its end points, no?
– uniquesolution
Jul 24 at 13:46
... after all, you are using the notation $(a,b)$ to designate an open interval, by means of the "end points" $a,b$.
– uniquesolution
Jul 24 at 13:56
add a comment |Â
up vote
0
down vote
This is essentially Paul Frost's answer, in a slighly less verbose form.
The range $f(I)$ contains the interval $(f(a),f(b))$, so every point in the interval $(f(a),f(b))$ is of the form $f(x)$ for some $xin I$. On the other hand, only one endpoint of the non-empty interval $J$ is a value of $f$ - because $f$ jumps from $y_0^-$ to $y_0^+$. That is the contradiction.
answered Jul 24 at 12:53
uniquesolution
7,630721
Don't we have $f((a,b))subseteq (f(a),f(b))$ because $f$ is monotone increasing? It loks lik you claim it the other way round? And what do you mean by endpoints of the interval? They are open...
– mathstackuser
Jul 24 at 13:38
I am assuming that $I$ is an interval containing the points $a$ and $b$. Therefore $f(a)$ and $f(b)$ are values of $f$ that belong to the range, $f(I)$. Since the range is an interval, it must contain the interval $(f(a),f(b))$. In order to prove the inclusion that you wrote down we need continuity, but we don't have it yet. Yes, the interval is open, but surely you understand what are its end points, no?
– uniquesolution
Jul 24 at 13:46
... after all, you are using the notation $(a,b)$ to designate an open interval, by means of the "end points" $a,b$.
– uniquesolution
Jul 24 at 13:56
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is essentially Paul Frost's answer, in a slighly less verbose form.
The range $f(I)$ contains the interval $(f(a),f(b))$, so every point in the interval $(f(a),f(b))$ is of the form $f(x)$ for some $xin I$. On the other hand, only one endpoint of the non-empty interval $J$ is a value of $f$ - because $f$ jumps from $y_0^-$ to $y_0^+$. That is the contradiction.
answered Jul 24 at 12:53
uniquesolution
7,630721
This is essentially Paul Frost's answer, in a slighly less verbose form.
The range $f(I)$ contains the interval $(f(a),f(b))$, so every point in the interval $(f(a),f(b))$ is of the form $f(x)$ for some $xin I$. On the other hand, only one endpoint of the non-empty interval $J$ is a value of $f$ - because $f$ jumps from $y_0^-$ to $y_0^+$. That is the contradiction.
answered Jul 24 at 12:53
uniquesolution
7,630721
edited Jul 24 at 13:52
answered Jul 24 at 12:53
uniquesolution
7,630721
answered Jul 24 at 12:53
uniquesolution
7,630721
answered Jul 24 at 12:53
uniquesolution
7,630721
7,630721
Don't we have $f((a,b))subseteq (f(a),f(b))$ because $f$ is monotone increasing? It loks lik you claim it the other way round? And what do you mean by endpoints of the interval? They are open...
– mathstackuser
Jul 24 at 13:38
I am assuming that $I$ is an interval containing the points $a$ and $b$. Therefore $f(a)$ and $f(b)$ are values of $f$ that belong to the range, $f(I)$. Since the range is an interval, it must contain the interval $(f(a),f(b))$. In order to prove the inclusion that you wrote down we need continuity, but we don't have it yet. Yes, the interval is open, but surely you understand what are its end points, no?
– uniquesolution
Jul 24 at 13:46
... after all, you are using the notation $(a,b)$ to designate an open interval, by means of the "end points" $a,b$.
– uniquesolution
Jul 24 at 13:56
add a comment |Â
Don't we have $f((a,b))subseteq (f(a),f(b))$ because $f$ is monotone increasing? It loks lik you claim it the other way round? And what do you mean by endpoints of the interval? They are open...
– mathstackuser
Jul 24 at 13:38
I am assuming that $I$ is an interval containing the points $a$ and $b$. Therefore $f(a)$ and $f(b)$ are values of $f$ that belong to the range, $f(I)$. Since the range is an interval, it must contain the interval $(f(a),f(b))$. In order to prove the inclusion that you wrote down we need continuity, but we don't have it yet. Yes, the interval is open, but surely you understand what are its end points, no?
– uniquesolution
Jul 24 at 13:46
... after all, you are using the notation $(a,b)$ to designate an open interval, by means of the "end points" $a,b$.
– uniquesolution
Jul 24 at 13:56
Don't we have $f((a,b))subseteq (f(a),f(b))$ because $f$ is monotone increasing? It loks lik you claim it the other way round? And what do you mean by endpoints of the interval? They are open...
– mathstackuser
Jul 24 at 13:38
Don't we have $f((a,b))subseteq (f(a),f(b))$ because $f$ is monotone increasing? It loks lik you claim it the other way round? And what do you mean by endpoints of the interval? They are open...
– mathstackuser
Jul 24 at 13:38
I am assuming that $I$ is an interval containing the points $a$ and $b$. Therefore $f(a)$ and $f(b)$ are values of $f$ that belong to the range, $f(I)$. Since the range is an interval, it must contain the interval $(f(a),f(b))$. In order to prove the inclusion that you wrote down we need continuity, but we don't have it yet. Yes, the interval is open, but surely you understand what are its end points, no?
– uniquesolution
Jul 24 at 13:46
I am assuming that $I$ is an interval containing the points $a$ and $b$. Therefore $f(a)$ and $f(b)$ are values of $f$ that belong to the range, $f(I)$. Since the range is an interval, it must contain the interval $(f(a),f(b))$. In order to prove the inclusion that you wrote down we need continuity, but we don't have it yet. Yes, the interval is open, but surely you understand what are its end points, no?
– uniquesolution
Jul 24 at 13:46
... after all, you are using the notation $(a,b)$ to designate an open interval, by means of the "end points" $a,b$.
– uniquesolution
Jul 24 at 13:56
... after all, you are using the notation $(a,b)$ to designate an open interval, by means of the "end points" $a,b$.
– uniquesolution
Jul 24 at 13:56
add a comment |Â
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The contradiction comes in that for $yin J$, the equation $y = f(x)$ has no solution.
– BindersFull
Jul 24 at 12:58
Why is there no solution?
– mathstackuser
Jul 24 at 13:16