If $f(x_1)=f(x_2)=y$, then prove that $f$ can not be continuous everywhere [duplicate]
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This question already has an answer here:
Can a continuous function from the reals to the reals assume each value an even number of times?
1 answer
Let $ f:[a,b]→mathbb R$ be a function with the following property:
For every $yin f([a,b])$, you have exactly two $ x_1, x_2 $,such as $f(x_1)=f(x_2)=y$.
Prove that f can not be continuous everywhere.
continuity definition
edited Dec 18 '18 at 23:23
Martund
1,667213
asked Dec 18 '18 at 22:44
Chris RafaelChris Rafael
106
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marked as duplicate by Theo Bendit, Cesareo, Shailesh, metamorphy, Will Fisher Dec 19 '18 at 1:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Can a continuous function from the reals to the reals assume each value an even number of times?
1 answer
Let $ f:[a,b]→mathbb R$ be a function with the following property:
For every $yin f([a,b])$, you have exactly two $ x_1, x_2 $,such as $f(x_1)=f(x_2)=y$.
Prove that f can not be continuous everywhere.
continuity definition
edited Dec 18 '18 at 23:23
Martund
1,667213
asked Dec 18 '18 at 22:44
Chris RafaelChris Rafael
106
$endgroup$
marked as duplicate by Theo Bendit, Cesareo, Shailesh, metamorphy, Will Fisher Dec 19 '18 at 1:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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@TheoBendit I think that the fact that the domain of $f$ is a closed interval in this question is relevant.
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– rafa11111
Dec 18 '18 at 23:35
$begingroup$
@rafa11111 That's a good point. However, in the step where an $a in mathbb{R}$ is fixed, you can salvage the proof by ensuring that $a$ is not chosen to be the image of either end point.
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– Theo Bendit
Dec 18 '18 at 23:39
add a comment |
$begingroup$
This question already has an answer here:
Can a continuous function from the reals to the reals assume each value an even number of times?
1 answer
Let $ f:[a,b]→mathbb R$ be a function with the following property:
For every $yin f([a,b])$, you have exactly two $ x_1, x_2 $,such as $f(x_1)=f(x_2)=y$.
Prove that f can not be continuous everywhere.
continuity definition
edited Dec 18 '18 at 23:23
Martund
1,667213
asked Dec 18 '18 at 22:44
Chris RafaelChris Rafael
106
$endgroup$
This question already has an answer here:
Can a continuous function from the reals to the reals assume each value an even number of times?
1 answer
Let $ f:[a,b]→mathbb R$ be a function with the following property:
For every $yin f([a,b])$, you have exactly two $ x_1, x_2 $,such as $f(x_1)=f(x_2)=y$.
Prove that f can not be continuous everywhere.
This question already has an answer here:
Can a continuous function from the reals to the reals assume each value an even number of times?
1 answer
continuity definition
continuity definition
edited Dec 18 '18 at 23:23
Martund
1,667213
asked Dec 18 '18 at 22:44
Chris RafaelChris Rafael
106
edited Dec 18 '18 at 23:23
Martund
1,667213
asked Dec 18 '18 at 22:44
Chris RafaelChris Rafael
106
edited Dec 18 '18 at 23:23
Martund
1,667213
edited Dec 18 '18 at 23:23
Martund
1,667213
edited Dec 18 '18 at 23:23
Martund
1,667213
1,667213
asked Dec 18 '18 at 22:44
Chris RafaelChris Rafael
106
asked Dec 18 '18 at 22:44
Chris RafaelChris Rafael
106
asked Dec 18 '18 at 22:44
Chris RafaelChris Rafael
106
106
marked as duplicate by Theo Bendit, Cesareo, Shailesh, metamorphy, Will Fisher Dec 19 '18 at 1:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Theo Bendit, Cesareo, Shailesh, metamorphy, Will Fisher Dec 19 '18 at 1:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
@TheoBendit I think that the fact that the domain of $f$ is a closed interval in this question is relevant.
$endgroup$
– rafa11111
Dec 18 '18 at 23:35
$begingroup$
@rafa11111 That's a good point. However, in the step where an $a in mathbb{R}$ is fixed, you can salvage the proof by ensuring that $a$ is not chosen to be the image of either end point.
$endgroup$
– Theo Bendit
Dec 18 '18 at 23:39
add a comment |
$begingroup$
@TheoBendit I think that the fact that the domain of $f$ is a closed interval in this question is relevant.
$endgroup$
– rafa11111
Dec 18 '18 at 23:35
$begingroup$
@rafa11111 That's a good point. However, in the step where an $a in mathbb{R}$ is fixed, you can salvage the proof by ensuring that $a$ is not chosen to be the image of either end point.
$endgroup$
– Theo Bendit
Dec 18 '18 at 23:39
$begingroup$
@TheoBendit I think that the fact that the domain of $f$ is a closed interval in this question is relevant.
$endgroup$
– rafa11111
Dec 18 '18 at 23:35
$begingroup$
@TheoBendit I think that the fact that the domain of $f$ is a closed interval in this question is relevant.
$endgroup$
– rafa11111
Dec 18 '18 at 23:35
$begingroup$
@rafa11111 That's a good point. However, in the step where an $a in mathbb{R}$ is fixed, you can salvage the proof by ensuring that $a$ is not chosen to be the image of either end point.
$endgroup$
– Theo Bendit
Dec 18 '18 at 23:39
$begingroup$
@rafa11111 That's a good point. However, in the step where an $a in mathbb{R}$ is fixed, you can salvage the proof by ensuring that $a$ is not chosen to be the image of either end point.
$endgroup$
– Theo Bendit
Dec 18 '18 at 23:39
add a comment |
2 Answers
2
active
oldest
votes
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By the extreme value theorem, $f$ attains its maximum at some point. Then it must attain its maximum at two points: let us name them $x_1 < x_2$. Now, $f$ restricted to the interval $[x_1, x_2]$ must attain its minimum at some point $ x_3 in (x_1, x_2)$. And there must be some other point $x_4 in [a, b]$ with $f(x_4)=f(x_3)$. If $x_4 notin (x_1,x_2)$, then by the intermediate value theorem $f$ must attain the value $frac{f(x_1)+f(x_3)}{2}$ at least three times. And if $x_4 in (x_1, x_2)$, let us pick any $x_5$ between $x_3$ and $x_4$: then $f$ must attain $f(x_5)$ at least three times (because it is between $f(x_1)$ and $f(x_3)$).
answered Dec 18 '18 at 23:42
Lucas Kenji MooriLucas Kenji Moori
384
$endgroup$
$begingroup$
Where do we must end up to show that f isn't continuous everywhere ? Can you post the full answer because i am stuck ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 10:36
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If I understood Lucas' answer, he showed in the end that there is a contradiction, because there will be three points with the same value instead of two. However, the summoning of the intermediate value theorem requires the hypothesis that $f$ is continuous. The contradiction arose due to an incorrect assumption, which, in the case, is the continuity of $f$.
$endgroup$
– rafa11111
Dec 19 '18 at 11:08
$begingroup$
@rafa11111First , why must f attain its maximum at two points ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:17
$begingroup$
@ChrisRafael because the function has each value at two different points, it's on the statement of your question.
$endgroup$
– rafa11111
Dec 19 '18 at 11:18
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because f(x1)=f(x2) ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:18
|
show 5 more comments
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Hint: assume there is such an f and then it has exactly two absolute minimum points and two absolute maximum points; if say one of the absolute minimum points is strictly inside the interval, show that there are small intervals to the right and left of it which are sent by f to the same interval and then you get a contradiction using the same property for the other absolute minimum point for which there is such a small interval either to the left or right (or both of course if it is inside); so the function must take its absolute minimum at the ends; in the same way it must take its absolute max at the ends and of course f is non-constant so that is impossible
Taking the absolute value or square functions shows that you cannot relax the condition to exclude one point in the interval
answered Dec 18 '18 at 23:44
ConradConrad
50924
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By the extreme value theorem, $f$ attains its maximum at some point. Then it must attain its maximum at two points: let us name them $x_1 < x_2$. Now, $f$ restricted to the interval $[x_1, x_2]$ must attain its minimum at some point $ x_3 in (x_1, x_2)$. And there must be some other point $x_4 in [a, b]$ with $f(x_4)=f(x_3)$. If $x_4 notin (x_1,x_2)$, then by the intermediate value theorem $f$ must attain the value $frac{f(x_1)+f(x_3)}{2}$ at least three times. And if $x_4 in (x_1, x_2)$, let us pick any $x_5$ between $x_3$ and $x_4$: then $f$ must attain $f(x_5)$ at least three times (because it is between $f(x_1)$ and $f(x_3)$).
answered Dec 18 '18 at 23:42
Lucas Kenji MooriLucas Kenji Moori
384
$endgroup$
$begingroup$
Where do we must end up to show that f isn't continuous everywhere ? Can you post the full answer because i am stuck ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 10:36
$begingroup$
If I understood Lucas' answer, he showed in the end that there is a contradiction, because there will be three points with the same value instead of two. However, the summoning of the intermediate value theorem requires the hypothesis that $f$ is continuous. The contradiction arose due to an incorrect assumption, which, in the case, is the continuity of $f$.
$endgroup$
– rafa11111
Dec 19 '18 at 11:08
$begingroup$
@rafa11111First , why must f attain its maximum at two points ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:17
$begingroup$
@ChrisRafael because the function has each value at two different points, it's on the statement of your question.
$endgroup$
– rafa11111
Dec 19 '18 at 11:18
$begingroup$
because f(x1)=f(x2) ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:18
|
show 5 more comments
$begingroup$
By the extreme value theorem, $f$ attains its maximum at some point. Then it must attain its maximum at two points: let us name them $x_1 < x_2$. Now, $f$ restricted to the interval $[x_1, x_2]$ must attain its minimum at some point $ x_3 in (x_1, x_2)$. And there must be some other point $x_4 in [a, b]$ with $f(x_4)=f(x_3)$. If $x_4 notin (x_1,x_2)$, then by the intermediate value theorem $f$ must attain the value $frac{f(x_1)+f(x_3)}{2}$ at least three times. And if $x_4 in (x_1, x_2)$, let us pick any $x_5$ between $x_3$ and $x_4$: then $f$ must attain $f(x_5)$ at least three times (because it is between $f(x_1)$ and $f(x_3)$).
answered Dec 18 '18 at 23:42
Lucas Kenji MooriLucas Kenji Moori
384
$endgroup$
$begingroup$
Where do we must end up to show that f isn't continuous everywhere ? Can you post the full answer because i am stuck ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 10:36
$begingroup$
If I understood Lucas' answer, he showed in the end that there is a contradiction, because there will be three points with the same value instead of two. However, the summoning of the intermediate value theorem requires the hypothesis that $f$ is continuous. The contradiction arose due to an incorrect assumption, which, in the case, is the continuity of $f$.
$endgroup$
– rafa11111
Dec 19 '18 at 11:08
$begingroup$
@rafa11111First , why must f attain its maximum at two points ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:17
$begingroup$
@ChrisRafael because the function has each value at two different points, it's on the statement of your question.
$endgroup$
– rafa11111
Dec 19 '18 at 11:18
$begingroup$
because f(x1)=f(x2) ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:18
|
show 5 more comments
$begingroup$
By the extreme value theorem, $f$ attains its maximum at some point. Then it must attain its maximum at two points: let us name them $x_1 < x_2$. Now, $f$ restricted to the interval $[x_1, x_2]$ must attain its minimum at some point $ x_3 in (x_1, x_2)$. And there must be some other point $x_4 in [a, b]$ with $f(x_4)=f(x_3)$. If $x_4 notin (x_1,x_2)$, then by the intermediate value theorem $f$ must attain the value $frac{f(x_1)+f(x_3)}{2}$ at least three times. And if $x_4 in (x_1, x_2)$, let us pick any $x_5$ between $x_3$ and $x_4$: then $f$ must attain $f(x_5)$ at least three times (because it is between $f(x_1)$ and $f(x_3)$).
answered Dec 18 '18 at 23:42
Lucas Kenji MooriLucas Kenji Moori
384
$endgroup$
By the extreme value theorem, $f$ attains its maximum at some point. Then it must attain its maximum at two points: let us name them $x_1 < x_2$. Now, $f$ restricted to the interval $[x_1, x_2]$ must attain its minimum at some point $ x_3 in (x_1, x_2)$. And there must be some other point $x_4 in [a, b]$ with $f(x_4)=f(x_3)$. If $x_4 notin (x_1,x_2)$, then by the intermediate value theorem $f$ must attain the value $frac{f(x_1)+f(x_3)}{2}$ at least three times. And if $x_4 in (x_1, x_2)$, let us pick any $x_5$ between $x_3$ and $x_4$: then $f$ must attain $f(x_5)$ at least three times (because it is between $f(x_1)$ and $f(x_3)$).
answered Dec 18 '18 at 23:42
Lucas Kenji MooriLucas Kenji Moori
384
edited Dec 18 '18 at 23:52
answered Dec 18 '18 at 23:42
Lucas Kenji MooriLucas Kenji Moori
384
answered Dec 18 '18 at 23:42
Lucas Kenji MooriLucas Kenji Moori
384
answered Dec 18 '18 at 23:42
Lucas Kenji MooriLucas Kenji Moori
384
384
$begingroup$
Where do we must end up to show that f isn't continuous everywhere ? Can you post the full answer because i am stuck ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 10:36
$begingroup$
If I understood Lucas' answer, he showed in the end that there is a contradiction, because there will be three points with the same value instead of two. However, the summoning of the intermediate value theorem requires the hypothesis that $f$ is continuous. The contradiction arose due to an incorrect assumption, which, in the case, is the continuity of $f$.
$endgroup$
– rafa11111
Dec 19 '18 at 11:08
$begingroup$
@rafa11111First , why must f attain its maximum at two points ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:17
$begingroup$
@ChrisRafael because the function has each value at two different points, it's on the statement of your question.
$endgroup$
– rafa11111
Dec 19 '18 at 11:18
$begingroup$
because f(x1)=f(x2) ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:18
|
show 5 more comments
$begingroup$
Where do we must end up to show that f isn't continuous everywhere ? Can you post the full answer because i am stuck ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 10:36
$begingroup$
If I understood Lucas' answer, he showed in the end that there is a contradiction, because there will be three points with the same value instead of two. However, the summoning of the intermediate value theorem requires the hypothesis that $f$ is continuous. The contradiction arose due to an incorrect assumption, which, in the case, is the continuity of $f$.
$endgroup$
– rafa11111
Dec 19 '18 at 11:08
$begingroup$
@rafa11111First , why must f attain its maximum at two points ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:17
$begingroup$
@ChrisRafael because the function has each value at two different points, it's on the statement of your question.
$endgroup$
– rafa11111
Dec 19 '18 at 11:18
$begingroup$
because f(x1)=f(x2) ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:18
$begingroup$
Where do we must end up to show that f isn't continuous everywhere ? Can you post the full answer because i am stuck ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 10:36
$begingroup$
Where do we must end up to show that f isn't continuous everywhere ? Can you post the full answer because i am stuck ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 10:36
$begingroup$
If I understood Lucas' answer, he showed in the end that there is a contradiction, because there will be three points with the same value instead of two. However, the summoning of the intermediate value theorem requires the hypothesis that $f$ is continuous. The contradiction arose due to an incorrect assumption, which, in the case, is the continuity of $f$.
$endgroup$
– rafa11111
Dec 19 '18 at 11:08
$begingroup$
If I understood Lucas' answer, he showed in the end that there is a contradiction, because there will be three points with the same value instead of two. However, the summoning of the intermediate value theorem requires the hypothesis that $f$ is continuous. The contradiction arose due to an incorrect assumption, which, in the case, is the continuity of $f$.
$endgroup$
– rafa11111
Dec 19 '18 at 11:08
$begingroup$
@rafa11111First , why must f attain its maximum at two points ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:17
$begingroup$
@rafa11111First , why must f attain its maximum at two points ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:17
$begingroup$
@ChrisRafael because the function has each value at two different points, it's on the statement of your question.
$endgroup$
– rafa11111
Dec 19 '18 at 11:18
$begingroup$
@ChrisRafael because the function has each value at two different points, it's on the statement of your question.
$endgroup$
– rafa11111
Dec 19 '18 at 11:18
$begingroup$
because f(x1)=f(x2) ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:18
$begingroup$
because f(x1)=f(x2) ?
$endgroup$
– Chris Rafael
Dec 19 '18 at 11:18
|
show 5 more comments
$begingroup$
Hint: assume there is such an f and then it has exactly two absolute minimum points and two absolute maximum points; if say one of the absolute minimum points is strictly inside the interval, show that there are small intervals to the right and left of it which are sent by f to the same interval and then you get a contradiction using the same property for the other absolute minimum point for which there is such a small interval either to the left or right (or both of course if it is inside); so the function must take its absolute minimum at the ends; in the same way it must take its absolute max at the ends and of course f is non-constant so that is impossible
Taking the absolute value or square functions shows that you cannot relax the condition to exclude one point in the interval
answered Dec 18 '18 at 23:44
ConradConrad
50924
$endgroup$
add a comment |
$begingroup$
Hint: assume there is such an f and then it has exactly two absolute minimum points and two absolute maximum points; if say one of the absolute minimum points is strictly inside the interval, show that there are small intervals to the right and left of it which are sent by f to the same interval and then you get a contradiction using the same property for the other absolute minimum point for which there is such a small interval either to the left or right (or both of course if it is inside); so the function must take its absolute minimum at the ends; in the same way it must take its absolute max at the ends and of course f is non-constant so that is impossible
Taking the absolute value or square functions shows that you cannot relax the condition to exclude one point in the interval
answered Dec 18 '18 at 23:44
ConradConrad
50924
$endgroup$
add a comment |
$begingroup$
Hint: assume there is such an f and then it has exactly two absolute minimum points and two absolute maximum points; if say one of the absolute minimum points is strictly inside the interval, show that there are small intervals to the right and left of it which are sent by f to the same interval and then you get a contradiction using the same property for the other absolute minimum point for which there is such a small interval either to the left or right (or both of course if it is inside); so the function must take its absolute minimum at the ends; in the same way it must take its absolute max at the ends and of course f is non-constant so that is impossible
Taking the absolute value or square functions shows that you cannot relax the condition to exclude one point in the interval
answered Dec 18 '18 at 23:44
ConradConrad
50924
$endgroup$
Hint: assume there is such an f and then it has exactly two absolute minimum points and two absolute maximum points; if say one of the absolute minimum points is strictly inside the interval, show that there are small intervals to the right and left of it which are sent by f to the same interval and then you get a contradiction using the same property for the other absolute minimum point for which there is such a small interval either to the left or right (or both of course if it is inside); so the function must take its absolute minimum at the ends; in the same way it must take its absolute max at the ends and of course f is non-constant so that is impossible
Taking the absolute value or square functions shows that you cannot relax the condition to exclude one point in the interval
answered Dec 18 '18 at 23:44
ConradConrad
50924
answered Dec 18 '18 at 23:44
ConradConrad
50924
answered Dec 18 '18 at 23:44
ConradConrad
50924
answered Dec 18 '18 at 23:44
ConradConrad
50924
50924
add a comment |
add a comment |
$begingroup$
@TheoBendit I think that the fact that the domain of $f$ is a closed interval in this question is relevant.
$endgroup$
– rafa11111
Dec 18 '18 at 23:35
$begingroup$
@rafa11111 That's a good point. However, in the step where an $a in mathbb{R}$ is fixed, you can salvage the proof by ensuring that $a$ is not chosen to be the image of either end point.
$endgroup$
– Theo Bendit
Dec 18 '18 at 23:39