If a function is continuous and one-to-one then it's strictly monotonic
$begingroup$
Let $f:(A,B)to mathbb{R}^1$ be continuous and one-to-one.
Prove that $f$ is strictly monotonic on $(A,B)$.
Proof:
Suppose $f$ is not strictly monotonic on $(A,B)$. Then there exist $a<b<c$ such that $f(a)geqslant f(b)$ and $f(c)geqslant f(b)$.
Suppose that $f(a)geqslant f(c)$. Since $[a,b]$ is connected and $f$ is continuous then $f([a,b])$ is connected.
It's easy to prove that $[f(b),f(a)]subset f([a,b])$.
If $f(b)=f(a)$, then $f$ is not one-to-one and we have contradiction.
If $f(b)<f(a)$, then $f(c)in f([a,b])$ and by the intermediate value theorem $exists c'in [a,b]$ such that $f(c')=f(c)$ and $f$ is not one-to-one.
Sorry if this topic repeated but I would like to know is my proof correct?
Thanks in advance.
real-analysis proof-verification proof-writing
edited Sep 28 '18 at 15:32
amWhy
192k28225439
asked Feb 23 '16 at 7:15
ZFRZFR
5,00631339
$endgroup$
|
show 5 more comments
$begingroup$
Let $f:(A,B)to mathbb{R}^1$ be continuous and one-to-one.
Prove that $f$ is strictly monotonic on $(A,B)$.
Proof:
Suppose $f$ is not strictly monotonic on $(A,B)$. Then there exist $a<b<c$ such that $f(a)geqslant f(b)$ and $f(c)geqslant f(b)$.
Suppose that $f(a)geqslant f(c)$. Since $[a,b]$ is connected and $f$ is continuous then $f([a,b])$ is connected.
It's easy to prove that $[f(b),f(a)]subset f([a,b])$.
If $f(b)=f(a)$, then $f$ is not one-to-one and we have contradiction.
If $f(b)<f(a)$, then $f(c)in f([a,b])$ and by the intermediate value theorem $exists c'in [a,b]$ such that $f(c')=f(c)$ and $f$ is not one-to-one.
Sorry if this topic repeated but I would like to know is my proof correct?
Thanks in advance.
real-analysis proof-verification proof-writing
edited Sep 28 '18 at 15:32
amWhy
192k28225439
asked Feb 23 '16 at 7:15
ZFRZFR
5,00631339
$endgroup$
$begingroup$
what if $f(a)$ or $f(b)$ are undefined?
$endgroup$
– uniquesolution
Feb 23 '16 at 8:14
$begingroup$
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:18
1
$begingroup$
Mostly right, but I don't get a few things. You suppose $f(a)ge f(c)$, but it seems you never consider $f(a)<f(c)$. Your initial statement of what it means for $f$ to not be monotonic seems odd — given that $f$ is 1-1, not monotonic means there are two pairs $a<b, c<d$ with $f(a)<f(b), f(c)>f(d)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:22
$begingroup$
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
$endgroup$
– ZFR
Feb 23 '16 at 15:49
1
$begingroup$
Because he's saying if it's not monotonic then there is a set of three points where the middle one is beneath the other two. Checking f(c) >= f(a) is silly. It's parallel so there's no benefit in reproving that half. We've already proved it with one half.
$endgroup$
– The Great Duck
Apr 24 '16 at 7:14
|
show 5 more comments
$begingroup$
Let $f:(A,B)to mathbb{R}^1$ be continuous and one-to-one.
Prove that $f$ is strictly monotonic on $(A,B)$.
Proof:
Suppose $f$ is not strictly monotonic on $(A,B)$. Then there exist $a<b<c$ such that $f(a)geqslant f(b)$ and $f(c)geqslant f(b)$.
Suppose that $f(a)geqslant f(c)$. Since $[a,b]$ is connected and $f$ is continuous then $f([a,b])$ is connected.
It's easy to prove that $[f(b),f(a)]subset f([a,b])$.
If $f(b)=f(a)$, then $f$ is not one-to-one and we have contradiction.
If $f(b)<f(a)$, then $f(c)in f([a,b])$ and by the intermediate value theorem $exists c'in [a,b]$ such that $f(c')=f(c)$ and $f$ is not one-to-one.
Sorry if this topic repeated but I would like to know is my proof correct?
Thanks in advance.
real-analysis proof-verification proof-writing
edited Sep 28 '18 at 15:32
amWhy
192k28225439
asked Feb 23 '16 at 7:15
ZFRZFR
5,00631339
$endgroup$
Let $f:(A,B)to mathbb{R}^1$ be continuous and one-to-one.
Prove that $f$ is strictly monotonic on $(A,B)$.
Proof:
Suppose $f$ is not strictly monotonic on $(A,B)$. Then there exist $a<b<c$ such that $f(a)geqslant f(b)$ and $f(c)geqslant f(b)$.
Suppose that $f(a)geqslant f(c)$. Since $[a,b]$ is connected and $f$ is continuous then $f([a,b])$ is connected.
It's easy to prove that $[f(b),f(a)]subset f([a,b])$.
If $f(b)=f(a)$, then $f$ is not one-to-one and we have contradiction.
If $f(b)<f(a)$, then $f(c)in f([a,b])$ and by the intermediate value theorem $exists c'in [a,b]$ such that $f(c')=f(c)$ and $f$ is not one-to-one.
Sorry if this topic repeated but I would like to know is my proof correct?
Thanks in advance.
real-analysis proof-verification proof-writing
real-analysis proof-verification proof-writing
edited Sep 28 '18 at 15:32
amWhy
192k28225439
asked Feb 23 '16 at 7:15
ZFRZFR
5,00631339
edited Sep 28 '18 at 15:32
amWhy
192k28225439
asked Feb 23 '16 at 7:15
ZFRZFR
5,00631339
edited Sep 28 '18 at 15:32
amWhy
192k28225439
edited Sep 28 '18 at 15:32
amWhy
192k28225439
edited Sep 28 '18 at 15:32
amWhy
192k28225439
192k28225439
asked Feb 23 '16 at 7:15
ZFRZFR
5,00631339
asked Feb 23 '16 at 7:15
ZFRZFR
5,00631339
asked Feb 23 '16 at 7:15
ZFRZFR
5,00631339
5,00631339
$begingroup$
what if $f(a)$ or $f(b)$ are undefined?
$endgroup$
– uniquesolution
Feb 23 '16 at 8:14
$begingroup$
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:18
1
$begingroup$
Mostly right, but I don't get a few things. You suppose $f(a)ge f(c)$, but it seems you never consider $f(a)<f(c)$. Your initial statement of what it means for $f$ to not be monotonic seems odd — given that $f$ is 1-1, not monotonic means there are two pairs $a<b, c<d$ with $f(a)<f(b), f(c)>f(d)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:22
$begingroup$
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
$endgroup$
– ZFR
Feb 23 '16 at 15:49
1
$begingroup$
Because he's saying if it's not monotonic then there is a set of three points where the middle one is beneath the other two. Checking f(c) >= f(a) is silly. It's parallel so there's no benefit in reproving that half. We've already proved it with one half.
$endgroup$
– The Great Duck
Apr 24 '16 at 7:14
|
show 5 more comments
$begingroup$
what if $f(a)$ or $f(b)$ are undefined?
$endgroup$
– uniquesolution
Feb 23 '16 at 8:14
$begingroup$
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:18
1
$begingroup$
Mostly right, but I don't get a few things. You suppose $f(a)ge f(c)$, but it seems you never consider $f(a)<f(c)$. Your initial statement of what it means for $f$ to not be monotonic seems odd — given that $f$ is 1-1, not monotonic means there are two pairs $a<b, c<d$ with $f(a)<f(b), f(c)>f(d)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:22
$begingroup$
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
$endgroup$
– ZFR
Feb 23 '16 at 15:49
1
$begingroup$
Because he's saying if it's not monotonic then there is a set of three points where the middle one is beneath the other two. Checking f(c) >= f(a) is silly. It's parallel so there's no benefit in reproving that half. We've already proved it with one half.
$endgroup$
– The Great Duck
Apr 24 '16 at 7:14
$begingroup$
what if $f(a)$ or $f(b)$ are undefined?
$endgroup$
– uniquesolution
Feb 23 '16 at 8:14
$begingroup$
what if $f(a)$ or $f(b)$ are undefined?
$endgroup$
– uniquesolution
Feb 23 '16 at 8:14
$begingroup$
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:18
$begingroup$
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:18
1
1
$begingroup$
Mostly right, but I don't get a few things. You suppose $f(a)ge f(c)$, but it seems you never consider $f(a)<f(c)$. Your initial statement of what it means for $f$ to not be monotonic seems odd — given that $f$ is 1-1, not monotonic means there are two pairs $a<b, c<d$ with $f(a)<f(b), f(c)>f(d)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:22
$begingroup$
Mostly right, but I don't get a few things. You suppose $f(a)ge f(c)$, but it seems you never consider $f(a)<f(c)$. Your initial statement of what it means for $f$ to not be monotonic seems odd — given that $f$ is 1-1, not monotonic means there are two pairs $a<b, c<d$ with $f(a)<f(b), f(c)>f(d)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:22
$begingroup$
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
$endgroup$
– ZFR
Feb 23 '16 at 15:49
$begingroup$
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
$endgroup$
– ZFR
Feb 23 '16 at 15:49
1
1
$begingroup$
Because he's saying if it's not monotonic then there is a set of three points where the middle one is beneath the other two. Checking f(c) >= f(a) is silly. It's parallel so there's no benefit in reproving that half. We've already proved it with one half.
$endgroup$
– The Great Duck
Apr 24 '16 at 7:14
$begingroup$
Because he's saying if it's not monotonic then there is a set of three points where the middle one is beneath the other two. Checking f(c) >= f(a) is silly. It's parallel so there's no benefit in reproving that half. We've already proved it with one half.
$endgroup$
– The Great Duck
Apr 24 '16 at 7:14
|
show 5 more comments
1 Answer
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$begingroup$
Like BrianO said, your reduction to studying how $f$ behaves with three points is a little quick. However, the reduction is elementary in the sense that it only uses ordered field axioms and not the LUB or other analytic theorems.
Let me just give the details of the reduction:
A map is monotone iff:
$(i)$: it is monotone on every set of four points or less.
This is just figuring out the negation of being monotone: if $f$ is not monotone then there are points $a leq b$, $c leq d$ in its domain such that $f(a) > f(b)$ (non increasing) and $f(c) < f(d)$ (non decreasing) and $f$ is not monotone on ${a;b;c;d}$.
$(ii)$: it is monotone on every set of three points or less.
Indeed, assume $f$ satisfies $(ii)$. Let $a < b < c < d$ be four points with for instance $f(a) leq f(b) leq f(c)$.
If $f(a) = f(b) = f(c)$ then $f$ is monotone on ${a;b;c;d}$: increasing if $f(d) geq f(a)$ and decreasing otherwise. Else, if for instance $f(a) < f(c)$, then applying $(ii)$ to ${a;c;d}$ yields $f(a) < f(c) leq f(d)$ so $f$ is increasing on ${a;b;c;d}$. The three other cases work the same.
answered Feb 24 '16 at 6:33
nombrenombre
2,634913
$endgroup$
add a comment |
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$begingroup$
Like BrianO said, your reduction to studying how $f$ behaves with three points is a little quick. However, the reduction is elementary in the sense that it only uses ordered field axioms and not the LUB or other analytic theorems.
Let me just give the details of the reduction:
A map is monotone iff:
$(i)$: it is monotone on every set of four points or less.
This is just figuring out the negation of being monotone: if $f$ is not monotone then there are points $a leq b$, $c leq d$ in its domain such that $f(a) > f(b)$ (non increasing) and $f(c) < f(d)$ (non decreasing) and $f$ is not monotone on ${a;b;c;d}$.
$(ii)$: it is monotone on every set of three points or less.
Indeed, assume $f$ satisfies $(ii)$. Let $a < b < c < d$ be four points with for instance $f(a) leq f(b) leq f(c)$.
If $f(a) = f(b) = f(c)$ then $f$ is monotone on ${a;b;c;d}$: increasing if $f(d) geq f(a)$ and decreasing otherwise. Else, if for instance $f(a) < f(c)$, then applying $(ii)$ to ${a;c;d}$ yields $f(a) < f(c) leq f(d)$ so $f$ is increasing on ${a;b;c;d}$. The three other cases work the same.
answered Feb 24 '16 at 6:33
nombrenombre
2,634913
$endgroup$
add a comment |
$begingroup$
Like BrianO said, your reduction to studying how $f$ behaves with three points is a little quick. However, the reduction is elementary in the sense that it only uses ordered field axioms and not the LUB or other analytic theorems.
Let me just give the details of the reduction:
A map is monotone iff:
$(i)$: it is monotone on every set of four points or less.
This is just figuring out the negation of being monotone: if $f$ is not monotone then there are points $a leq b$, $c leq d$ in its domain such that $f(a) > f(b)$ (non increasing) and $f(c) < f(d)$ (non decreasing) and $f$ is not monotone on ${a;b;c;d}$.
$(ii)$: it is monotone on every set of three points or less.
Indeed, assume $f$ satisfies $(ii)$. Let $a < b < c < d$ be four points with for instance $f(a) leq f(b) leq f(c)$.
If $f(a) = f(b) = f(c)$ then $f$ is monotone on ${a;b;c;d}$: increasing if $f(d) geq f(a)$ and decreasing otherwise. Else, if for instance $f(a) < f(c)$, then applying $(ii)$ to ${a;c;d}$ yields $f(a) < f(c) leq f(d)$ so $f$ is increasing on ${a;b;c;d}$. The three other cases work the same.
answered Feb 24 '16 at 6:33
nombrenombre
2,634913
$endgroup$
add a comment |
$begingroup$
Like BrianO said, your reduction to studying how $f$ behaves with three points is a little quick. However, the reduction is elementary in the sense that it only uses ordered field axioms and not the LUB or other analytic theorems.
Let me just give the details of the reduction:
A map is monotone iff:
$(i)$: it is monotone on every set of four points or less.
This is just figuring out the negation of being monotone: if $f$ is not monotone then there are points $a leq b$, $c leq d$ in its domain such that $f(a) > f(b)$ (non increasing) and $f(c) < f(d)$ (non decreasing) and $f$ is not monotone on ${a;b;c;d}$.
$(ii)$: it is monotone on every set of three points or less.
Indeed, assume $f$ satisfies $(ii)$. Let $a < b < c < d$ be four points with for instance $f(a) leq f(b) leq f(c)$.
If $f(a) = f(b) = f(c)$ then $f$ is monotone on ${a;b;c;d}$: increasing if $f(d) geq f(a)$ and decreasing otherwise. Else, if for instance $f(a) < f(c)$, then applying $(ii)$ to ${a;c;d}$ yields $f(a) < f(c) leq f(d)$ so $f$ is increasing on ${a;b;c;d}$. The three other cases work the same.
answered Feb 24 '16 at 6:33
nombrenombre
2,634913
$endgroup$
Like BrianO said, your reduction to studying how $f$ behaves with three points is a little quick. However, the reduction is elementary in the sense that it only uses ordered field axioms and not the LUB or other analytic theorems.
Let me just give the details of the reduction:
A map is monotone iff:
$(i)$: it is monotone on every set of four points or less.
This is just figuring out the negation of being monotone: if $f$ is not monotone then there are points $a leq b$, $c leq d$ in its domain such that $f(a) > f(b)$ (non increasing) and $f(c) < f(d)$ (non decreasing) and $f$ is not monotone on ${a;b;c;d}$.
$(ii)$: it is monotone on every set of three points or less.
Indeed, assume $f$ satisfies $(ii)$. Let $a < b < c < d$ be four points with for instance $f(a) leq f(b) leq f(c)$.
If $f(a) = f(b) = f(c)$ then $f$ is monotone on ${a;b;c;d}$: increasing if $f(d) geq f(a)$ and decreasing otherwise. Else, if for instance $f(a) < f(c)$, then applying $(ii)$ to ${a;c;d}$ yields $f(a) < f(c) leq f(d)$ so $f$ is increasing on ${a;b;c;d}$. The three other cases work the same.
answered Feb 24 '16 at 6:33
nombrenombre
2,634913
answered Feb 24 '16 at 6:33
nombrenombre
2,634913
answered Feb 24 '16 at 6:33
nombrenombre
2,634913
answered Feb 24 '16 at 6:33
nombrenombre
2,634913
2,634913
add a comment |
add a comment |
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$begingroup$
what if $f(a)$ or $f(b)$ are undefined?
$endgroup$
– uniquesolution
Feb 23 '16 at 8:14
$begingroup$
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:18
1
$begingroup$
Mostly right, but I don't get a few things. You suppose $f(a)ge f(c)$, but it seems you never consider $f(a)<f(c)$. Your initial statement of what it means for $f$ to not be monotonic seems odd — given that $f$ is 1-1, not monotonic means there are two pairs $a<b, c<d$ with $f(a)<f(b), f(c)>f(d)$.
$endgroup$
– BrianO
Feb 23 '16 at 9:22
$begingroup$
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
$endgroup$
– ZFR
Feb 23 '16 at 15:49
1
$begingroup$
Because he's saying if it's not monotonic then there is a set of three points where the middle one is beneath the other two. Checking f(c) >= f(a) is silly. It's parallel so there's no benefit in reproving that half. We've already proved it with one half.
$endgroup$
– The Great Duck
Apr 24 '16 at 7:14