Mixing
Supremum/Infimum of a Function
$begingroup$
Let a function $f:[-3,3] rightarrow mathbb{R}$ be defined by:
$$
f(x) = left{
begin{array}{ll}
2|x|+1 & quad x in mathbb{Q} \
0 & quad x notin mathbb{Q}
end{array}
right.
$$
Let $P = {x_{0}, x_{1}, ..., x_{n}}$ be any partition of $[-3,3]$.
Then, since each subinterval $[x_{k-1}, x_{k}]$ contains both rational and irrational numbers,
$sup {f(x) | x in [x_{k-1}, x_{k}]=1$
and
$inf {f(x) | x in [x_{k-1}, x_{k}]=0$.
Am I interpreting the supremum and infimum correctly?
real-analysis
asked Jan 19 at 17:55
PerpetualStudentPerpetualStudent
9817
$endgroup$
|
show 6 more comments
$begingroup$
Let a function $f:[-3,3] rightarrow mathbb{R}$ be defined by:
$$
f(x) = left{
begin{array}{ll}
2|x|+1 & quad x in mathbb{Q} \
0 & quad x notin mathbb{Q}
end{array}
right.
$$
Let $P = {x_{0}, x_{1}, ..., x_{n}}$ be any partition of $[-3,3]$.
Then, since each subinterval $[x_{k-1}, x_{k}]$ contains both rational and irrational numbers,
$sup {f(x) | x in [x_{k-1}, x_{k}]=1$
and
$inf {f(x) | x in [x_{k-1}, x_{k}]=0$.
Am I interpreting the supremum and infimum correctly?
real-analysis
asked Jan 19 at 17:55
PerpetualStudentPerpetualStudent
9817
$endgroup$
$begingroup$
Why do you say that the $sup$ is equal to $1$? It depends on the way $f$ is defined.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:01
$begingroup$
My thinking is that over an arbitrary interval the least upper bound would be 1 since if x is rational, the function value will be at least 1. I am having a hard time thinking of the supremum with this function.
$endgroup$
– PerpetualStudent
Jan 19 at 18:06
$begingroup$
Your supremum definition is wrong
$endgroup$
– Andrei
Jan 19 at 18:08
$begingroup$
Take the interval $I=[0,1/10]$. What are the values of $f$ if $x in I$ is irrational? If $x in I$ is rational? The best is to start taking some examples like $1/10$ or $pi/100$.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:09
$begingroup$
@mathcounterexamples.net Regardless of which interval $I=[x_{k-1}, x_{k}]$ that I choose, isn't the least upper bound $2|x_{k}|+1$? Or I am attributing the supremum to the max?
$endgroup$
– PerpetualStudent
Jan 19 at 18:28
|
show 6 more comments
$begingroup$
Let a function $f:[-3,3] rightarrow mathbb{R}$ be defined by:
$$
f(x) = left{
begin{array}{ll}
2|x|+1 & quad x in mathbb{Q} \
0 & quad x notin mathbb{Q}
end{array}
right.
$$
Let $P = {x_{0}, x_{1}, ..., x_{n}}$ be any partition of $[-3,3]$.
Then, since each subinterval $[x_{k-1}, x_{k}]$ contains both rational and irrational numbers,
$sup {f(x) | x in [x_{k-1}, x_{k}]=1$
and
$inf {f(x) | x in [x_{k-1}, x_{k}]=0$.
Am I interpreting the supremum and infimum correctly?
real-analysis
asked Jan 19 at 17:55
PerpetualStudentPerpetualStudent
9817
$endgroup$
Let a function $f:[-3,3] rightarrow mathbb{R}$ be defined by:
$$
f(x) = left{
begin{array}{ll}
2|x|+1 & quad x in mathbb{Q} \
0 & quad x notin mathbb{Q}
end{array}
right.
$$
Let $P = {x_{0}, x_{1}, ..., x_{n}}$ be any partition of $[-3,3]$.
Then, since each subinterval $[x_{k-1}, x_{k}]$ contains both rational and irrational numbers,
$sup {f(x) | x in [x_{k-1}, x_{k}]=1$
and
$inf {f(x) | x in [x_{k-1}, x_{k}]=0$.
Am I interpreting the supremum and infimum correctly?
real-analysis
real-analysis
asked Jan 19 at 17:55
PerpetualStudentPerpetualStudent
9817
asked Jan 19 at 17:55
PerpetualStudentPerpetualStudent
9817
edited Jan 19 at 18:30
PerpetualStudent
asked Jan 19 at 17:55
PerpetualStudentPerpetualStudent
9817
asked Jan 19 at 17:55
PerpetualStudentPerpetualStudent
9817
asked Jan 19 at 17:55
PerpetualStudentPerpetualStudent
9817
9817
$begingroup$
Why do you say that the $sup$ is equal to $1$? It depends on the way $f$ is defined.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:01
$begingroup$
My thinking is that over an arbitrary interval the least upper bound would be 1 since if x is rational, the function value will be at least 1. I am having a hard time thinking of the supremum with this function.
$endgroup$
– PerpetualStudent
Jan 19 at 18:06
$begingroup$
Your supremum definition is wrong
$endgroup$
– Andrei
Jan 19 at 18:08
$begingroup$
Take the interval $I=[0,1/10]$. What are the values of $f$ if $x in I$ is irrational? If $x in I$ is rational? The best is to start taking some examples like $1/10$ or $pi/100$.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:09
$begingroup$
@mathcounterexamples.net Regardless of which interval $I=[x_{k-1}, x_{k}]$ that I choose, isn't the least upper bound $2|x_{k}|+1$? Or I am attributing the supremum to the max?
$endgroup$
– PerpetualStudent
Jan 19 at 18:28
|
show 6 more comments
$begingroup$
Why do you say that the $sup$ is equal to $1$? It depends on the way $f$ is defined.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:01
$begingroup$
My thinking is that over an arbitrary interval the least upper bound would be 1 since if x is rational, the function value will be at least 1. I am having a hard time thinking of the supremum with this function.
$endgroup$
– PerpetualStudent
Jan 19 at 18:06
$begingroup$
Your supremum definition is wrong
$endgroup$
– Andrei
Jan 19 at 18:08
$begingroup$
Take the interval $I=[0,1/10]$. What are the values of $f$ if $x in I$ is irrational? If $x in I$ is rational? The best is to start taking some examples like $1/10$ or $pi/100$.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:09
$begingroup$
@mathcounterexamples.net Regardless of which interval $I=[x_{k-1}, x_{k}]$ that I choose, isn't the least upper bound $2|x_{k}|+1$? Or I am attributing the supremum to the max?
$endgroup$
– PerpetualStudent
Jan 19 at 18:28
$begingroup$
Why do you say that the $sup$ is equal to $1$? It depends on the way $f$ is defined.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:01
$begingroup$
Why do you say that the $sup$ is equal to $1$? It depends on the way $f$ is defined.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:01
$begingroup$
My thinking is that over an arbitrary interval the least upper bound would be 1 since if x is rational, the function value will be at least 1. I am having a hard time thinking of the supremum with this function.
$endgroup$
– PerpetualStudent
Jan 19 at 18:06
$begingroup$
My thinking is that over an arbitrary interval the least upper bound would be 1 since if x is rational, the function value will be at least 1. I am having a hard time thinking of the supremum with this function.
$endgroup$
– PerpetualStudent
Jan 19 at 18:06
$begingroup$
Your supremum definition is wrong
$endgroup$
– Andrei
Jan 19 at 18:08
$begingroup$
Your supremum definition is wrong
$endgroup$
– Andrei
Jan 19 at 18:08
$begingroup$
Take the interval $I=[0,1/10]$. What are the values of $f$ if $x in I$ is irrational? If $x in I$ is rational? The best is to start taking some examples like $1/10$ or $pi/100$.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:09
$begingroup$
Take the interval $I=[0,1/10]$. What are the values of $f$ if $x in I$ is irrational? If $x in I$ is rational? The best is to start taking some examples like $1/10$ or $pi/100$.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:09
$begingroup$
@mathcounterexamples.net Regardless of which interval $I=[x_{k-1}, x_{k}]$ that I choose, isn't the least upper bound $2|x_{k}|+1$? Or I am attributing the supremum to the max?
$endgroup$
– PerpetualStudent
Jan 19 at 18:28
$begingroup$
@mathcounterexamples.net Regardless of which interval $I=[x_{k-1}, x_{k}]$ that I choose, isn't the least upper bound $2|x_{k}|+1$? Or I am attributing the supremum to the max?
$endgroup$
– PerpetualStudent
Jan 19 at 18:28
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If your aim is to prove that $f$ is not Riemann integrable, then it is simpler.
For any partition $P equiv -3 =x_0 < x_1 < dots < x_n=3$ of $[-3,3]$, you have $L(f,P)=0$ as $inflimits_{x in [x_i,x_{i+1}]} f(x)=0$ for $0 le i <n$. Therefore if $f $ was Riemann integrable on $[-3,3]$ it’s integral $s$ would be equal to $0$.
However for $x in [2,3] cap mathbb Q$ when have $f(x) ge 2vert 2 vert +1 =5$. Hence $$suplimits_{x in [x_i,x_{i+1}]} f(x)ge 5$$ for all the integers $0 le i <n$ for which $[x_i,x_{i+1}] cap [2,3] neq emptyset$.
This implies $U(f,P) ge 5*(3-2)=5$, proving that $f$ is not Riemann integrable.
answered Jan 19 at 19:50
mathcounterexamples.netmathcounterexamples.net
27k22157
$endgroup$
$begingroup$
Thank you for this...my main goal is to understand what is happening all together at a deeper level, so I wanted to address "any" partition and be able to see it and explain it. This makes perfect sense though.
$endgroup$
– PerpetualStudent
Jan 19 at 19:57
add a comment |
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1 Answer
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$begingroup$
If your aim is to prove that $f$ is not Riemann integrable, then it is simpler.
For any partition $P equiv -3 =x_0 < x_1 < dots < x_n=3$ of $[-3,3]$, you have $L(f,P)=0$ as $inflimits_{x in [x_i,x_{i+1}]} f(x)=0$ for $0 le i <n$. Therefore if $f $ was Riemann integrable on $[-3,3]$ it’s integral $s$ would be equal to $0$.
However for $x in [2,3] cap mathbb Q$ when have $f(x) ge 2vert 2 vert +1 =5$. Hence $$suplimits_{x in [x_i,x_{i+1}]} f(x)ge 5$$ for all the integers $0 le i <n$ for which $[x_i,x_{i+1}] cap [2,3] neq emptyset$.
This implies $U(f,P) ge 5*(3-2)=5$, proving that $f$ is not Riemann integrable.
answered Jan 19 at 19:50
mathcounterexamples.netmathcounterexamples.net
27k22157
$endgroup$
$begingroup$
Thank you for this...my main goal is to understand what is happening all together at a deeper level, so I wanted to address "any" partition and be able to see it and explain it. This makes perfect sense though.
$endgroup$
– PerpetualStudent
Jan 19 at 19:57
add a comment |
$begingroup$
If your aim is to prove that $f$ is not Riemann integrable, then it is simpler.
For any partition $P equiv -3 =x_0 < x_1 < dots < x_n=3$ of $[-3,3]$, you have $L(f,P)=0$ as $inflimits_{x in [x_i,x_{i+1}]} f(x)=0$ for $0 le i <n$. Therefore if $f $ was Riemann integrable on $[-3,3]$ it’s integral $s$ would be equal to $0$.
However for $x in [2,3] cap mathbb Q$ when have $f(x) ge 2vert 2 vert +1 =5$. Hence $$suplimits_{x in [x_i,x_{i+1}]} f(x)ge 5$$ for all the integers $0 le i <n$ for which $[x_i,x_{i+1}] cap [2,3] neq emptyset$.
This implies $U(f,P) ge 5*(3-2)=5$, proving that $f$ is not Riemann integrable.
answered Jan 19 at 19:50
mathcounterexamples.netmathcounterexamples.net
27k22157
$endgroup$
$begingroup$
Thank you for this...my main goal is to understand what is happening all together at a deeper level, so I wanted to address "any" partition and be able to see it and explain it. This makes perfect sense though.
$endgroup$
– PerpetualStudent
Jan 19 at 19:57
add a comment |
$begingroup$
If your aim is to prove that $f$ is not Riemann integrable, then it is simpler.
For any partition $P equiv -3 =x_0 < x_1 < dots < x_n=3$ of $[-3,3]$, you have $L(f,P)=0$ as $inflimits_{x in [x_i,x_{i+1}]} f(x)=0$ for $0 le i <n$. Therefore if $f $ was Riemann integrable on $[-3,3]$ it’s integral $s$ would be equal to $0$.
However for $x in [2,3] cap mathbb Q$ when have $f(x) ge 2vert 2 vert +1 =5$. Hence $$suplimits_{x in [x_i,x_{i+1}]} f(x)ge 5$$ for all the integers $0 le i <n$ for which $[x_i,x_{i+1}] cap [2,3] neq emptyset$.
This implies $U(f,P) ge 5*(3-2)=5$, proving that $f$ is not Riemann integrable.
answered Jan 19 at 19:50
mathcounterexamples.netmathcounterexamples.net
27k22157
$endgroup$
If your aim is to prove that $f$ is not Riemann integrable, then it is simpler.
For any partition $P equiv -3 =x_0 < x_1 < dots < x_n=3$ of $[-3,3]$, you have $L(f,P)=0$ as $inflimits_{x in [x_i,x_{i+1}]} f(x)=0$ for $0 le i <n$. Therefore if $f $ was Riemann integrable on $[-3,3]$ it’s integral $s$ would be equal to $0$.
However for $x in [2,3] cap mathbb Q$ when have $f(x) ge 2vert 2 vert +1 =5$. Hence $$suplimits_{x in [x_i,x_{i+1}]} f(x)ge 5$$ for all the integers $0 le i <n$ for which $[x_i,x_{i+1}] cap [2,3] neq emptyset$.
This implies $U(f,P) ge 5*(3-2)=5$, proving that $f$ is not Riemann integrable.
answered Jan 19 at 19:50
mathcounterexamples.netmathcounterexamples.net
27k22157
edited Jan 19 at 20:01
answered Jan 19 at 19:50
mathcounterexamples.netmathcounterexamples.net
27k22157
answered Jan 19 at 19:50
mathcounterexamples.netmathcounterexamples.net
27k22157
answered Jan 19 at 19:50
mathcounterexamples.netmathcounterexamples.net
27k22157
27k22157
$begingroup$
Thank you for this...my main goal is to understand what is happening all together at a deeper level, so I wanted to address "any" partition and be able to see it and explain it. This makes perfect sense though.
$endgroup$
– PerpetualStudent
Jan 19 at 19:57
add a comment |
$begingroup$
Thank you for this...my main goal is to understand what is happening all together at a deeper level, so I wanted to address "any" partition and be able to see it and explain it. This makes perfect sense though.
$endgroup$
– PerpetualStudent
Jan 19 at 19:57
$begingroup$
Thank you for this...my main goal is to understand what is happening all together at a deeper level, so I wanted to address "any" partition and be able to see it and explain it. This makes perfect sense though.
$endgroup$
– PerpetualStudent
Jan 19 at 19:57
$begingroup$
Thank you for this...my main goal is to understand what is happening all together at a deeper level, so I wanted to address "any" partition and be able to see it and explain it. This makes perfect sense though.
$endgroup$
– PerpetualStudent
Jan 19 at 19:57
add a comment |
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$begingroup$
Why do you say that the $sup$ is equal to $1$? It depends on the way $f$ is defined.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:01
$begingroup$
My thinking is that over an arbitrary interval the least upper bound would be 1 since if x is rational, the function value will be at least 1. I am having a hard time thinking of the supremum with this function.
$endgroup$
– PerpetualStudent
Jan 19 at 18:06
$begingroup$
Your supremum definition is wrong
$endgroup$
– Andrei
Jan 19 at 18:08
$begingroup$
Take the interval $I=[0,1/10]$. What are the values of $f$ if $x in I$ is irrational? If $x in I$ is rational? The best is to start taking some examples like $1/10$ or $pi/100$.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:09
$begingroup$
@mathcounterexamples.net Regardless of which interval $I=[x_{k-1}, x_{k}]$ that I choose, isn't the least upper bound $2|x_{k}|+1$? Or I am attributing the supremum to the max?
$endgroup$
– PerpetualStudent
Jan 19 at 18:28