If a function is continuous and one-to-one then it's strictly monotonic
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
4,97631338
|
show 5 more comments
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
4,97631338
what if $f(a)$ or $f(b)$ are undefined?
– uniquesolution
Feb 23 '16 at 8:14
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
– BrianO
Feb 23 '16 at 9:18
1
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 $f(a)<f(b), f(c)>f(d)$.
– BrianO
Feb 23 '16 at 9:22
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
– ZFR
Feb 23 '16 at 15:49
1
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.
– The Great Duck
Apr 24 '16 at 7:14
|
show 5 more comments
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
4,97631338
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
4,97631338
edited Sep 28 '18 at 15:32
amWhy
192k28225439
asked Feb 23 '16 at 7:15
ZFRZFR
4,97631338
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
4,97631338
asked Feb 23 '16 at 7:15
ZFRZFR
4,97631338
asked Feb 23 '16 at 7:15
ZFRZFR
4,97631338
4,97631338
what if $f(a)$ or $f(b)$ are undefined?
– uniquesolution
Feb 23 '16 at 8:14
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
– BrianO
Feb 23 '16 at 9:18
1
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 $f(a)<f(b), f(c)>f(d)$.
– BrianO
Feb 23 '16 at 9:22
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
– ZFR
Feb 23 '16 at 15:49
1
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.
– The Great Duck
Apr 24 '16 at 7:14
|
show 5 more comments
what if $f(a)$ or $f(b)$ are undefined?
– uniquesolution
Feb 23 '16 at 8:14
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
– BrianO
Feb 23 '16 at 9:18
1
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 $f(a)<f(b), f(c)>f(d)$.
– BrianO
Feb 23 '16 at 9:22
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
– ZFR
Feb 23 '16 at 15:49
1
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.
– The Great Duck
Apr 24 '16 at 7:14
what if $f(a)$ or $f(b)$ are undefined?
– uniquesolution
Feb 23 '16 at 8:14
what if $f(a)$ or $f(b)$ are undefined?
– uniquesolution
Feb 23 '16 at 8:14
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
– BrianO
Feb 23 '16 at 9:18
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
– BrianO
Feb 23 '16 at 9:18
1
1
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 $f(a)<f(b), f(c)>f(d)$.
– BrianO
Feb 23 '16 at 9:22
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 $f(a)<f(b), f(c)>f(d)$.
– BrianO
Feb 23 '16 at 9:22
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
– ZFR
Feb 23 '16 at 15:49
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
– ZFR
Feb 23 '16 at 15:49
1
1
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.
– The Great Duck
Apr 24 '16 at 7:14
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.
– The Great Duck
Apr 24 '16 at 7:14
|
show 5 more comments
1 Answer
1
active
oldest
votes
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1668257%2fif-a-function-is-continuous-and-one-to-one-then-its-strictly-monotonic%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
add a comment |
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
add a comment |
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
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1668257%2fif-a-function-is-continuous-and-one-to-one-then-its-strictly-monotonic%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
what if $f(a)$ or $f(b)$ are undefined?
– uniquesolution
Feb 23 '16 at 8:14
@uniquesolution Not possible, as $a,bin (A,B) = dom(f)$.
– BrianO
Feb 23 '16 at 9:18
1
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 $f(a)<f(b), f(c)>f(d)$.
– BrianO
Feb 23 '16 at 9:22
@BrianO, the case $f(c)geqslant f(a)$ is analogous.The initial statement which I wrote also mean that $f$ is not STRICTLY monotonic.
– ZFR
Feb 23 '16 at 15:49
1
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.
– The Great Duck
Apr 24 '16 at 7:14