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Why the rationals do not form a countable intersection of open sets? [duplicate]
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How would one go about proving that the rationals are not the countable intersection of open sets?
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Why is $mathbbQ$ not a countable intersection of open sets in $mathbbR$?
The hint shown in the book is to recall the Baire Category Theorem: A countable intersection of open dense sets in $mathbbR$ is again dense.
general-topology
edited Jul 18 at 19:30
mechanodroid
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asked Jul 18 at 19:19
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marked as duplicate by Mostafa Ayaz, José Carlos Santos, Henno Brandsma general-topology
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Jul 18 at 23:05
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.
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This question already has an answer here:
How would one go about proving that the rationals are not the countable intersection of open sets?
2 answers
Why is $mathbbQ$ not a countable intersection of open sets in $mathbbR$?
The hint shown in the book is to recall the Baire Category Theorem: A countable intersection of open dense sets in $mathbbR$ is again dense.
general-topology
edited Jul 18 at 19:30
mechanodroid
22.2k52041
asked Jul 18 at 19:19
Shen Chong
386
marked as duplicate by Mostafa Ayaz, José Carlos Santos, Henno Brandsma general-topology
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Jul 18 at 23:05
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.
Do you see why, if such an intersection existed, all of the sets being intersected must be dense?
– Mark
Jul 18 at 19:25
Q is already dense, so what is the contradiction?
– Shen Chong
Jul 18 at 19:28
add a comment |Â
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up vote
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down vote
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This question already has an answer here:
How would one go about proving that the rationals are not the countable intersection of open sets?
2 answers
Why is $mathbbQ$ not a countable intersection of open sets in $mathbbR$?
The hint shown in the book is to recall the Baire Category Theorem: A countable intersection of open dense sets in $mathbbR$ is again dense.
general-topology
edited Jul 18 at 19:30
mechanodroid
22.2k52041
asked Jul 18 at 19:19
Shen Chong
386
This question already has an answer here:
How would one go about proving that the rationals are not the countable intersection of open sets?
2 answers
Why is $mathbbQ$ not a countable intersection of open sets in $mathbbR$?
The hint shown in the book is to recall the Baire Category Theorem: A countable intersection of open dense sets in $mathbbR$ is again dense.
This question already has an answer here:
How would one go about proving that the rationals are not the countable intersection of open sets?
2 answers
general-topology
edited Jul 18 at 19:30
mechanodroid
22.2k52041
asked Jul 18 at 19:19
Shen Chong
386
edited Jul 18 at 19:30
mechanodroid
22.2k52041
edited Jul 18 at 19:30
mechanodroid
22.2k52041
edited Jul 18 at 19:30
mechanodroid
22.2k52041
22.2k52041
asked Jul 18 at 19:19
Shen Chong
386
asked Jul 18 at 19:19
Shen Chong
386
asked Jul 18 at 19:19
Shen Chong
386
386
marked as duplicate by Mostafa Ayaz, José Carlos Santos, Henno Brandsma general-topology
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Jul 18 at 23:05
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.
Do you see why, if such an intersection existed, all of the sets being intersected must be dense?
– Mark
Jul 18 at 19:25
Q is already dense, so what is the contradiction?
– Shen Chong
Jul 18 at 19:28
add a comment |Â
Do you see why, if such an intersection existed, all of the sets being intersected must be dense?
– Mark
Jul 18 at 19:25
Q is already dense, so what is the contradiction?
– Shen Chong
Jul 18 at 19:28
Do you see why, if such an intersection existed, all of the sets being intersected must be dense?
– Mark
Jul 18 at 19:25
Do you see why, if such an intersection existed, all of the sets being intersected must be dense?
– Mark
Jul 18 at 19:25
Q is already dense, so what is the contradiction?
– Shen Chong
Jul 18 at 19:28
Q is already dense, so what is the contradiction?
– Shen Chong
Jul 18 at 19:28
add a comment |Â
1 Answer
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Assume $mathbbQ = bigcap_n=1^infty U_n$ where $U_n$ are open. $mathbbQ subseteq U_n$ so $U_n$ are dense.
We have
$$emptyset = (mathbbR setminus mathbbQ) cap mathbbQ = left(bigcap_q in mathbbQ mathbbR setminus qright) cap left( bigcap_n=1^infty U_nright)$$
so $emptyset$ is a countable intersection of dense open sets. This is a contradiction with your hint.
answered Jul 18 at 19:28
mechanodroid
22.2k52041
thank very much
– Shen Chong
Jul 18 at 19:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Assume $mathbbQ = bigcap_n=1^infty U_n$ where $U_n$ are open. $mathbbQ subseteq U_n$ so $U_n$ are dense.
We have
$$emptyset = (mathbbR setminus mathbbQ) cap mathbbQ = left(bigcap_q in mathbbQ mathbbR setminus qright) cap left( bigcap_n=1^infty U_nright)$$
so $emptyset$ is a countable intersection of dense open sets. This is a contradiction with your hint.
answered Jul 18 at 19:28
mechanodroid
22.2k52041
thank very much
– Shen Chong
Jul 18 at 19:30
add a comment |Â
up vote
3
down vote
accepted
Assume $mathbbQ = bigcap_n=1^infty U_n$ where $U_n$ are open. $mathbbQ subseteq U_n$ so $U_n$ are dense.
We have
$$emptyset = (mathbbR setminus mathbbQ) cap mathbbQ = left(bigcap_q in mathbbQ mathbbR setminus qright) cap left( bigcap_n=1^infty U_nright)$$
so $emptyset$ is a countable intersection of dense open sets. This is a contradiction with your hint.
answered Jul 18 at 19:28
mechanodroid
22.2k52041
thank very much
– Shen Chong
Jul 18 at 19:30
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Assume $mathbbQ = bigcap_n=1^infty U_n$ where $U_n$ are open. $mathbbQ subseteq U_n$ so $U_n$ are dense.
We have
$$emptyset = (mathbbR setminus mathbbQ) cap mathbbQ = left(bigcap_q in mathbbQ mathbbR setminus qright) cap left( bigcap_n=1^infty U_nright)$$
so $emptyset$ is a countable intersection of dense open sets. This is a contradiction with your hint.
answered Jul 18 at 19:28
mechanodroid
22.2k52041
Assume $mathbbQ = bigcap_n=1^infty U_n$ where $U_n$ are open. $mathbbQ subseteq U_n$ so $U_n$ are dense.
We have
$$emptyset = (mathbbR setminus mathbbQ) cap mathbbQ = left(bigcap_q in mathbbQ mathbbR setminus qright) cap left( bigcap_n=1^infty U_nright)$$
so $emptyset$ is a countable intersection of dense open sets. This is a contradiction with your hint.
answered Jul 18 at 19:28
mechanodroid
22.2k52041
answered Jul 18 at 19:28
mechanodroid
22.2k52041
answered Jul 18 at 19:28
mechanodroid
22.2k52041
answered Jul 18 at 19:28
mechanodroid
22.2k52041
22.2k52041
thank very much
– Shen Chong
Jul 18 at 19:30
add a comment |Â
thank very much
– Shen Chong
Jul 18 at 19:30
thank very much
– Shen Chong
Jul 18 at 19:30
thank very much
– Shen Chong
Jul 18 at 19:30
add a comment |Â
Do you see why, if such an intersection existed, all of the sets being intersected must be dense?
– Mark
Jul 18 at 19:25
Q is already dense, so what is the contradiction?
– Shen Chong
Jul 18 at 19:28