Showing that an injective Darboux function is strictly monotone.
$begingroup$
I was hoping someone could tell me how to prove the following problem I was given:
Let $f:[a,b]tomathbb{R}$ be a function such that for every $yin[f(a),f(b)]$ there exists $xin[a,b]$ such that $y=f(x)$. Then the function is strictly monotone.
Attemp 1:
Suppose $f(a)<f(b)$, and suppose that the function is not strictly increasing.
Then there exists $a',b'in{(a,b)}$ such that $a'lt b'$ and $ f(b')le{f(a')}$. First, note that $f(b')not=f(a')$, for otherwise the function is not one-to-one, so suppose $f(b')lt{f(a')}$.
Since $f$ is Darboux and since $f(a)<f(b')<f(a')$, there exists $a''in{(a,a')}$ such that $f(a'')=f(b')$, but this is a contradiction since $b'notin{(a,a'')}$ and $f$ is injective.
Attempt 2:
Suppose $f(a)<f(b)$, and suppose that the function is not strictly increasing.
Then there exists $a',b'in{(a,b)}$ such that $a'lt b'$ and $ f(b')le{f(a')}$. First, note that $f(b')not=f(a')$, for otherwise the function is not one-to-one, so suppose $f(b')lt{f(a')}$.
First, suppose $f(b')>f(b)$.
Since $f$ is Darboux and since $f(a)<f(b')<f(a')$, there exists $a''in{(a,a')}$ such that $f(a'')=f(b')$, but this is a contradiction since $b'notin{(a,a'')}$ and $f$ is injective.
Now, suppose $f(b')<f(b)$.
Without loss of generality we can assume $f(a')<f(b)$. Similarly, since $f$ is Darboux and since $f(b')<f(a')<f(b)$, there exists $b''in{(b',b)}$ such that $f(b'')=f(a')$, but this is a contradiction since $a''notin{(b',b)}$ and $f$ is injective.
real-analysis calculus proof-verification proof-writing
asked Dec 9 '18 at 20:36
Alex DAlex D
508219
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|
show 1 more comment
$begingroup$
I was hoping someone could tell me how to prove the following problem I was given:
Let $f:[a,b]tomathbb{R}$ be a function such that for every $yin[f(a),f(b)]$ there exists $xin[a,b]$ such that $y=f(x)$. Then the function is strictly monotone.
Attemp 1:
Suppose $f(a)<f(b)$, and suppose that the function is not strictly increasing.
Then there exists $a',b'in{(a,b)}$ such that $a'lt b'$ and $ f(b')le{f(a')}$. First, note that $f(b')not=f(a')$, for otherwise the function is not one-to-one, so suppose $f(b')lt{f(a')}$.
Since $f$ is Darboux and since $f(a)<f(b')<f(a')$, there exists $a''in{(a,a')}$ such that $f(a'')=f(b')$, but this is a contradiction since $b'notin{(a,a'')}$ and $f$ is injective.
Attempt 2:
Suppose $f(a)<f(b)$, and suppose that the function is not strictly increasing.
Then there exists $a',b'in{(a,b)}$ such that $a'lt b'$ and $ f(b')le{f(a')}$. First, note that $f(b')not=f(a')$, for otherwise the function is not one-to-one, so suppose $f(b')lt{f(a')}$.
First, suppose $f(b')>f(b)$.
Since $f$ is Darboux and since $f(a)<f(b')<f(a')$, there exists $a''in{(a,a')}$ such that $f(a'')=f(b')$, but this is a contradiction since $b'notin{(a,a'')}$ and $f$ is injective.
Now, suppose $f(b')<f(b)$.
Without loss of generality we can assume $f(a')<f(b)$. Similarly, since $f$ is Darboux and since $f(b')<f(a')<f(b)$, there exists $b''in{(b',b)}$ such that $f(b'')=f(a')$, but this is a contradiction since $a''notin{(b',b)}$ and $f$ is injective.
real-analysis calculus proof-verification proof-writing
asked Dec 9 '18 at 20:36
Alex DAlex D
508219
$endgroup$
$begingroup$
@egreg What would you want clarified, if you were being extremely critical?
$endgroup$
– Alex D
Dec 9 '18 at 21:02
1
$begingroup$
How do you know that $f(a)<f(b')<f(a')$? We know that $f(a)<f(b)$, but we don't know that $f(a)$ is an infimum.
$endgroup$
– Melody
Dec 9 '18 at 21:04
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@AlexD Actually, Melody is raising a good point.
$endgroup$
– egreg
Dec 9 '18 at 21:09
$begingroup$
@egreg Melody Ok guys I think I got it. Let me rewrite it. Thank you so much
$endgroup$
– Alex D
Dec 9 '18 at 21:20
$begingroup$
@Melody Is it better?
$endgroup$
– Alex D
Dec 9 '18 at 21:45
|
show 1 more comment
$begingroup$
I was hoping someone could tell me how to prove the following problem I was given:
Let $f:[a,b]tomathbb{R}$ be a function such that for every $yin[f(a),f(b)]$ there exists $xin[a,b]$ such that $y=f(x)$. Then the function is strictly monotone.
Attemp 1:
Suppose $f(a)<f(b)$, and suppose that the function is not strictly increasing.
Then there exists $a',b'in{(a,b)}$ such that $a'lt b'$ and $ f(b')le{f(a')}$. First, note that $f(b')not=f(a')$, for otherwise the function is not one-to-one, so suppose $f(b')lt{f(a')}$.
Since $f$ is Darboux and since $f(a)<f(b')<f(a')$, there exists $a''in{(a,a')}$ such that $f(a'')=f(b')$, but this is a contradiction since $b'notin{(a,a'')}$ and $f$ is injective.
Attempt 2:
Suppose $f(a)<f(b)$, and suppose that the function is not strictly increasing.
Then there exists $a',b'in{(a,b)}$ such that $a'lt b'$ and $ f(b')le{f(a')}$. First, note that $f(b')not=f(a')$, for otherwise the function is not one-to-one, so suppose $f(b')lt{f(a')}$.
First, suppose $f(b')>f(b)$.
Since $f$ is Darboux and since $f(a)<f(b')<f(a')$, there exists $a''in{(a,a')}$ such that $f(a'')=f(b')$, but this is a contradiction since $b'notin{(a,a'')}$ and $f$ is injective.
Now, suppose $f(b')<f(b)$.
Without loss of generality we can assume $f(a')<f(b)$. Similarly, since $f$ is Darboux and since $f(b')<f(a')<f(b)$, there exists $b''in{(b',b)}$ such that $f(b'')=f(a')$, but this is a contradiction since $a''notin{(b',b)}$ and $f$ is injective.
real-analysis calculus proof-verification proof-writing
asked Dec 9 '18 at 20:36
Alex DAlex D
508219
$endgroup$
I was hoping someone could tell me how to prove the following problem I was given:
Let $f:[a,b]tomathbb{R}$ be a function such that for every $yin[f(a),f(b)]$ there exists $xin[a,b]$ such that $y=f(x)$. Then the function is strictly monotone.
Attemp 1:
Suppose $f(a)<f(b)$, and suppose that the function is not strictly increasing.
Then there exists $a',b'in{(a,b)}$ such that $a'lt b'$ and $ f(b')le{f(a')}$. First, note that $f(b')not=f(a')$, for otherwise the function is not one-to-one, so suppose $f(b')lt{f(a')}$.
Since $f$ is Darboux and since $f(a)<f(b')<f(a')$, there exists $a''in{(a,a')}$ such that $f(a'')=f(b')$, but this is a contradiction since $b'notin{(a,a'')}$ and $f$ is injective.
Attempt 2:
Suppose $f(a)<f(b)$, and suppose that the function is not strictly increasing.
Then there exists $a',b'in{(a,b)}$ such that $a'lt b'$ and $ f(b')le{f(a')}$. First, note that $f(b')not=f(a')$, for otherwise the function is not one-to-one, so suppose $f(b')lt{f(a')}$.
First, suppose $f(b')>f(b)$.
Since $f$ is Darboux and since $f(a)<f(b')<f(a')$, there exists $a''in{(a,a')}$ such that $f(a'')=f(b')$, but this is a contradiction since $b'notin{(a,a'')}$ and $f$ is injective.
Now, suppose $f(b')<f(b)$.
Without loss of generality we can assume $f(a')<f(b)$. Similarly, since $f$ is Darboux and since $f(b')<f(a')<f(b)$, there exists $b''in{(b',b)}$ such that $f(b'')=f(a')$, but this is a contradiction since $a''notin{(b',b)}$ and $f$ is injective.
real-analysis calculus proof-verification proof-writing
real-analysis calculus proof-verification proof-writing
asked Dec 9 '18 at 20:36
Alex DAlex D
508219
asked Dec 9 '18 at 20:36
Alex DAlex D
508219
edited Dec 9 '18 at 21:36
Alex D
asked Dec 9 '18 at 20:36
Alex DAlex D
508219
asked Dec 9 '18 at 20:36
Alex DAlex D
508219
asked Dec 9 '18 at 20:36
Alex DAlex D
508219
508219
$begingroup$
@egreg What would you want clarified, if you were being extremely critical?
$endgroup$
– Alex D
Dec 9 '18 at 21:02
1
$begingroup$
How do you know that $f(a)<f(b')<f(a')$? We know that $f(a)<f(b)$, but we don't know that $f(a)$ is an infimum.
$endgroup$
– Melody
Dec 9 '18 at 21:04
$begingroup$
@AlexD Actually, Melody is raising a good point.
$endgroup$
– egreg
Dec 9 '18 at 21:09
$begingroup$
@egreg Melody Ok guys I think I got it. Let me rewrite it. Thank you so much
$endgroup$
– Alex D
Dec 9 '18 at 21:20
$begingroup$
@Melody Is it better?
$endgroup$
– Alex D
Dec 9 '18 at 21:45
|
show 1 more comment
$begingroup$
@egreg What would you want clarified, if you were being extremely critical?
$endgroup$
– Alex D
Dec 9 '18 at 21:02
1
$begingroup$
How do you know that $f(a)<f(b')<f(a')$? We know that $f(a)<f(b)$, but we don't know that $f(a)$ is an infimum.
$endgroup$
– Melody
Dec 9 '18 at 21:04
$begingroup$
@AlexD Actually, Melody is raising a good point.
$endgroup$
– egreg
Dec 9 '18 at 21:09
$begingroup$
@egreg Melody Ok guys I think I got it. Let me rewrite it. Thank you so much
$endgroup$
– Alex D
Dec 9 '18 at 21:20
$begingroup$
@Melody Is it better?
$endgroup$
– Alex D
Dec 9 '18 at 21:45
$begingroup$
@egreg What would you want clarified, if you were being extremely critical?
$endgroup$
– Alex D
Dec 9 '18 at 21:02
$begingroup$
@egreg What would you want clarified, if you were being extremely critical?
$endgroup$
– Alex D
Dec 9 '18 at 21:02
1
1
$begingroup$
How do you know that $f(a)<f(b')<f(a')$? We know that $f(a)<f(b)$, but we don't know that $f(a)$ is an infimum.
$endgroup$
– Melody
Dec 9 '18 at 21:04
$begingroup$
How do you know that $f(a)<f(b')<f(a')$? We know that $f(a)<f(b)$, but we don't know that $f(a)$ is an infimum.
$endgroup$
– Melody
Dec 9 '18 at 21:04
$begingroup$
@AlexD Actually, Melody is raising a good point.
$endgroup$
– egreg
Dec 9 '18 at 21:09
$begingroup$
@AlexD Actually, Melody is raising a good point.
$endgroup$
– egreg
Dec 9 '18 at 21:09
$begingroup$
@egreg Melody Ok guys I think I got it. Let me rewrite it. Thank you so much
$endgroup$
– Alex D
Dec 9 '18 at 21:20
$begingroup$
@egreg Melody Ok guys I think I got it. Let me rewrite it. Thank you so much
$endgroup$
– Alex D
Dec 9 '18 at 21:20
$begingroup$
@Melody Is it better?
$endgroup$
– Alex D
Dec 9 '18 at 21:45
$begingroup$
@Melody Is it better?
$endgroup$
– Alex D
Dec 9 '18 at 21:45
|
show 1 more comment
1 Answer
1
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$begingroup$
Without loss of generality we can assume $f(a)<f(b)$.
Assume $f$ is not strictly increasing. Then there are $c,din[a,b]$ with $c<d$ and $f(c)ge f(d)$. Since $f$ is injective, we can say that $f(c)>f(d)$.
At this point we don't know whether $f(a)le f(c)$ or $f(a)>f(c)$.
Suppose $f(a)>f(c)$. Then there is $a'in[c,b]$ with $f(a)=f(a')$, contradicting injectivity. Similarly we can exclude $f(d)>f(b)$. Hence we have
$$
f(a)le f(c),quad f(c)>f(d),quad f(d)le f(b)
$$
and your argument carries over after noticing that we must also have $f(c)le f(b)$ and $f(a)le f(d)$ (same reasoning as before).
answered Dec 9 '18 at 21:16
egregegreg
181k1485202
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add a comment |
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$begingroup$
Without loss of generality we can assume $f(a)<f(b)$.
Assume $f$ is not strictly increasing. Then there are $c,din[a,b]$ with $c<d$ and $f(c)ge f(d)$. Since $f$ is injective, we can say that $f(c)>f(d)$.
At this point we don't know whether $f(a)le f(c)$ or $f(a)>f(c)$.
Suppose $f(a)>f(c)$. Then there is $a'in[c,b]$ with $f(a)=f(a')$, contradicting injectivity. Similarly we can exclude $f(d)>f(b)$. Hence we have
$$
f(a)le f(c),quad f(c)>f(d),quad f(d)le f(b)
$$
and your argument carries over after noticing that we must also have $f(c)le f(b)$ and $f(a)le f(d)$ (same reasoning as before).
answered Dec 9 '18 at 21:16
egregegreg
181k1485202
$endgroup$
add a comment |
$begingroup$
Without loss of generality we can assume $f(a)<f(b)$.
Assume $f$ is not strictly increasing. Then there are $c,din[a,b]$ with $c<d$ and $f(c)ge f(d)$. Since $f$ is injective, we can say that $f(c)>f(d)$.
At this point we don't know whether $f(a)le f(c)$ or $f(a)>f(c)$.
Suppose $f(a)>f(c)$. Then there is $a'in[c,b]$ with $f(a)=f(a')$, contradicting injectivity. Similarly we can exclude $f(d)>f(b)$. Hence we have
$$
f(a)le f(c),quad f(c)>f(d),quad f(d)le f(b)
$$
and your argument carries over after noticing that we must also have $f(c)le f(b)$ and $f(a)le f(d)$ (same reasoning as before).
answered Dec 9 '18 at 21:16
egregegreg
181k1485202
$endgroup$
add a comment |
$begingroup$
Without loss of generality we can assume $f(a)<f(b)$.
Assume $f$ is not strictly increasing. Then there are $c,din[a,b]$ with $c<d$ and $f(c)ge f(d)$. Since $f$ is injective, we can say that $f(c)>f(d)$.
At this point we don't know whether $f(a)le f(c)$ or $f(a)>f(c)$.
Suppose $f(a)>f(c)$. Then there is $a'in[c,b]$ with $f(a)=f(a')$, contradicting injectivity. Similarly we can exclude $f(d)>f(b)$. Hence we have
$$
f(a)le f(c),quad f(c)>f(d),quad f(d)le f(b)
$$
and your argument carries over after noticing that we must also have $f(c)le f(b)$ and $f(a)le f(d)$ (same reasoning as before).
answered Dec 9 '18 at 21:16
egregegreg
181k1485202
$endgroup$
Without loss of generality we can assume $f(a)<f(b)$.
Assume $f$ is not strictly increasing. Then there are $c,din[a,b]$ with $c<d$ and $f(c)ge f(d)$. Since $f$ is injective, we can say that $f(c)>f(d)$.
At this point we don't know whether $f(a)le f(c)$ or $f(a)>f(c)$.
Suppose $f(a)>f(c)$. Then there is $a'in[c,b]$ with $f(a)=f(a')$, contradicting injectivity. Similarly we can exclude $f(d)>f(b)$. Hence we have
$$
f(a)le f(c),quad f(c)>f(d),quad f(d)le f(b)
$$
and your argument carries over after noticing that we must also have $f(c)le f(b)$ and $f(a)le f(d)$ (same reasoning as before).
answered Dec 9 '18 at 21:16
egregegreg
181k1485202
answered Dec 9 '18 at 21:16
egregegreg
181k1485202
answered Dec 9 '18 at 21:16
egregegreg
181k1485202
answered Dec 9 '18 at 21:16
egregegreg
181k1485202
181k1485202
add a comment |
add a comment |
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$begingroup$
@egreg What would you want clarified, if you were being extremely critical?
$endgroup$
– Alex D
Dec 9 '18 at 21:02
1
$begingroup$
How do you know that $f(a)<f(b')<f(a')$? We know that $f(a)<f(b)$, but we don't know that $f(a)$ is an infimum.
$endgroup$
– Melody
Dec 9 '18 at 21:04
$begingroup$
@AlexD Actually, Melody is raising a good point.
$endgroup$
– egreg
Dec 9 '18 at 21:09
$begingroup$
@egreg Melody Ok guys I think I got it. Let me rewrite it. Thank you so much
$endgroup$
– Alex D
Dec 9 '18 at 21:20
$begingroup$
@Melody Is it better?
$endgroup$
– Alex D
Dec 9 '18 at 21:45