$ bigcup_{n=1}^{infty}{zinBbb C:z^n=1}={zinBbb C:|z|=1}$?
$begingroup$
Is $$ bigcup_{n=1}^{infty}{zinBbb C:z^n=1}={zinBbb C:|z|=1}$$my argument is that the argument of the elements of the first set are rational multiples of $pi$ whereas the second set also consists of elements with irrational multiples of $pi$.
But I am not so sure, there is still some doubt.
complex-numbers roots-of-unity
$endgroup$
add a comment |
$begingroup$
Is $$ bigcup_{n=1}^{infty}{zinBbb C:z^n=1}={zinBbb C:|z|=1}$$my argument is that the argument of the elements of the first set are rational multiples of $pi$ whereas the second set also consists of elements with irrational multiples of $pi$.
But I am not so sure, there is still some doubt.
complex-numbers roots-of-unity
$endgroup$
add a comment |
$begingroup$
Is $$ bigcup_{n=1}^{infty}{zinBbb C:z^n=1}={zinBbb C:|z|=1}$$my argument is that the argument of the elements of the first set are rational multiples of $pi$ whereas the second set also consists of elements with irrational multiples of $pi$.
But I am not so sure, there is still some doubt.
complex-numbers roots-of-unity
$endgroup$
Is $$ bigcup_{n=1}^{infty}{zinBbb C:z^n=1}={zinBbb C:|z|=1}$$my argument is that the argument of the elements of the first set are rational multiples of $pi$ whereas the second set also consists of elements with irrational multiples of $pi$.
But I am not so sure, there is still some doubt.
complex-numbers roots-of-unity
complex-numbers roots-of-unity
edited Feb 2 at 2:52
user21820
40.1k544162
40.1k544162
asked Feb 1 at 15:14
KarthikKarthik
13212
13212
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add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Those two sets are certainly not equal.
$displaystylebigcup_{n=1}^{infty}{z,|,z^n=1,nin mathbb N}$ is countable as it is a countable union of finite sets, while ${z/|z|=1}$ has the cardinality of the continuum.
$endgroup$
1
$begingroup$
What does "power of a continuum" mean?
$endgroup$
– Karthik
Feb 1 at 15:21
4
$begingroup$
I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:23
1
$begingroup$
"power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
$endgroup$
– Paul Sinclair
Feb 1 at 17:22
1
$begingroup$
@PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
$endgroup$
– mathcounterexamples.net
Feb 1 at 17:35
$begingroup$
This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
$endgroup$
– drjpizzle
Feb 1 at 18:22
|
show 1 more comment
$begingroup$
Your argument is correct.
For example, $e^{sqrt{2} pi i}$ is not in the first set.
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Those two sets are certainly not equal.
$displaystylebigcup_{n=1}^{infty}{z,|,z^n=1,nin mathbb N}$ is countable as it is a countable union of finite sets, while ${z/|z|=1}$ has the cardinality of the continuum.
$endgroup$
1
$begingroup$
What does "power of a continuum" mean?
$endgroup$
– Karthik
Feb 1 at 15:21
4
$begingroup$
I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:23
1
$begingroup$
"power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
$endgroup$
– Paul Sinclair
Feb 1 at 17:22
1
$begingroup$
@PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
$endgroup$
– mathcounterexamples.net
Feb 1 at 17:35
$begingroup$
This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
$endgroup$
– drjpizzle
Feb 1 at 18:22
|
show 1 more comment
$begingroup$
Those two sets are certainly not equal.
$displaystylebigcup_{n=1}^{infty}{z,|,z^n=1,nin mathbb N}$ is countable as it is a countable union of finite sets, while ${z/|z|=1}$ has the cardinality of the continuum.
$endgroup$
1
$begingroup$
What does "power of a continuum" mean?
$endgroup$
– Karthik
Feb 1 at 15:21
4
$begingroup$
I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:23
1
$begingroup$
"power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
$endgroup$
– Paul Sinclair
Feb 1 at 17:22
1
$begingroup$
@PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
$endgroup$
– mathcounterexamples.net
Feb 1 at 17:35
$begingroup$
This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
$endgroup$
– drjpizzle
Feb 1 at 18:22
|
show 1 more comment
$begingroup$
Those two sets are certainly not equal.
$displaystylebigcup_{n=1}^{infty}{z,|,z^n=1,nin mathbb N}$ is countable as it is a countable union of finite sets, while ${z/|z|=1}$ has the cardinality of the continuum.
$endgroup$
Those two sets are certainly not equal.
$displaystylebigcup_{n=1}^{infty}{z,|,z^n=1,nin mathbb N}$ is countable as it is a countable union of finite sets, while ${z/|z|=1}$ has the cardinality of the continuum.
edited Feb 1 at 15:56


El borito
664216
664216
answered Feb 1 at 15:19


mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
1
$begingroup$
What does "power of a continuum" mean?
$endgroup$
– Karthik
Feb 1 at 15:21
4
$begingroup$
I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:23
1
$begingroup$
"power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
$endgroup$
– Paul Sinclair
Feb 1 at 17:22
1
$begingroup$
@PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
$endgroup$
– mathcounterexamples.net
Feb 1 at 17:35
$begingroup$
This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
$endgroup$
– drjpizzle
Feb 1 at 18:22
|
show 1 more comment
1
$begingroup$
What does "power of a continuum" mean?
$endgroup$
– Karthik
Feb 1 at 15:21
4
$begingroup$
I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:23
1
$begingroup$
"power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
$endgroup$
– Paul Sinclair
Feb 1 at 17:22
1
$begingroup$
@PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
$endgroup$
– mathcounterexamples.net
Feb 1 at 17:35
$begingroup$
This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
$endgroup$
– drjpizzle
Feb 1 at 18:22
1
1
$begingroup$
What does "power of a continuum" mean?
$endgroup$
– Karthik
Feb 1 at 15:21
$begingroup$
What does "power of a continuum" mean?
$endgroup$
– Karthik
Feb 1 at 15:21
4
4
$begingroup$
I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:23
$begingroup$
I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $mathbb R$.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:23
1
1
$begingroup$
"power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
$endgroup$
– Paul Sinclair
Feb 1 at 17:22
$begingroup$
"power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I).
$endgroup$
– Paul Sinclair
Feb 1 at 17:22
1
1
$begingroup$
@PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
$endgroup$
– mathcounterexamples.net
Feb 1 at 17:35
$begingroup$
@PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste).
$endgroup$
– mathcounterexamples.net
Feb 1 at 17:35
$begingroup$
This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
$endgroup$
– drjpizzle
Feb 1 at 18:22
$begingroup$
This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow.
$endgroup$
– drjpizzle
Feb 1 at 18:22
|
show 1 more comment
$begingroup$
Your argument is correct.
For example, $e^{sqrt{2} pi i}$ is not in the first set.
$endgroup$
add a comment |
$begingroup$
Your argument is correct.
For example, $e^{sqrt{2} pi i}$ is not in the first set.
$endgroup$
add a comment |
$begingroup$
Your argument is correct.
For example, $e^{sqrt{2} pi i}$ is not in the first set.
$endgroup$
Your argument is correct.
For example, $e^{sqrt{2} pi i}$ is not in the first set.
answered Feb 1 at 15:22


DubsDubs
59426
59426
add a comment |
add a comment |
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