Given least upper bound $alpha$ for ${ f(x) : x in [a,b] }$, $forall epsilon > 0 exists x$ s.t. $alpha -...
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I can't figure out how all of this follows. Taken from Ch.8 of Spivak's Calculus.
If $alpha$ is the least upper bound of ${ f(x) : x in [a,b] }$ then, $$forall epsilon > 0 exists xin [a,b] alpha - f(x) < epsilon$$
This, in turn, means that
$$ frac{1}{epsilon} < frac{1}{alpha - f(x)}$$
real-analysis
asked Jan 14 at 21:34
user_hello1user_hello1
987
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add a comment |
$begingroup$
I can't figure out how all of this follows. Taken from Ch.8 of Spivak's Calculus.
If $alpha$ is the least upper bound of ${ f(x) : x in [a,b] }$ then, $$forall epsilon > 0 exists xin [a,b] alpha - f(x) < epsilon$$
This, in turn, means that
$$ frac{1}{epsilon} < frac{1}{alpha - f(x)}$$
real-analysis
asked Jan 14 at 21:34
user_hello1user_hello1
987
$endgroup$
add a comment |
$begingroup$
I can't figure out how all of this follows. Taken from Ch.8 of Spivak's Calculus.
If $alpha$ is the least upper bound of ${ f(x) : x in [a,b] }$ then, $$forall epsilon > 0 exists xin [a,b] alpha - f(x) < epsilon$$
This, in turn, means that
$$ frac{1}{epsilon} < frac{1}{alpha - f(x)}$$
real-analysis
asked Jan 14 at 21:34
user_hello1user_hello1
987
$endgroup$
I can't figure out how all of this follows. Taken from Ch.8 of Spivak's Calculus.
If $alpha$ is the least upper bound of ${ f(x) : x in [a,b] }$ then, $$forall epsilon > 0 exists xin [a,b] alpha - f(x) < epsilon$$
This, in turn, means that
$$ frac{1}{epsilon} < frac{1}{alpha - f(x)}$$
real-analysis
real-analysis
asked Jan 14 at 21:34
user_hello1user_hello1
987
asked Jan 14 at 21:34
user_hello1user_hello1
987
asked Jan 14 at 21:34
user_hello1user_hello1
987
asked Jan 14 at 21:34
user_hello1user_hello1
987
asked Jan 14 at 21:34
user_hello1user_hello1
987
987
add a comment |
add a comment |
2 Answers
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$begingroup$
By definition of a least upper bound $alpha$, any smaller value than $a$ is not an upper bound. So take $epsilon>0$. $alpha-epsilon$ is not an upper bound of $S={ f(x) : x in [a,b] }$. Which means that it exists $x in [a,b]$ such that $f(x)>alpha -epsilon$ or $alpha -epsilon <f(x)$ as desired.
answered Jan 14 at 21:43
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
If $alpha$ is a least upper bound, it is an upper bound. So $alpha$ is greater or equal to all the elements of $S$.
$endgroup$
– mathcounterexamples.net
Jan 14 at 22:05
add a comment |
$begingroup$
The definition of $alpha $ is the least upperbound of ${f(x)| xin [a,b]}$.
1) $alpha$ is an upper bound of ${f(x)| xin [a,b]}$
That means for all $x in [a,b]$, $alpha ge f(x)$.
2) If $y < alpha$ then $y$ is not an upper bound of ${f(x)| xin [a,b]}$
Alternatively that means.
2') If $y < alpha$ then there is an $xin [a,b]$ so that $y < f(x)$.
Now for every $epsilon > 0$ then $alpha - epsilon < alpha$ so by 2') it follows that:
For every $epsilon > 0$ the $alpha - epsilon < alpha$ and so there is an $xin[a,b]$ so that $alpha - epsilon < f(x) le alpha$ which in turn means:
$alpha - f(x) < epsilon$.
Now $alpha ge f(x)$ so if $alpha - f(x) ge 0$. and if $f(x)ne 0$ then
$alpha - f(x) < epsilon implies frac 1{alpha - f(x)} > frac 1{epsilon}$.
However your book made an error. If $f(x) = alpha$ this is not true.
...
A counter example would be $f(x) =x$ if $x < b$ but $f(x)= b+1$ if $x ge b$.
Then ${f(x)|x in [a,b]} = [a,b)cup {b+1}$ and $alpha = b+1$.
For any $epsilon: 0 < epsilon < 1$ we have $bin [a,b]$ and $b-epsilon < f(b) le b+1$ but $b$ is the only possible $x in [a,b]$ where $b < b-epsilon < f(x) le b+1$ (because if $x < b$ then $f(x) = x < b$).
So it is true that there is an $x in [a,b]$ where $(b+1) - f(x) < epsilon$ but there isn't any $x in [a,b]$ where $frac{1}{(b+1) - f(x)} > frac 1{epsilon}$ because $frac{1}{(b+1) - f(x)} = frac{1}{(b+1) - f(b)} =frac 10$ is not defined.
answered Jan 14 at 22:32
fleabloodfleablood
73.4k22891
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
By definition of a least upper bound $alpha$, any smaller value than $a$ is not an upper bound. So take $epsilon>0$. $alpha-epsilon$ is not an upper bound of $S={ f(x) : x in [a,b] }$. Which means that it exists $x in [a,b]$ such that $f(x)>alpha -epsilon$ or $alpha -epsilon <f(x)$ as desired.
answered Jan 14 at 21:43
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
If $alpha$ is a least upper bound, it is an upper bound. So $alpha$ is greater or equal to all the elements of $S$.
$endgroup$
– mathcounterexamples.net
Jan 14 at 22:05
add a comment |
$begingroup$
By definition of a least upper bound $alpha$, any smaller value than $a$ is not an upper bound. So take $epsilon>0$. $alpha-epsilon$ is not an upper bound of $S={ f(x) : x in [a,b] }$. Which means that it exists $x in [a,b]$ such that $f(x)>alpha -epsilon$ or $alpha -epsilon <f(x)$ as desired.
answered Jan 14 at 21:43
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
If $alpha$ is a least upper bound, it is an upper bound. So $alpha$ is greater or equal to all the elements of $S$.
$endgroup$
– mathcounterexamples.net
Jan 14 at 22:05
add a comment |
$begingroup$
By definition of a least upper bound $alpha$, any smaller value than $a$ is not an upper bound. So take $epsilon>0$. $alpha-epsilon$ is not an upper bound of $S={ f(x) : x in [a,b] }$. Which means that it exists $x in [a,b]$ such that $f(x)>alpha -epsilon$ or $alpha -epsilon <f(x)$ as desired.
answered Jan 14 at 21:43
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
By definition of a least upper bound $alpha$, any smaller value than $a$ is not an upper bound. So take $epsilon>0$. $alpha-epsilon$ is not an upper bound of $S={ f(x) : x in [a,b] }$. Which means that it exists $x in [a,b]$ such that $f(x)>alpha -epsilon$ or $alpha -epsilon <f(x)$ as desired.
answered Jan 14 at 21:43
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 14 at 21:43
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 14 at 21:43
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 14 at 21:43
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
$begingroup$
If $alpha$ is a least upper bound, it is an upper bound. So $alpha$ is greater or equal to all the elements of $S$.
$endgroup$
– mathcounterexamples.net
Jan 14 at 22:05
add a comment |
$begingroup$
If $alpha$ is a least upper bound, it is an upper bound. So $alpha$ is greater or equal to all the elements of $S$.
$endgroup$
– mathcounterexamples.net
Jan 14 at 22:05
$begingroup$
If $alpha$ is a least upper bound, it is an upper bound. So $alpha$ is greater or equal to all the elements of $S$.
$endgroup$
– mathcounterexamples.net
Jan 14 at 22:05
$begingroup$
If $alpha$ is a least upper bound, it is an upper bound. So $alpha$ is greater or equal to all the elements of $S$.
$endgroup$
– mathcounterexamples.net
Jan 14 at 22:05
add a comment |
$begingroup$
The definition of $alpha $ is the least upperbound of ${f(x)| xin [a,b]}$.
1) $alpha$ is an upper bound of ${f(x)| xin [a,b]}$
That means for all $x in [a,b]$, $alpha ge f(x)$.
2) If $y < alpha$ then $y$ is not an upper bound of ${f(x)| xin [a,b]}$
Alternatively that means.
2') If $y < alpha$ then there is an $xin [a,b]$ so that $y < f(x)$.
Now for every $epsilon > 0$ then $alpha - epsilon < alpha$ so by 2') it follows that:
For every $epsilon > 0$ the $alpha - epsilon < alpha$ and so there is an $xin[a,b]$ so that $alpha - epsilon < f(x) le alpha$ which in turn means:
$alpha - f(x) < epsilon$.
Now $alpha ge f(x)$ so if $alpha - f(x) ge 0$. and if $f(x)ne 0$ then
$alpha - f(x) < epsilon implies frac 1{alpha - f(x)} > frac 1{epsilon}$.
However your book made an error. If $f(x) = alpha$ this is not true.
...
A counter example would be $f(x) =x$ if $x < b$ but $f(x)= b+1$ if $x ge b$.
Then ${f(x)|x in [a,b]} = [a,b)cup {b+1}$ and $alpha = b+1$.
For any $epsilon: 0 < epsilon < 1$ we have $bin [a,b]$ and $b-epsilon < f(b) le b+1$ but $b$ is the only possible $x in [a,b]$ where $b < b-epsilon < f(x) le b+1$ (because if $x < b$ then $f(x) = x < b$).
So it is true that there is an $x in [a,b]$ where $(b+1) - f(x) < epsilon$ but there isn't any $x in [a,b]$ where $frac{1}{(b+1) - f(x)} > frac 1{epsilon}$ because $frac{1}{(b+1) - f(x)} = frac{1}{(b+1) - f(b)} =frac 10$ is not defined.
answered Jan 14 at 22:32
fleabloodfleablood
73.4k22891
$endgroup$
add a comment |
$begingroup$
The definition of $alpha $ is the least upperbound of ${f(x)| xin [a,b]}$.
1) $alpha$ is an upper bound of ${f(x)| xin [a,b]}$
That means for all $x in [a,b]$, $alpha ge f(x)$.
2) If $y < alpha$ then $y$ is not an upper bound of ${f(x)| xin [a,b]}$
Alternatively that means.
2') If $y < alpha$ then there is an $xin [a,b]$ so that $y < f(x)$.
Now for every $epsilon > 0$ then $alpha - epsilon < alpha$ so by 2') it follows that:
For every $epsilon > 0$ the $alpha - epsilon < alpha$ and so there is an $xin[a,b]$ so that $alpha - epsilon < f(x) le alpha$ which in turn means:
$alpha - f(x) < epsilon$.
Now $alpha ge f(x)$ so if $alpha - f(x) ge 0$. and if $f(x)ne 0$ then
$alpha - f(x) < epsilon implies frac 1{alpha - f(x)} > frac 1{epsilon}$.
However your book made an error. If $f(x) = alpha$ this is not true.
...
A counter example would be $f(x) =x$ if $x < b$ but $f(x)= b+1$ if $x ge b$.
Then ${f(x)|x in [a,b]} = [a,b)cup {b+1}$ and $alpha = b+1$.
For any $epsilon: 0 < epsilon < 1$ we have $bin [a,b]$ and $b-epsilon < f(b) le b+1$ but $b$ is the only possible $x in [a,b]$ where $b < b-epsilon < f(x) le b+1$ (because if $x < b$ then $f(x) = x < b$).
So it is true that there is an $x in [a,b]$ where $(b+1) - f(x) < epsilon$ but there isn't any $x in [a,b]$ where $frac{1}{(b+1) - f(x)} > frac 1{epsilon}$ because $frac{1}{(b+1) - f(x)} = frac{1}{(b+1) - f(b)} =frac 10$ is not defined.
answered Jan 14 at 22:32
fleabloodfleablood
73.4k22891
$endgroup$
add a comment |
$begingroup$
The definition of $alpha $ is the least upperbound of ${f(x)| xin [a,b]}$.
1) $alpha$ is an upper bound of ${f(x)| xin [a,b]}$
That means for all $x in [a,b]$, $alpha ge f(x)$.
2) If $y < alpha$ then $y$ is not an upper bound of ${f(x)| xin [a,b]}$
Alternatively that means.
2') If $y < alpha$ then there is an $xin [a,b]$ so that $y < f(x)$.
Now for every $epsilon > 0$ then $alpha - epsilon < alpha$ so by 2') it follows that:
For every $epsilon > 0$ the $alpha - epsilon < alpha$ and so there is an $xin[a,b]$ so that $alpha - epsilon < f(x) le alpha$ which in turn means:
$alpha - f(x) < epsilon$.
Now $alpha ge f(x)$ so if $alpha - f(x) ge 0$. and if $f(x)ne 0$ then
$alpha - f(x) < epsilon implies frac 1{alpha - f(x)} > frac 1{epsilon}$.
However your book made an error. If $f(x) = alpha$ this is not true.
...
A counter example would be $f(x) =x$ if $x < b$ but $f(x)= b+1$ if $x ge b$.
Then ${f(x)|x in [a,b]} = [a,b)cup {b+1}$ and $alpha = b+1$.
For any $epsilon: 0 < epsilon < 1$ we have $bin [a,b]$ and $b-epsilon < f(b) le b+1$ but $b$ is the only possible $x in [a,b]$ where $b < b-epsilon < f(x) le b+1$ (because if $x < b$ then $f(x) = x < b$).
So it is true that there is an $x in [a,b]$ where $(b+1) - f(x) < epsilon$ but there isn't any $x in [a,b]$ where $frac{1}{(b+1) - f(x)} > frac 1{epsilon}$ because $frac{1}{(b+1) - f(x)} = frac{1}{(b+1) - f(b)} =frac 10$ is not defined.
answered Jan 14 at 22:32
fleabloodfleablood
73.4k22891
$endgroup$
The definition of $alpha $ is the least upperbound of ${f(x)| xin [a,b]}$.
1) $alpha$ is an upper bound of ${f(x)| xin [a,b]}$
That means for all $x in [a,b]$, $alpha ge f(x)$.
2) If $y < alpha$ then $y$ is not an upper bound of ${f(x)| xin [a,b]}$
Alternatively that means.
2') If $y < alpha$ then there is an $xin [a,b]$ so that $y < f(x)$.
Now for every $epsilon > 0$ then $alpha - epsilon < alpha$ so by 2') it follows that:
For every $epsilon > 0$ the $alpha - epsilon < alpha$ and so there is an $xin[a,b]$ so that $alpha - epsilon < f(x) le alpha$ which in turn means:
$alpha - f(x) < epsilon$.
Now $alpha ge f(x)$ so if $alpha - f(x) ge 0$. and if $f(x)ne 0$ then
$alpha - f(x) < epsilon implies frac 1{alpha - f(x)} > frac 1{epsilon}$.
However your book made an error. If $f(x) = alpha$ this is not true.
...
A counter example would be $f(x) =x$ if $x < b$ but $f(x)= b+1$ if $x ge b$.
Then ${f(x)|x in [a,b]} = [a,b)cup {b+1}$ and $alpha = b+1$.
For any $epsilon: 0 < epsilon < 1$ we have $bin [a,b]$ and $b-epsilon < f(b) le b+1$ but $b$ is the only possible $x in [a,b]$ where $b < b-epsilon < f(x) le b+1$ (because if $x < b$ then $f(x) = x < b$).
So it is true that there is an $x in [a,b]$ where $(b+1) - f(x) < epsilon$ but there isn't any $x in [a,b]$ where $frac{1}{(b+1) - f(x)} > frac 1{epsilon}$ because $frac{1}{(b+1) - f(x)} = frac{1}{(b+1) - f(b)} =frac 10$ is not defined.
answered Jan 14 at 22:32
fleabloodfleablood
73.4k22891
answered Jan 14 at 22:32
fleabloodfleablood
73.4k22891
answered Jan 14 at 22:32
fleabloodfleablood
73.4k22891
answered Jan 14 at 22:32
fleabloodfleablood
73.4k22891
73.4k22891
add a comment |
add a comment |
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