Supremum Infimum argument: what did I do wrong?
up vote
2
down vote
favorite
Suppose $f:[a,b]tomathbb R$ be a function such that $|f(x)-f(y)|<epsilon_0$ for all $x,yin[a,b]$. Then $M-mleepsilon_0$, where $M=sup{f(x):xin[a,b]}$ and $m=inf{f(x):xin[a,b]}$.
I went this way: suppose $M-m>epsilon_0$, hence $M>epsilon_0+m$, hence there is an $xin[a,b]$ such that$$M>f(x)>epsilon_0+m,tag1$$and similarly $M-epsilon_0>m$ implies$$M-epsilon_0>f(y)>m.tag2$$Subtracting $(1)$ from $(2)$ says $-epsilon_0>f(y)-f(x)>-epsilon_0$, which is absurd.
But on a second thought, I realized that my argument says that $M-m>varepsilon$ is false for any $varepsilon>0$, not just for $epsilon_0$, implying $M-m=0$. What did I do wrong?
real-analysis proof-verification supremum-and-infimum
edited Dec 4 at 9:48
Glorfindel
3,41981730
asked Dec 4 at 8:51
Silent
2,64932050
add a comment |
up vote
2
down vote
favorite
Suppose $f:[a,b]tomathbb R$ be a function such that $|f(x)-f(y)|<epsilon_0$ for all $x,yin[a,b]$. Then $M-mleepsilon_0$, where $M=sup{f(x):xin[a,b]}$ and $m=inf{f(x):xin[a,b]}$.
I went this way: suppose $M-m>epsilon_0$, hence $M>epsilon_0+m$, hence there is an $xin[a,b]$ such that$$M>f(x)>epsilon_0+m,tag1$$and similarly $M-epsilon_0>m$ implies$$M-epsilon_0>f(y)>m.tag2$$Subtracting $(1)$ from $(2)$ says $-epsilon_0>f(y)-f(x)>-epsilon_0$, which is absurd.
But on a second thought, I realized that my argument says that $M-m>varepsilon$ is false for any $varepsilon>0$, not just for $epsilon_0$, implying $M-m=0$. What did I do wrong?
real-analysis proof-verification supremum-and-infimum
edited Dec 4 at 9:48
Glorfindel
3,41981730
asked Dec 4 at 8:51
Silent
2,64932050
I hope that you don't mind about the way I've edited your question.
– José Carlos Santos
Dec 4 at 8:58
Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
– Surb
Dec 4 at 8:58
@JoséCarlosSantos, no, in fact, this seems awesome. Thank you
– Silent
Dec 4 at 8:59
An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
– Paul Sinclair
Dec 4 at 17:24
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $f:[a,b]tomathbb R$ be a function such that $|f(x)-f(y)|<epsilon_0$ for all $x,yin[a,b]$. Then $M-mleepsilon_0$, where $M=sup{f(x):xin[a,b]}$ and $m=inf{f(x):xin[a,b]}$.
I went this way: suppose $M-m>epsilon_0$, hence $M>epsilon_0+m$, hence there is an $xin[a,b]$ such that$$M>f(x)>epsilon_0+m,tag1$$and similarly $M-epsilon_0>m$ implies$$M-epsilon_0>f(y)>m.tag2$$Subtracting $(1)$ from $(2)$ says $-epsilon_0>f(y)-f(x)>-epsilon_0$, which is absurd.
But on a second thought, I realized that my argument says that $M-m>varepsilon$ is false for any $varepsilon>0$, not just for $epsilon_0$, implying $M-m=0$. What did I do wrong?
real-analysis proof-verification supremum-and-infimum
edited Dec 4 at 9:48
Glorfindel
3,41981730
asked Dec 4 at 8:51
Silent
2,64932050
Suppose $f:[a,b]tomathbb R$ be a function such that $|f(x)-f(y)|<epsilon_0$ for all $x,yin[a,b]$. Then $M-mleepsilon_0$, where $M=sup{f(x):xin[a,b]}$ and $m=inf{f(x):xin[a,b]}$.
I went this way: suppose $M-m>epsilon_0$, hence $M>epsilon_0+m$, hence there is an $xin[a,b]$ such that$$M>f(x)>epsilon_0+m,tag1$$and similarly $M-epsilon_0>m$ implies$$M-epsilon_0>f(y)>m.tag2$$Subtracting $(1)$ from $(2)$ says $-epsilon_0>f(y)-f(x)>-epsilon_0$, which is absurd.
But on a second thought, I realized that my argument says that $M-m>varepsilon$ is false for any $varepsilon>0$, not just for $epsilon_0$, implying $M-m=0$. What did I do wrong?
real-analysis proof-verification supremum-and-infimum
real-analysis proof-verification supremum-and-infimum
edited Dec 4 at 9:48
Glorfindel
3,41981730
asked Dec 4 at 8:51
Silent
2,64932050
edited Dec 4 at 9:48
Glorfindel
3,41981730
asked Dec 4 at 8:51
Silent
2,64932050
edited Dec 4 at 9:48
Glorfindel
3,41981730
edited Dec 4 at 9:48
Glorfindel
3,41981730
edited Dec 4 at 9:48
Glorfindel
3,41981730
3,41981730
asked Dec 4 at 8:51
Silent
2,64932050
asked Dec 4 at 8:51
Silent
2,64932050
asked Dec 4 at 8:51
Silent
2,64932050
2,64932050
I hope that you don't mind about the way I've edited your question.
– José Carlos Santos
Dec 4 at 8:58
Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
– Surb
Dec 4 at 8:58
@JoséCarlosSantos, no, in fact, this seems awesome. Thank you
– Silent
Dec 4 at 8:59
An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
– Paul Sinclair
Dec 4 at 17:24
add a comment |
I hope that you don't mind about the way I've edited your question.
– José Carlos Santos
Dec 4 at 8:58
Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
– Surb
Dec 4 at 8:58
@JoséCarlosSantos, no, in fact, this seems awesome. Thank you
– Silent
Dec 4 at 8:59
An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
– Paul Sinclair
Dec 4 at 17:24
I hope that you don't mind about the way I've edited your question.
– José Carlos Santos
Dec 4 at 8:58
I hope that you don't mind about the way I've edited your question.
– José Carlos Santos
Dec 4 at 8:58
Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
– Surb
Dec 4 at 8:58
Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
– Surb
Dec 4 at 8:58
@JoséCarlosSantos, no, in fact, this seems awesome. Thank you
– Silent
Dec 4 at 8:59
@JoséCarlosSantos, no, in fact, this seems awesome. Thank you
– Silent
Dec 4 at 8:59
An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
– Paul Sinclair
Dec 4 at 17:24
An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
– Paul Sinclair
Dec 4 at 17:24
add a comment |
3 Answers
3
active
oldest
votes
up vote
8
down vote
accepted
Your mistake is thinking you can just subtract inequalities. You can't do that. For example,
$$1>0$$
is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.
The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.
For an actual proof, a sketch of it would be this:
- Find some $x$ for which $f(x)$ is "near" $M$
- Find some $y$ for which $f(y)$ is "near" $m$
- Use the fact that $|f(x)-f(y)|<epsilon_0$ and the fact that $$|f(x)-f(y)| = |f(x)-M+M-f(y)+m-m|leq |f(x)-M| + |f(y)-m| + |M-m|$$ to reach a conclusion.
Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!
answered Dec 4 at 8:57
5xum
89.4k393161
add a comment |
up vote
3
down vote
Last step is wrong. $aleq bleq c$ and $a'leq b'leq c'$ do not imply $a-a'leq b-b'leq c-c'$.
answered Dec 4 at 8:57
Kavi Rama Murthy
47.7k31854
add a comment |
up vote
1
down vote
You may also proceed as follows:
- Note that $|x|$ is continuous.
- Choose sequences $(x_n), (y_n)$ with $lim_{n to infty} f(x_n) = M$ and $lim_{n to infty} f(y_n) = m$.
It follows:
$$|f(x_n) - f(y_n)| stackrel{n to infty}{longrightarrow} M-m$$
Now, as $|f(x_n) - f(y_n)| < epsilon_0 Rightarrow M-m leq epsilon_0$.
answered Dec 4 at 10:13
trancelocation
8,8801521
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%2f3025304%2fsupremum-infimum-argument-what-did-i-do-wrong%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Your mistake is thinking you can just subtract inequalities. You can't do that. For example,
$$1>0$$
is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.
The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.
For an actual proof, a sketch of it would be this:
- Find some $x$ for which $f(x)$ is "near" $M$
- Find some $y$ for which $f(y)$ is "near" $m$
- Use the fact that $|f(x)-f(y)|<epsilon_0$ and the fact that $$|f(x)-f(y)| = |f(x)-M+M-f(y)+m-m|leq |f(x)-M| + |f(y)-m| + |M-m|$$ to reach a conclusion.
Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!
answered Dec 4 at 8:57
5xum
89.4k393161
add a comment |
up vote
8
down vote
accepted
Your mistake is thinking you can just subtract inequalities. You can't do that. For example,
$$1>0$$
is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.
The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.
For an actual proof, a sketch of it would be this:
- Find some $x$ for which $f(x)$ is "near" $M$
- Find some $y$ for which $f(y)$ is "near" $m$
- Use the fact that $|f(x)-f(y)|<epsilon_0$ and the fact that $$|f(x)-f(y)| = |f(x)-M+M-f(y)+m-m|leq |f(x)-M| + |f(y)-m| + |M-m|$$ to reach a conclusion.
Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!
answered Dec 4 at 8:57
5xum
89.4k393161
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Your mistake is thinking you can just subtract inequalities. You can't do that. For example,
$$1>0$$
is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.
The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.
For an actual proof, a sketch of it would be this:
- Find some $x$ for which $f(x)$ is "near" $M$
- Find some $y$ for which $f(y)$ is "near" $m$
- Use the fact that $|f(x)-f(y)|<epsilon_0$ and the fact that $$|f(x)-f(y)| = |f(x)-M+M-f(y)+m-m|leq |f(x)-M| + |f(y)-m| + |M-m|$$ to reach a conclusion.
Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!
answered Dec 4 at 8:57
5xum
89.4k393161
Your mistake is thinking you can just subtract inequalities. You can't do that. For example,
$$1>0$$
is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.
The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.
For an actual proof, a sketch of it would be this:
- Find some $x$ for which $f(x)$ is "near" $M$
- Find some $y$ for which $f(y)$ is "near" $m$
- Use the fact that $|f(x)-f(y)|<epsilon_0$ and the fact that $$|f(x)-f(y)| = |f(x)-M+M-f(y)+m-m|leq |f(x)-M| + |f(y)-m| + |M-m|$$ to reach a conclusion.
Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!
answered Dec 4 at 8:57
5xum
89.4k393161
edited Dec 4 at 9:02
answered Dec 4 at 8:57
5xum
89.4k393161
answered Dec 4 at 8:57
5xum
89.4k393161
answered Dec 4 at 8:57
5xum
89.4k393161
89.4k393161
add a comment |
add a comment |
up vote
3
down vote
Last step is wrong. $aleq bleq c$ and $a'leq b'leq c'$ do not imply $a-a'leq b-b'leq c-c'$.
answered Dec 4 at 8:57
Kavi Rama Murthy
47.7k31854
add a comment |
up vote
3
down vote
Last step is wrong. $aleq bleq c$ and $a'leq b'leq c'$ do not imply $a-a'leq b-b'leq c-c'$.
answered Dec 4 at 8:57
Kavi Rama Murthy
47.7k31854
add a comment |
up vote
3
down vote
up vote
3
down vote
Last step is wrong. $aleq bleq c$ and $a'leq b'leq c'$ do not imply $a-a'leq b-b'leq c-c'$.
answered Dec 4 at 8:57
Kavi Rama Murthy
47.7k31854
Last step is wrong. $aleq bleq c$ and $a'leq b'leq c'$ do not imply $a-a'leq b-b'leq c-c'$.
answered Dec 4 at 8:57
Kavi Rama Murthy
47.7k31854
answered Dec 4 at 8:57
Kavi Rama Murthy
47.7k31854
answered Dec 4 at 8:57
Kavi Rama Murthy
47.7k31854
answered Dec 4 at 8:57
Kavi Rama Murthy
47.7k31854
47.7k31854
add a comment |
add a comment |
up vote
1
down vote
You may also proceed as follows:
- Note that $|x|$ is continuous.
- Choose sequences $(x_n), (y_n)$ with $lim_{n to infty} f(x_n) = M$ and $lim_{n to infty} f(y_n) = m$.
It follows:
$$|f(x_n) - f(y_n)| stackrel{n to infty}{longrightarrow} M-m$$
Now, as $|f(x_n) - f(y_n)| < epsilon_0 Rightarrow M-m leq epsilon_0$.
answered Dec 4 at 10:13
trancelocation
8,8801521
add a comment |
up vote
1
down vote
You may also proceed as follows:
- Note that $|x|$ is continuous.
- Choose sequences $(x_n), (y_n)$ with $lim_{n to infty} f(x_n) = M$ and $lim_{n to infty} f(y_n) = m$.
It follows:
$$|f(x_n) - f(y_n)| stackrel{n to infty}{longrightarrow} M-m$$
Now, as $|f(x_n) - f(y_n)| < epsilon_0 Rightarrow M-m leq epsilon_0$.
answered Dec 4 at 10:13
trancelocation
8,8801521
add a comment |
up vote
1
down vote
up vote
1
down vote
You may also proceed as follows:
- Note that $|x|$ is continuous.
- Choose sequences $(x_n), (y_n)$ with $lim_{n to infty} f(x_n) = M$ and $lim_{n to infty} f(y_n) = m$.
It follows:
$$|f(x_n) - f(y_n)| stackrel{n to infty}{longrightarrow} M-m$$
Now, as $|f(x_n) - f(y_n)| < epsilon_0 Rightarrow M-m leq epsilon_0$.
answered Dec 4 at 10:13
trancelocation
8,8801521
You may also proceed as follows:
- Note that $|x|$ is continuous.
- Choose sequences $(x_n), (y_n)$ with $lim_{n to infty} f(x_n) = M$ and $lim_{n to infty} f(y_n) = m$.
It follows:
$$|f(x_n) - f(y_n)| stackrel{n to infty}{longrightarrow} M-m$$
Now, as $|f(x_n) - f(y_n)| < epsilon_0 Rightarrow M-m leq epsilon_0$.
answered Dec 4 at 10:13
trancelocation
8,8801521
answered Dec 4 at 10:13
trancelocation
8,8801521
answered Dec 4 at 10:13
trancelocation
8,8801521
answered Dec 4 at 10:13
trancelocation
8,8801521
8,8801521
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%2f3025304%2fsupremum-infimum-argument-what-did-i-do-wrong%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
I hope that you don't mind about the way I've edited your question.
– José Carlos Santos
Dec 4 at 8:58
Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
– Surb
Dec 4 at 8:58
@JoséCarlosSantos, no, in fact, this seems awesome. Thank you
– Silent
Dec 4 at 8:59
An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
– Paul Sinclair
Dec 4 at 17:24