Mixing
Continuity of the modified Dirichlet function
$begingroup$
Let
$$
f:(0,1) to mathbb{R}, x mapsto begin{cases}
frac{1}{q}, & text{if } x = frac{p}{q} in mathbb{Q} text{ with gcd}(p,q) = 1, \
0, & text{elsewhere.}
end{cases}
$$
This function is semicontinuous for $x in mathbb{Q}cap (0,1)$ and continuous for $x in (mathbb{R} setminus mathbb{Q}) cap (0,1)$.
Now our professor wrote:
- Every rational number is the limit of a sequence of irrational numbers.
- Therefore, $f$ is not continuous for $x in mathbb{Q} cap (0,1)$
How does 1. imply 2.? By the same reasoning, couldn't you also say that every irrational number is the limit of a sequence of rational numbers, and therefore, $f$ is not continuous for all $x in (mathbb{R} setminus mathbb{Q}) cap (0,1)$.
He continues:
- On the other hand, for every $q in mathbb{N}$ there exist maximum $q$ rational numbers of the form $tfrac{p}{q}$ in $(0,1)$.
- Therefore, for each sequence $(tfrac{p_n}{q_n})_{n in mathbb{N}}$, which converges to an irrational number $x$, we have $lim_{n to infty} frac{1}{q_n} = 0$ and so $f$ is continuous in $x$
Here, I understand 2., but what does 1. have to do with it?
real-analysis continuity
asked Jan 25 at 22:17
Viktor GlombikViktor Glombik
1,1821528
$endgroup$
add a comment |
$begingroup$
Let
$$
f:(0,1) to mathbb{R}, x mapsto begin{cases}
frac{1}{q}, & text{if } x = frac{p}{q} in mathbb{Q} text{ with gcd}(p,q) = 1, \
0, & text{elsewhere.}
end{cases}
$$
This function is semicontinuous for $x in mathbb{Q}cap (0,1)$ and continuous for $x in (mathbb{R} setminus mathbb{Q}) cap (0,1)$.
Now our professor wrote:
- Every rational number is the limit of a sequence of irrational numbers.
- Therefore, $f$ is not continuous for $x in mathbb{Q} cap (0,1)$
How does 1. imply 2.? By the same reasoning, couldn't you also say that every irrational number is the limit of a sequence of rational numbers, and therefore, $f$ is not continuous for all $x in (mathbb{R} setminus mathbb{Q}) cap (0,1)$.
He continues:
- On the other hand, for every $q in mathbb{N}$ there exist maximum $q$ rational numbers of the form $tfrac{p}{q}$ in $(0,1)$.
- Therefore, for each sequence $(tfrac{p_n}{q_n})_{n in mathbb{N}}$, which converges to an irrational number $x$, we have $lim_{n to infty} frac{1}{q_n} = 0$ and so $f$ is continuous in $x$
Here, I understand 2., but what does 1. have to do with it?
real-analysis continuity
asked Jan 25 at 22:17
Viktor GlombikViktor Glombik
1,1821528
$endgroup$
add a comment |
$begingroup$
Let
$$
f:(0,1) to mathbb{R}, x mapsto begin{cases}
frac{1}{q}, & text{if } x = frac{p}{q} in mathbb{Q} text{ with gcd}(p,q) = 1, \
0, & text{elsewhere.}
end{cases}
$$
This function is semicontinuous for $x in mathbb{Q}cap (0,1)$ and continuous for $x in (mathbb{R} setminus mathbb{Q}) cap (0,1)$.
Now our professor wrote:
- Every rational number is the limit of a sequence of irrational numbers.
- Therefore, $f$ is not continuous for $x in mathbb{Q} cap (0,1)$
How does 1. imply 2.? By the same reasoning, couldn't you also say that every irrational number is the limit of a sequence of rational numbers, and therefore, $f$ is not continuous for all $x in (mathbb{R} setminus mathbb{Q}) cap (0,1)$.
He continues:
- On the other hand, for every $q in mathbb{N}$ there exist maximum $q$ rational numbers of the form $tfrac{p}{q}$ in $(0,1)$.
- Therefore, for each sequence $(tfrac{p_n}{q_n})_{n in mathbb{N}}$, which converges to an irrational number $x$, we have $lim_{n to infty} frac{1}{q_n} = 0$ and so $f$ is continuous in $x$
Here, I understand 2., but what does 1. have to do with it?
real-analysis continuity
asked Jan 25 at 22:17
Viktor GlombikViktor Glombik
1,1821528
$endgroup$
Let
$$
f:(0,1) to mathbb{R}, x mapsto begin{cases}
frac{1}{q}, & text{if } x = frac{p}{q} in mathbb{Q} text{ with gcd}(p,q) = 1, \
0, & text{elsewhere.}
end{cases}
$$
This function is semicontinuous for $x in mathbb{Q}cap (0,1)$ and continuous for $x in (mathbb{R} setminus mathbb{Q}) cap (0,1)$.
Now our professor wrote:
- Every rational number is the limit of a sequence of irrational numbers.
- Therefore, $f$ is not continuous for $x in mathbb{Q} cap (0,1)$
How does 1. imply 2.? By the same reasoning, couldn't you also say that every irrational number is the limit of a sequence of rational numbers, and therefore, $f$ is not continuous for all $x in (mathbb{R} setminus mathbb{Q}) cap (0,1)$.
He continues:
- On the other hand, for every $q in mathbb{N}$ there exist maximum $q$ rational numbers of the form $tfrac{p}{q}$ in $(0,1)$.
- Therefore, for each sequence $(tfrac{p_n}{q_n})_{n in mathbb{N}}$, which converges to an irrational number $x$, we have $lim_{n to infty} frac{1}{q_n} = 0$ and so $f$ is continuous in $x$
Here, I understand 2., but what does 1. have to do with it?
real-analysis continuity
real-analysis continuity
asked Jan 25 at 22:17
Viktor GlombikViktor Glombik
1,1821528
asked Jan 25 at 22:17
Viktor GlombikViktor Glombik
1,1821528
asked Jan 25 at 22:17
Viktor GlombikViktor Glombik
1,1821528
asked Jan 25 at 22:17
Viktor GlombikViktor Glombik
1,1821528
asked Jan 25 at 22:17
Viktor GlombikViktor Glombik
1,1821528
1,1821528
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Now our professor wrote:
1. Every rational number is the limit of a sequence of irrational numbers.
2. Therefore, $f$ is not continuous for $xinmathbb Qcap(0,1)$.
How does 1. imply 2.?
Take some rational number $rinmathbb Qcap (0,1)$. There exists a representation $r=frac{p}{q}$ such that $gcd(p,q)=1$. So, there holds $f(r)=frac1{q}>0$.
Now, you can use 1. to get a sequence $(x_n)_n$ in $(mathbb Rsetminus mathbb Q)cap (0,1)$ such that $x_nto r$. But $x_n$ are all irrational, hence $f(x_n)=0$ for all $n$!. You get $$
f(x_n)=0to 0$$
But $f(r)neq 0$. You have got a contradition to the continuity at $r$.
By the same reasoning, couldn't you also say that every irrational number is the limit of a sequence of rational numbers,
Yes you can.
and therefore, f is not continuous for all $xin(mathbb Rsetminus mathbb Q)cap (0,1)$.
No, here you have to be careful since the situation changed!
If you approximate the irrational number by a sequence of rational numbers, you have to to a bit more work to see what the values of the sequence are.
Let us pick an irrational number $xin(mathbb Rsetminus mathbb Q)cap (0,1)$. Then you now, that $f(x)=0$.
If you like to get a contradiction using a sequence $(r_n)_n$ of rational numbers in $(0,1)$, then you have to find a sequence such that $f(r_n)notto 0$.
But in that case, there exists $varepsilon>0$ such that your sequence has infinitely many elements such that $f(r_n)>varepsilon$.
Therefore, we can choose a subsequence $(r_{n_k})_k$ such that $f(r_{n_k})>varepsilon$ for all $k$.
Next, we choose a $qinmathbb N$, such that $varepsilon>frac1q$. But $f(r_{n_k})>frac1q$ will lead to the contradition $r_{n_k}notto x$.
He continues: [...]
If you just have a sequence $left(frac{p_n}{q_n}right)_n$, why should you get $frac1{q_n}to 0$?
It is because of 1.
If $frac1{q_n}notto 0$, then $q_n$ is bounded from above. But then $left(frac{p_n}{q_n}right)_n$ has, because of 1., just finitely many different elements and $frac{p_n}{q_n}to x$ can't hold, since the distance from $x$ to finitely many rational points is positive.
edited Jan 26 at 0:11
Viktor Glombik
1,1821528
answered Jan 25 at 22:51
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
add a comment |
$begingroup$
Let $q$ be a rational number.
Then $f(q)neq0$.
On the other hand, $q$ is the limit of a sequence $(x_n)_{ninmathbb N}$ of irrational numbers.
But then
$$
0=lim_{ntoinfty}f(x_n)
neq fleft(lim_{ntoinfty}x_nright)=f(q).
$$
And, no, this argument does not work if you exchange “rational” with “irrational”.
Concerning the rest, my guess is that what your teach said was that, for every $varepsilon>0$, there are only finitely many rationals $q$ such that $f(q)>varepsilon$. It is not hard to deduce from this that if $lim_{ntoinfty}frac{p_n}{q_n}=x$, then $lim_{ntoinfty}fleft(frac{p_n}{q_n}right)=0$.
edited Jan 25 at 23:33
Viktor Glombik
1,1821528
answered Jan 25 at 22:29
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Now our professor wrote:
1. Every rational number is the limit of a sequence of irrational numbers.
2. Therefore, $f$ is not continuous for $xinmathbb Qcap(0,1)$.
How does 1. imply 2.?
Take some rational number $rinmathbb Qcap (0,1)$. There exists a representation $r=frac{p}{q}$ such that $gcd(p,q)=1$. So, there holds $f(r)=frac1{q}>0$.
Now, you can use 1. to get a sequence $(x_n)_n$ in $(mathbb Rsetminus mathbb Q)cap (0,1)$ such that $x_nto r$. But $x_n$ are all irrational, hence $f(x_n)=0$ for all $n$!. You get $$
f(x_n)=0to 0$$
But $f(r)neq 0$. You have got a contradition to the continuity at $r$.
By the same reasoning, couldn't you also say that every irrational number is the limit of a sequence of rational numbers,
Yes you can.
and therefore, f is not continuous for all $xin(mathbb Rsetminus mathbb Q)cap (0,1)$.
No, here you have to be careful since the situation changed!
If you approximate the irrational number by a sequence of rational numbers, you have to to a bit more work to see what the values of the sequence are.
Let us pick an irrational number $xin(mathbb Rsetminus mathbb Q)cap (0,1)$. Then you now, that $f(x)=0$.
If you like to get a contradiction using a sequence $(r_n)_n$ of rational numbers in $(0,1)$, then you have to find a sequence such that $f(r_n)notto 0$.
But in that case, there exists $varepsilon>0$ such that your sequence has infinitely many elements such that $f(r_n)>varepsilon$.
Therefore, we can choose a subsequence $(r_{n_k})_k$ such that $f(r_{n_k})>varepsilon$ for all $k$.
Next, we choose a $qinmathbb N$, such that $varepsilon>frac1q$. But $f(r_{n_k})>frac1q$ will lead to the contradition $r_{n_k}notto x$.
He continues: [...]
If you just have a sequence $left(frac{p_n}{q_n}right)_n$, why should you get $frac1{q_n}to 0$?
It is because of 1.
If $frac1{q_n}notto 0$, then $q_n$ is bounded from above. But then $left(frac{p_n}{q_n}right)_n$ has, because of 1., just finitely many different elements and $frac{p_n}{q_n}to x$ can't hold, since the distance from $x$ to finitely many rational points is positive.
edited Jan 26 at 0:11
Viktor Glombik
1,1821528
answered Jan 25 at 22:51
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
add a comment |
$begingroup$
Now our professor wrote:
1. Every rational number is the limit of a sequence of irrational numbers.
2. Therefore, $f$ is not continuous for $xinmathbb Qcap(0,1)$.
How does 1. imply 2.?
Take some rational number $rinmathbb Qcap (0,1)$. There exists a representation $r=frac{p}{q}$ such that $gcd(p,q)=1$. So, there holds $f(r)=frac1{q}>0$.
Now, you can use 1. to get a sequence $(x_n)_n$ in $(mathbb Rsetminus mathbb Q)cap (0,1)$ such that $x_nto r$. But $x_n$ are all irrational, hence $f(x_n)=0$ for all $n$!. You get $$
f(x_n)=0to 0$$
But $f(r)neq 0$. You have got a contradition to the continuity at $r$.
By the same reasoning, couldn't you also say that every irrational number is the limit of a sequence of rational numbers,
Yes you can.
and therefore, f is not continuous for all $xin(mathbb Rsetminus mathbb Q)cap (0,1)$.
No, here you have to be careful since the situation changed!
If you approximate the irrational number by a sequence of rational numbers, you have to to a bit more work to see what the values of the sequence are.
Let us pick an irrational number $xin(mathbb Rsetminus mathbb Q)cap (0,1)$. Then you now, that $f(x)=0$.
If you like to get a contradiction using a sequence $(r_n)_n$ of rational numbers in $(0,1)$, then you have to find a sequence such that $f(r_n)notto 0$.
But in that case, there exists $varepsilon>0$ such that your sequence has infinitely many elements such that $f(r_n)>varepsilon$.
Therefore, we can choose a subsequence $(r_{n_k})_k$ such that $f(r_{n_k})>varepsilon$ for all $k$.
Next, we choose a $qinmathbb N$, such that $varepsilon>frac1q$. But $f(r_{n_k})>frac1q$ will lead to the contradition $r_{n_k}notto x$.
He continues: [...]
If you just have a sequence $left(frac{p_n}{q_n}right)_n$, why should you get $frac1{q_n}to 0$?
It is because of 1.
If $frac1{q_n}notto 0$, then $q_n$ is bounded from above. But then $left(frac{p_n}{q_n}right)_n$ has, because of 1., just finitely many different elements and $frac{p_n}{q_n}to x$ can't hold, since the distance from $x$ to finitely many rational points is positive.
edited Jan 26 at 0:11
Viktor Glombik
1,1821528
answered Jan 25 at 22:51
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
add a comment |
$begingroup$
Now our professor wrote:
1. Every rational number is the limit of a sequence of irrational numbers.
2. Therefore, $f$ is not continuous for $xinmathbb Qcap(0,1)$.
How does 1. imply 2.?
Take some rational number $rinmathbb Qcap (0,1)$. There exists a representation $r=frac{p}{q}$ such that $gcd(p,q)=1$. So, there holds $f(r)=frac1{q}>0$.
Now, you can use 1. to get a sequence $(x_n)_n$ in $(mathbb Rsetminus mathbb Q)cap (0,1)$ such that $x_nto r$. But $x_n$ are all irrational, hence $f(x_n)=0$ for all $n$!. You get $$
f(x_n)=0to 0$$
But $f(r)neq 0$. You have got a contradition to the continuity at $r$.
By the same reasoning, couldn't you also say that every irrational number is the limit of a sequence of rational numbers,
Yes you can.
and therefore, f is not continuous for all $xin(mathbb Rsetminus mathbb Q)cap (0,1)$.
No, here you have to be careful since the situation changed!
If you approximate the irrational number by a sequence of rational numbers, you have to to a bit more work to see what the values of the sequence are.
Let us pick an irrational number $xin(mathbb Rsetminus mathbb Q)cap (0,1)$. Then you now, that $f(x)=0$.
If you like to get a contradiction using a sequence $(r_n)_n$ of rational numbers in $(0,1)$, then you have to find a sequence such that $f(r_n)notto 0$.
But in that case, there exists $varepsilon>0$ such that your sequence has infinitely many elements such that $f(r_n)>varepsilon$.
Therefore, we can choose a subsequence $(r_{n_k})_k$ such that $f(r_{n_k})>varepsilon$ for all $k$.
Next, we choose a $qinmathbb N$, such that $varepsilon>frac1q$. But $f(r_{n_k})>frac1q$ will lead to the contradition $r_{n_k}notto x$.
He continues: [...]
If you just have a sequence $left(frac{p_n}{q_n}right)_n$, why should you get $frac1{q_n}to 0$?
It is because of 1.
If $frac1{q_n}notto 0$, then $q_n$ is bounded from above. But then $left(frac{p_n}{q_n}right)_n$ has, because of 1., just finitely many different elements and $frac{p_n}{q_n}to x$ can't hold, since the distance from $x$ to finitely many rational points is positive.
edited Jan 26 at 0:11
Viktor Glombik
1,1821528
answered Jan 25 at 22:51
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
Now our professor wrote:
1. Every rational number is the limit of a sequence of irrational numbers.
2. Therefore, $f$ is not continuous for $xinmathbb Qcap(0,1)$.
How does 1. imply 2.?
Take some rational number $rinmathbb Qcap (0,1)$. There exists a representation $r=frac{p}{q}$ such that $gcd(p,q)=1$. So, there holds $f(r)=frac1{q}>0$.
Now, you can use 1. to get a sequence $(x_n)_n$ in $(mathbb Rsetminus mathbb Q)cap (0,1)$ such that $x_nto r$. But $x_n$ are all irrational, hence $f(x_n)=0$ for all $n$!. You get $$
f(x_n)=0to 0$$
But $f(r)neq 0$. You have got a contradition to the continuity at $r$.
By the same reasoning, couldn't you also say that every irrational number is the limit of a sequence of rational numbers,
Yes you can.
and therefore, f is not continuous for all $xin(mathbb Rsetminus mathbb Q)cap (0,1)$.
No, here you have to be careful since the situation changed!
If you approximate the irrational number by a sequence of rational numbers, you have to to a bit more work to see what the values of the sequence are.
Let us pick an irrational number $xin(mathbb Rsetminus mathbb Q)cap (0,1)$. Then you now, that $f(x)=0$.
If you like to get a contradiction using a sequence $(r_n)_n$ of rational numbers in $(0,1)$, then you have to find a sequence such that $f(r_n)notto 0$.
But in that case, there exists $varepsilon>0$ such that your sequence has infinitely many elements such that $f(r_n)>varepsilon$.
Therefore, we can choose a subsequence $(r_{n_k})_k$ such that $f(r_{n_k})>varepsilon$ for all $k$.
Next, we choose a $qinmathbb N$, such that $varepsilon>frac1q$. But $f(r_{n_k})>frac1q$ will lead to the contradition $r_{n_k}notto x$.
He continues: [...]
If you just have a sequence $left(frac{p_n}{q_n}right)_n$, why should you get $frac1{q_n}to 0$?
It is because of 1.
If $frac1{q_n}notto 0$, then $q_n$ is bounded from above. But then $left(frac{p_n}{q_n}right)_n$ has, because of 1., just finitely many different elements and $frac{p_n}{q_n}to x$ can't hold, since the distance from $x$ to finitely many rational points is positive.
edited Jan 26 at 0:11
Viktor Glombik
1,1821528
answered Jan 25 at 22:51
Mundron SchmidtMundron Schmidt
7,5042729
edited Jan 26 at 0:11
Viktor Glombik
1,1821528
edited Jan 26 at 0:11
Viktor Glombik
1,1821528
edited Jan 26 at 0:11
Viktor Glombik
1,1821528
1,1821528
answered Jan 25 at 22:51
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 25 at 22:51
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 25 at 22:51
Mundron SchmidtMundron Schmidt
7,5042729
7,5042729
add a comment |
add a comment |
$begingroup$
Let $q$ be a rational number.
Then $f(q)neq0$.
On the other hand, $q$ is the limit of a sequence $(x_n)_{ninmathbb N}$ of irrational numbers.
But then
$$
0=lim_{ntoinfty}f(x_n)
neq fleft(lim_{ntoinfty}x_nright)=f(q).
$$
And, no, this argument does not work if you exchange “rational” with “irrational”.
Concerning the rest, my guess is that what your teach said was that, for every $varepsilon>0$, there are only finitely many rationals $q$ such that $f(q)>varepsilon$. It is not hard to deduce from this that if $lim_{ntoinfty}frac{p_n}{q_n}=x$, then $lim_{ntoinfty}fleft(frac{p_n}{q_n}right)=0$.
edited Jan 25 at 23:33
Viktor Glombik
1,1821528
answered Jan 25 at 22:29
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
$begingroup$
Let $q$ be a rational number.
Then $f(q)neq0$.
On the other hand, $q$ is the limit of a sequence $(x_n)_{ninmathbb N}$ of irrational numbers.
But then
$$
0=lim_{ntoinfty}f(x_n)
neq fleft(lim_{ntoinfty}x_nright)=f(q).
$$
And, no, this argument does not work if you exchange “rational” with “irrational”.
Concerning the rest, my guess is that what your teach said was that, for every $varepsilon>0$, there are only finitely many rationals $q$ such that $f(q)>varepsilon$. It is not hard to deduce from this that if $lim_{ntoinfty}frac{p_n}{q_n}=x$, then $lim_{ntoinfty}fleft(frac{p_n}{q_n}right)=0$.
edited Jan 25 at 23:33
Viktor Glombik
1,1821528
answered Jan 25 at 22:29
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
$begingroup$
Let $q$ be a rational number.
Then $f(q)neq0$.
On the other hand, $q$ is the limit of a sequence $(x_n)_{ninmathbb N}$ of irrational numbers.
But then
$$
0=lim_{ntoinfty}f(x_n)
neq fleft(lim_{ntoinfty}x_nright)=f(q).
$$
And, no, this argument does not work if you exchange “rational” with “irrational”.
Concerning the rest, my guess is that what your teach said was that, for every $varepsilon>0$, there are only finitely many rationals $q$ such that $f(q)>varepsilon$. It is not hard to deduce from this that if $lim_{ntoinfty}frac{p_n}{q_n}=x$, then $lim_{ntoinfty}fleft(frac{p_n}{q_n}right)=0$.
edited Jan 25 at 23:33
Viktor Glombik
1,1821528
answered Jan 25 at 22:29
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
Let $q$ be a rational number.
Then $f(q)neq0$.
On the other hand, $q$ is the limit of a sequence $(x_n)_{ninmathbb N}$ of irrational numbers.
But then
$$
0=lim_{ntoinfty}f(x_n)
neq fleft(lim_{ntoinfty}x_nright)=f(q).
$$
And, no, this argument does not work if you exchange “rational” with “irrational”.
Concerning the rest, my guess is that what your teach said was that, for every $varepsilon>0$, there are only finitely many rationals $q$ such that $f(q)>varepsilon$. It is not hard to deduce from this that if $lim_{ntoinfty}frac{p_n}{q_n}=x$, then $lim_{ntoinfty}fleft(frac{p_n}{q_n}right)=0$.
edited Jan 25 at 23:33
Viktor Glombik
1,1821528
answered Jan 25 at 22:29
José Carlos SantosJosé Carlos Santos
169k23132237
edited Jan 25 at 23:33
Viktor Glombik
1,1821528
edited Jan 25 at 23:33
Viktor Glombik
1,1821528
edited Jan 25 at 23:33
Viktor Glombik
1,1821528
1,1821528
answered Jan 25 at 22:29
José Carlos SantosJosé Carlos Santos
169k23132237
answered Jan 25 at 22:29
José Carlos SantosJosé Carlos Santos
169k23132237
answered Jan 25 at 22:29
José Carlos SantosJosé Carlos Santos
169k23132237
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