Supremum Infimum argument: what did I do wrong?
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
3,41981830
asked Dec 4 at 8:51
Silent
2,65632050
add a comment |
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
3,41981830
asked Dec 4 at 8:51
Silent
2,65632050
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 |
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
3,41981830
asked Dec 4 at 8:51
Silent
2,65632050
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
3,41981830
asked Dec 4 at 8:51
Silent
2,65632050
edited Dec 4 at 9:48
Glorfindel
3,41981830
asked Dec 4 at 8:51
Silent
2,65632050
edited Dec 4 at 9:48
Glorfindel
3,41981830
edited Dec 4 at 9:48
Glorfindel
3,41981830
edited Dec 4 at 9:48
Glorfindel
3,41981830
3,41981830
asked Dec 4 at 8:51
Silent
2,65632050
asked Dec 4 at 8:51
Silent
2,65632050
asked Dec 4 at 8:51
Silent
2,65632050
2,65632050
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
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votes
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
89.5k393161
add a comment |
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
49.4k31854
add a comment |
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
9,0951521
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
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
89.5k393161
add a comment |
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
89.5k393161
add a comment |
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
89.5k393161
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
89.5k393161
edited Dec 4 at 9:02
answered Dec 4 at 8:57
5xum
89.5k393161
answered Dec 4 at 8:57
5xum
89.5k393161
answered Dec 4 at 8:57
5xum
89.5k393161
89.5k393161
add a comment |
add a comment |
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
49.4k31854
add a comment |
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
49.4k31854
add a comment |
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
49.4k31854
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
49.4k31854
answered Dec 4 at 8:57
Kavi Rama Murthy
49.4k31854
answered Dec 4 at 8:57
Kavi Rama Murthy
49.4k31854
answered Dec 4 at 8:57
Kavi Rama Murthy
49.4k31854
49.4k31854
add a comment |
add a comment |
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
9,0951521
add a comment |
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
9,0951521
add a comment |
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
9,0951521
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
9,0951521
answered Dec 4 at 10:13
trancelocation
9,0951521
answered Dec 4 at 10:13
trancelocation
9,0951521
answered Dec 4 at 10:13
trancelocation
9,0951521
9,0951521
add a comment |
add a comment |
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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