Mixing
Prove that ${e^{kalpha pi i}: kin mathbb N }$ is dense in $S^1 $ by using(or modifying the proof of) the...
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This question already has an answer here:
Let $q in mathbb C$, $|q|=1$ and $q^n neq 1, forall n in mathbb N$. Show that ${q^n: n in mathbb N}$ is dense in $S^1$
3 answers
Let $alpha$ be an irrational number.
There is a well-known result saying that ${e^{kalpha pi i}: kin mathbb Z }$ is dense in $S^1 $. In the proof of this result, the Pigeonhole Principle can't guarantee that we can always choose $i>j$. So I wonder how we can prove that ${e^{kalpha pi i}: kin mathbb N }$ is dense in $S^1 $ by using that result or modifying its proof($mathbb N$ is the set of natural numbers). Thanks in advance.
real-analysis general-topology
edited Apr 13 '17 at 12:20
Community♦
1
asked Mar 19 '16 at 19:18
No OneNo One
2,0661519
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marked as duplicate by rtybase, Cesareo, Kemono Chen, Lord Shark the Unknown, metamorphy Jan 24 at 5:42
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Let $q in mathbb C$, $|q|=1$ and $q^n neq 1, forall n in mathbb N$. Show that ${q^n: n in mathbb N}$ is dense in $S^1$
3 answers
Let $alpha$ be an irrational number.
There is a well-known result saying that ${e^{kalpha pi i}: kin mathbb Z }$ is dense in $S^1 $. In the proof of this result, the Pigeonhole Principle can't guarantee that we can always choose $i>j$. So I wonder how we can prove that ${e^{kalpha pi i}: kin mathbb N }$ is dense in $S^1 $ by using that result or modifying its proof($mathbb N$ is the set of natural numbers). Thanks in advance.
real-analysis general-topology
edited Apr 13 '17 at 12:20
Community♦
1
asked Mar 19 '16 at 19:18
No OneNo One
2,0661519
$endgroup$
marked as duplicate by rtybase, Cesareo, Kemono Chen, Lord Shark the Unknown, metamorphy Jan 24 at 5:42
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
A minor modification works; see this answer.
$endgroup$
– Brian M. Scott
Mar 19 '16 at 19:47
add a comment |
$begingroup$
This question already has an answer here:
Let $q in mathbb C$, $|q|=1$ and $q^n neq 1, forall n in mathbb N$. Show that ${q^n: n in mathbb N}$ is dense in $S^1$
3 answers
Let $alpha$ be an irrational number.
There is a well-known result saying that ${e^{kalpha pi i}: kin mathbb Z }$ is dense in $S^1 $. In the proof of this result, the Pigeonhole Principle can't guarantee that we can always choose $i>j$. So I wonder how we can prove that ${e^{kalpha pi i}: kin mathbb N }$ is dense in $S^1 $ by using that result or modifying its proof($mathbb N$ is the set of natural numbers). Thanks in advance.
real-analysis general-topology
edited Apr 13 '17 at 12:20
Community♦
1
asked Mar 19 '16 at 19:18
No OneNo One
2,0661519
$endgroup$
This question already has an answer here:
Let $q in mathbb C$, $|q|=1$ and $q^n neq 1, forall n in mathbb N$. Show that ${q^n: n in mathbb N}$ is dense in $S^1$
3 answers
Let $alpha$ be an irrational number.
There is a well-known result saying that ${e^{kalpha pi i}: kin mathbb Z }$ is dense in $S^1 $. In the proof of this result, the Pigeonhole Principle can't guarantee that we can always choose $i>j$. So I wonder how we can prove that ${e^{kalpha pi i}: kin mathbb N }$ is dense in $S^1 $ by using that result or modifying its proof($mathbb N$ is the set of natural numbers). Thanks in advance.
This question already has an answer here:
Let $q in mathbb C$, $|q|=1$ and $q^n neq 1, forall n in mathbb N$. Show that ${q^n: n in mathbb N}$ is dense in $S^1$
3 answers
real-analysis general-topology
real-analysis general-topology
edited Apr 13 '17 at 12:20
Community♦
1
asked Mar 19 '16 at 19:18
No OneNo One
2,0661519
edited Apr 13 '17 at 12:20
Community♦
1
asked Mar 19 '16 at 19:18
No OneNo One
2,0661519
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Mar 19 '16 at 19:18
No OneNo One
2,0661519
asked Mar 19 '16 at 19:18
No OneNo One
2,0661519
asked Mar 19 '16 at 19:18
No OneNo One
2,0661519
2,0661519
marked as duplicate by rtybase, Cesareo, Kemono Chen, Lord Shark the Unknown, metamorphy Jan 24 at 5:42
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by rtybase, Cesareo, Kemono Chen, Lord Shark the Unknown, metamorphy Jan 24 at 5:42
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
A minor modification works; see this answer.
$endgroup$
– Brian M. Scott
Mar 19 '16 at 19:47
add a comment |
1
$begingroup$
A minor modification works; see this answer.
$endgroup$
– Brian M. Scott
Mar 19 '16 at 19:47
1
1
$begingroup$
A minor modification works; see this answer.
$endgroup$
– Brian M. Scott
Mar 19 '16 at 19:47
$begingroup$
A minor modification works; see this answer.
$endgroup$
– Brian M. Scott
Mar 19 '16 at 19:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Proof assuming the $mathbb Z$ result: Suppose the result for $mathbb Z$ holds, which implies the result for $mathbb Zsetminus {0}$ holds.
Claim: There exists a sequence of positive integers $k_j$ such that $e^{k_jalpha pi i} to 1.$
Proof: There is a sequence $e^{k_jalpha pi i} to 1,$ where each $k_jin mathbb Zsetminus {0}.$ If infinitely many of these $k_j$ are positive, we're done. If not, note $e^{-k_jalpha pi i} to 1$ as well, and infinitely many $-k_j$ are positive, giving the claim.
Now suppose $e^{it}in S^1.$ There is a sequence $e^{k_jalpha pi i} to 1$ where each $k_jin mathbb Zsetminus {0}.$ By the claim we can choose $k_j'>|k_j|$ such that $e^{k_j'alphapi i} to 1.$ We then have $e^{(k_j+k_j')alphapi i} to zeta$ and each $k_j+k_j'>0.$ This gives the result.
Brief proof without using the $mathbb Z$ result: Because the circle is compact, the sequence of distinct points $e^{kalpha pi i},kin mathbb N,$ has a limit point $zeta$ on the circle. Thus there is a subsequence $0< k_1 < k_2 < cdots $ such that $e^{k_jalphapi i} to zeta.$ We then have $e^{(k_{j+1}-k_j)alphapi i} to 1.$ Thus for any $epsilon>0,$ there exists $k_epsilonin mathbb N$ such that $0<|1-e^{k_epsilonalphapi i}|<epsilon.$ The set $A_epsilon = {e^{alpha k_epsilon k pi i}:kin mathbb N}$ then has the property that every point on the circle is within $epsilon$ of some point in $A_epsilon.$ Since $epsilon>0$ is arbitrary, we have the desired density.
answered Mar 19 '16 at 19:40
zhw.zhw.
74.2k43175
$endgroup$
$begingroup$
Thank you! But I can't see why $A_epsilon = {e^{ialpha n_epsilon k}:kin mathbb N}$ then has the property that every point on the circle is within ϵ of some point in $A_ϵ$....
$endgroup$
– No One
Mar 19 '16 at 19:57
$begingroup$
It's because $e^{ialpha n_epsilon} = e^{it}$ for some small $t.$ Think about the points $ e^{it}, e^{i2t}, e^{i3t}, dots ,e^{ikt}, dots $
$endgroup$
– zhw.
Mar 19 '16 at 20:18
$begingroup$
Dear Sir,Nice solution and Can we argue like following: 1 is limit point by above construction. Now we rotate that to make any point on $S^1$ to be limit point as $0<|e^{ix}||1-e^{ialpha n_epsilon}|=|e^{ix}-e^{i(x+alpha n_epsilon}|$<$epsilon$
$endgroup$
– SRJ
Jan 22 at 10:33
$begingroup$
@SRJ I'm not sure what you're doing in the last line.
$endgroup$
– zhw.
Jan 23 at 0:36
$begingroup$
Sir I wanted to show for any point on S1 that is of form e^{ix} there is sequence converging to that point . As you had show this happen. For 1. I just multiply by e^ix to show it for other. Is my argument looking correct
$endgroup$
– SRJ
Jan 23 at 2:58
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Proof assuming the $mathbb Z$ result: Suppose the result for $mathbb Z$ holds, which implies the result for $mathbb Zsetminus {0}$ holds.
Claim: There exists a sequence of positive integers $k_j$ such that $e^{k_jalpha pi i} to 1.$
Proof: There is a sequence $e^{k_jalpha pi i} to 1,$ where each $k_jin mathbb Zsetminus {0}.$ If infinitely many of these $k_j$ are positive, we're done. If not, note $e^{-k_jalpha pi i} to 1$ as well, and infinitely many $-k_j$ are positive, giving the claim.
Now suppose $e^{it}in S^1.$ There is a sequence $e^{k_jalpha pi i} to 1$ where each $k_jin mathbb Zsetminus {0}.$ By the claim we can choose $k_j'>|k_j|$ such that $e^{k_j'alphapi i} to 1.$ We then have $e^{(k_j+k_j')alphapi i} to zeta$ and each $k_j+k_j'>0.$ This gives the result.
Brief proof without using the $mathbb Z$ result: Because the circle is compact, the sequence of distinct points $e^{kalpha pi i},kin mathbb N,$ has a limit point $zeta$ on the circle. Thus there is a subsequence $0< k_1 < k_2 < cdots $ such that $e^{k_jalphapi i} to zeta.$ We then have $e^{(k_{j+1}-k_j)alphapi i} to 1.$ Thus for any $epsilon>0,$ there exists $k_epsilonin mathbb N$ such that $0<|1-e^{k_epsilonalphapi i}|<epsilon.$ The set $A_epsilon = {e^{alpha k_epsilon k pi i}:kin mathbb N}$ then has the property that every point on the circle is within $epsilon$ of some point in $A_epsilon.$ Since $epsilon>0$ is arbitrary, we have the desired density.
answered Mar 19 '16 at 19:40
zhw.zhw.
74.2k43175
$endgroup$
$begingroup$
Thank you! But I can't see why $A_epsilon = {e^{ialpha n_epsilon k}:kin mathbb N}$ then has the property that every point on the circle is within ϵ of some point in $A_ϵ$....
$endgroup$
– No One
Mar 19 '16 at 19:57
$begingroup$
It's because $e^{ialpha n_epsilon} = e^{it}$ for some small $t.$ Think about the points $ e^{it}, e^{i2t}, e^{i3t}, dots ,e^{ikt}, dots $
$endgroup$
– zhw.
Mar 19 '16 at 20:18
$begingroup$
Dear Sir,Nice solution and Can we argue like following: 1 is limit point by above construction. Now we rotate that to make any point on $S^1$ to be limit point as $0<|e^{ix}||1-e^{ialpha n_epsilon}|=|e^{ix}-e^{i(x+alpha n_epsilon}|$<$epsilon$
$endgroup$
– SRJ
Jan 22 at 10:33
$begingroup$
@SRJ I'm not sure what you're doing in the last line.
$endgroup$
– zhw.
Jan 23 at 0:36
$begingroup$
Sir I wanted to show for any point on S1 that is of form e^{ix} there is sequence converging to that point . As you had show this happen. For 1. I just multiply by e^ix to show it for other. Is my argument looking correct
$endgroup$
– SRJ
Jan 23 at 2:58
add a comment |
$begingroup$
Proof assuming the $mathbb Z$ result: Suppose the result for $mathbb Z$ holds, which implies the result for $mathbb Zsetminus {0}$ holds.
Claim: There exists a sequence of positive integers $k_j$ such that $e^{k_jalpha pi i} to 1.$
Proof: There is a sequence $e^{k_jalpha pi i} to 1,$ where each $k_jin mathbb Zsetminus {0}.$ If infinitely many of these $k_j$ are positive, we're done. If not, note $e^{-k_jalpha pi i} to 1$ as well, and infinitely many $-k_j$ are positive, giving the claim.
Now suppose $e^{it}in S^1.$ There is a sequence $e^{k_jalpha pi i} to 1$ where each $k_jin mathbb Zsetminus {0}.$ By the claim we can choose $k_j'>|k_j|$ such that $e^{k_j'alphapi i} to 1.$ We then have $e^{(k_j+k_j')alphapi i} to zeta$ and each $k_j+k_j'>0.$ This gives the result.
Brief proof without using the $mathbb Z$ result: Because the circle is compact, the sequence of distinct points $e^{kalpha pi i},kin mathbb N,$ has a limit point $zeta$ on the circle. Thus there is a subsequence $0< k_1 < k_2 < cdots $ such that $e^{k_jalphapi i} to zeta.$ We then have $e^{(k_{j+1}-k_j)alphapi i} to 1.$ Thus for any $epsilon>0,$ there exists $k_epsilonin mathbb N$ such that $0<|1-e^{k_epsilonalphapi i}|<epsilon.$ The set $A_epsilon = {e^{alpha k_epsilon k pi i}:kin mathbb N}$ then has the property that every point on the circle is within $epsilon$ of some point in $A_epsilon.$ Since $epsilon>0$ is arbitrary, we have the desired density.
answered Mar 19 '16 at 19:40
zhw.zhw.
74.2k43175
$endgroup$
$begingroup$
Thank you! But I can't see why $A_epsilon = {e^{ialpha n_epsilon k}:kin mathbb N}$ then has the property that every point on the circle is within ϵ of some point in $A_ϵ$....
$endgroup$
– No One
Mar 19 '16 at 19:57
$begingroup$
It's because $e^{ialpha n_epsilon} = e^{it}$ for some small $t.$ Think about the points $ e^{it}, e^{i2t}, e^{i3t}, dots ,e^{ikt}, dots $
$endgroup$
– zhw.
Mar 19 '16 at 20:18
$begingroup$
Dear Sir,Nice solution and Can we argue like following: 1 is limit point by above construction. Now we rotate that to make any point on $S^1$ to be limit point as $0<|e^{ix}||1-e^{ialpha n_epsilon}|=|e^{ix}-e^{i(x+alpha n_epsilon}|$<$epsilon$
$endgroup$
– SRJ
Jan 22 at 10:33
$begingroup$
@SRJ I'm not sure what you're doing in the last line.
$endgroup$
– zhw.
Jan 23 at 0:36
$begingroup$
Sir I wanted to show for any point on S1 that is of form e^{ix} there is sequence converging to that point . As you had show this happen. For 1. I just multiply by e^ix to show it for other. Is my argument looking correct
$endgroup$
– SRJ
Jan 23 at 2:58
add a comment |
$begingroup$
Proof assuming the $mathbb Z$ result: Suppose the result for $mathbb Z$ holds, which implies the result for $mathbb Zsetminus {0}$ holds.
Claim: There exists a sequence of positive integers $k_j$ such that $e^{k_jalpha pi i} to 1.$
Proof: There is a sequence $e^{k_jalpha pi i} to 1,$ where each $k_jin mathbb Zsetminus {0}.$ If infinitely many of these $k_j$ are positive, we're done. If not, note $e^{-k_jalpha pi i} to 1$ as well, and infinitely many $-k_j$ are positive, giving the claim.
Now suppose $e^{it}in S^1.$ There is a sequence $e^{k_jalpha pi i} to 1$ where each $k_jin mathbb Zsetminus {0}.$ By the claim we can choose $k_j'>|k_j|$ such that $e^{k_j'alphapi i} to 1.$ We then have $e^{(k_j+k_j')alphapi i} to zeta$ and each $k_j+k_j'>0.$ This gives the result.
Brief proof without using the $mathbb Z$ result: Because the circle is compact, the sequence of distinct points $e^{kalpha pi i},kin mathbb N,$ has a limit point $zeta$ on the circle. Thus there is a subsequence $0< k_1 < k_2 < cdots $ such that $e^{k_jalphapi i} to zeta.$ We then have $e^{(k_{j+1}-k_j)alphapi i} to 1.$ Thus for any $epsilon>0,$ there exists $k_epsilonin mathbb N$ such that $0<|1-e^{k_epsilonalphapi i}|<epsilon.$ The set $A_epsilon = {e^{alpha k_epsilon k pi i}:kin mathbb N}$ then has the property that every point on the circle is within $epsilon$ of some point in $A_epsilon.$ Since $epsilon>0$ is arbitrary, we have the desired density.
answered Mar 19 '16 at 19:40
zhw.zhw.
74.2k43175
$endgroup$
Proof assuming the $mathbb Z$ result: Suppose the result for $mathbb Z$ holds, which implies the result for $mathbb Zsetminus {0}$ holds.
Claim: There exists a sequence of positive integers $k_j$ such that $e^{k_jalpha pi i} to 1.$
Proof: There is a sequence $e^{k_jalpha pi i} to 1,$ where each $k_jin mathbb Zsetminus {0}.$ If infinitely many of these $k_j$ are positive, we're done. If not, note $e^{-k_jalpha pi i} to 1$ as well, and infinitely many $-k_j$ are positive, giving the claim.
Now suppose $e^{it}in S^1.$ There is a sequence $e^{k_jalpha pi i} to 1$ where each $k_jin mathbb Zsetminus {0}.$ By the claim we can choose $k_j'>|k_j|$ such that $e^{k_j'alphapi i} to 1.$ We then have $e^{(k_j+k_j')alphapi i} to zeta$ and each $k_j+k_j'>0.$ This gives the result.
Brief proof without using the $mathbb Z$ result: Because the circle is compact, the sequence of distinct points $e^{kalpha pi i},kin mathbb N,$ has a limit point $zeta$ on the circle. Thus there is a subsequence $0< k_1 < k_2 < cdots $ such that $e^{k_jalphapi i} to zeta.$ We then have $e^{(k_{j+1}-k_j)alphapi i} to 1.$ Thus for any $epsilon>0,$ there exists $k_epsilonin mathbb N$ such that $0<|1-e^{k_epsilonalphapi i}|<epsilon.$ The set $A_epsilon = {e^{alpha k_epsilon k pi i}:kin mathbb N}$ then has the property that every point on the circle is within $epsilon$ of some point in $A_epsilon.$ Since $epsilon>0$ is arbitrary, we have the desired density.
answered Mar 19 '16 at 19:40
zhw.zhw.
74.2k43175
edited Jan 23 at 21:27
answered Mar 19 '16 at 19:40
zhw.zhw.
74.2k43175
answered Mar 19 '16 at 19:40
zhw.zhw.
74.2k43175
answered Mar 19 '16 at 19:40
zhw.zhw.
74.2k43175
74.2k43175
$begingroup$
Thank you! But I can't see why $A_epsilon = {e^{ialpha n_epsilon k}:kin mathbb N}$ then has the property that every point on the circle is within ϵ of some point in $A_ϵ$....
$endgroup$
– No One
Mar 19 '16 at 19:57
$begingroup$
It's because $e^{ialpha n_epsilon} = e^{it}$ for some small $t.$ Think about the points $ e^{it}, e^{i2t}, e^{i3t}, dots ,e^{ikt}, dots $
$endgroup$
– zhw.
Mar 19 '16 at 20:18
$begingroup$
Dear Sir,Nice solution and Can we argue like following: 1 is limit point by above construction. Now we rotate that to make any point on $S^1$ to be limit point as $0<|e^{ix}||1-e^{ialpha n_epsilon}|=|e^{ix}-e^{i(x+alpha n_epsilon}|$<$epsilon$
$endgroup$
– SRJ
Jan 22 at 10:33
$begingroup$
@SRJ I'm not sure what you're doing in the last line.
$endgroup$
– zhw.
Jan 23 at 0:36
$begingroup$
Sir I wanted to show for any point on S1 that is of form e^{ix} there is sequence converging to that point . As you had show this happen. For 1. I just multiply by e^ix to show it for other. Is my argument looking correct
$endgroup$
– SRJ
Jan 23 at 2:58
add a comment |
$begingroup$
Thank you! But I can't see why $A_epsilon = {e^{ialpha n_epsilon k}:kin mathbb N}$ then has the property that every point on the circle is within ϵ of some point in $A_ϵ$....
$endgroup$
– No One
Mar 19 '16 at 19:57
$begingroup$
It's because $e^{ialpha n_epsilon} = e^{it}$ for some small $t.$ Think about the points $ e^{it}, e^{i2t}, e^{i3t}, dots ,e^{ikt}, dots $
$endgroup$
– zhw.
Mar 19 '16 at 20:18
$begingroup$
Dear Sir,Nice solution and Can we argue like following: 1 is limit point by above construction. Now we rotate that to make any point on $S^1$ to be limit point as $0<|e^{ix}||1-e^{ialpha n_epsilon}|=|e^{ix}-e^{i(x+alpha n_epsilon}|$<$epsilon$
$endgroup$
– SRJ
Jan 22 at 10:33
$begingroup$
@SRJ I'm not sure what you're doing in the last line.
$endgroup$
– zhw.
Jan 23 at 0:36
$begingroup$
Sir I wanted to show for any point on S1 that is of form e^{ix} there is sequence converging to that point . As you had show this happen. For 1. I just multiply by e^ix to show it for other. Is my argument looking correct
$endgroup$
– SRJ
Jan 23 at 2:58
$begingroup$
Thank you! But I can't see why $A_epsilon = {e^{ialpha n_epsilon k}:kin mathbb N}$ then has the property that every point on the circle is within ϵ of some point in $A_ϵ$....
$endgroup$
– No One
Mar 19 '16 at 19:57
$begingroup$
Thank you! But I can't see why $A_epsilon = {e^{ialpha n_epsilon k}:kin mathbb N}$ then has the property that every point on the circle is within ϵ of some point in $A_ϵ$....
$endgroup$
– No One
Mar 19 '16 at 19:57
$begingroup$
It's because $e^{ialpha n_epsilon} = e^{it}$ for some small $t.$ Think about the points $ e^{it}, e^{i2t}, e^{i3t}, dots ,e^{ikt}, dots $
$endgroup$
– zhw.
Mar 19 '16 at 20:18
$begingroup$
It's because $e^{ialpha n_epsilon} = e^{it}$ for some small $t.$ Think about the points $ e^{it}, e^{i2t}, e^{i3t}, dots ,e^{ikt}, dots $
$endgroup$
– zhw.
Mar 19 '16 at 20:18
$begingroup$
Dear Sir,Nice solution and Can we argue like following: 1 is limit point by above construction. Now we rotate that to make any point on $S^1$ to be limit point as $0<|e^{ix}||1-e^{ialpha n_epsilon}|=|e^{ix}-e^{i(x+alpha n_epsilon}|$<$epsilon$
$endgroup$
– SRJ
Jan 22 at 10:33
$begingroup$
Dear Sir,Nice solution and Can we argue like following: 1 is limit point by above construction. Now we rotate that to make any point on $S^1$ to be limit point as $0<|e^{ix}||1-e^{ialpha n_epsilon}|=|e^{ix}-e^{i(x+alpha n_epsilon}|$<$epsilon$
$endgroup$
– SRJ
Jan 22 at 10:33
$begingroup$
@SRJ I'm not sure what you're doing in the last line.
$endgroup$
– zhw.
Jan 23 at 0:36
$begingroup$
@SRJ I'm not sure what you're doing in the last line.
$endgroup$
– zhw.
Jan 23 at 0:36
$begingroup$
Sir I wanted to show for any point on S1 that is of form e^{ix} there is sequence converging to that point . As you had show this happen. For 1. I just multiply by e^ix to show it for other. Is my argument looking correct
$endgroup$
– SRJ
Jan 23 at 2:58
$begingroup$
Sir I wanted to show for any point on S1 that is of form e^{ix} there is sequence converging to that point . As you had show this happen. For 1. I just multiply by e^ix to show it for other. Is my argument looking correct
$endgroup$
– SRJ
Jan 23 at 2:58
add a comment |
1
$begingroup$
A minor modification works; see this answer.
$endgroup$
– Brian M. Scott
Mar 19 '16 at 19:47