Truetone comjugate symmetry [closed]

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Truetone comjugate symmetry. What is it?




borel-sets




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closed as off-topic by Shailesh, Siong Thye Goh, Jendrik Stelzner, Brahadeesh, user91500 Aug 25 at 9:41


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Siong Thye Goh, Jendrik Stelzner, Brahadeesh, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.












  • You should also give what have you tried. However, I have given you the answer below.
    – Aniruddha Deshmukh
    Aug 25 at 5:07














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Truetone comjugate symmetry. What is it?




borel-sets




share|cite|improve this question














closed as off-topic by Shailesh, Siong Thye Goh, Jendrik Stelzner, Brahadeesh, user91500 Aug 25 at 9:41


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Siong Thye Goh, Jendrik Stelzner, Brahadeesh, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.












  • You should also give what have you tried. However, I have given you the answer below.
    – Aniruddha Deshmukh
    Aug 25 at 5:07












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Truetone comjugate symmetry. What is it?




borel-sets




share|cite|improve this question














Truetone comjugate symmetry. What is it?





borel-sets





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked Aug 25 at 4:53









mathnewbie

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274




closed as off-topic by Shailesh, Siong Thye Goh, Jendrik Stelzner, Brahadeesh, user91500 Aug 25 at 9:41


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Siong Thye Goh, Jendrik Stelzner, Brahadeesh, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Shailesh, Siong Thye Goh, Jendrik Stelzner, Brahadeesh, user91500 Aug 25 at 9:41


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Siong Thye Goh, Jendrik Stelzner, Brahadeesh, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.











  • You should also give what have you tried. However, I have given you the answer below.
    – Aniruddha Deshmukh
    Aug 25 at 5:07
















  • You should also give what have you tried. However, I have given you the answer below.
    – Aniruddha Deshmukh
    Aug 25 at 5:07















You should also give what have you tried. However, I have given you the answer below.
– Aniruddha Deshmukh
Aug 25 at 5:07




You should also give what have you tried. However, I have given you the answer below.
– Aniruddha Deshmukh
Aug 25 at 5:07










1 Answer
1






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First, observe that $forall m geq n$, $left( a - dfrac1m, a + dfrac1m right) subseteq left( a - dfrac1n, a + dfrac1n right)$. Therefore, $forall n in mathbbN$, $bigcuplimits_m = n^infty left( a - dfrac1m, a + dfrac1m right) = left( a - dfrac1n, a + dfrac1n right)$.



Now, $bigcaplimits_n = 1^infty left( a - dfrac1n, a + dfrac1n right) = leftlbrace a rightrbrace$ because no other point can be common to all these infinitely many open intervals (We can also prove this using the Archemedian Property).



Therefore,
$$bigcaplimits_n = 1^infty bigcuplimits_m = n^infty left( a - dfrac1m, a + dfrac1m right) = leftlbrace a rightrbrace$$






share|cite|improve this answer




















  • Thanks for the answer. If I interchange the union and intersection in this question. What would the answer be? The expression would reduce to $bigcuplimits_n=1^infty a$, and finally the answer would be a?
    – mathnewbie
    Aug 25 at 12:24











  • Yes. That will be the case. The expression would become $bigcuplimits_n = 1^infty leftlbrace a rightrbrace = leftlbrace a rightrbrace$.
    – Aniruddha Deshmukh
    Aug 25 at 12:41

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote













First, observe that $forall m geq n$, $left( a - dfrac1m, a + dfrac1m right) subseteq left( a - dfrac1n, a + dfrac1n right)$. Therefore, $forall n in mathbbN$, $bigcuplimits_m = n^infty left( a - dfrac1m, a + dfrac1m right) = left( a - dfrac1n, a + dfrac1n right)$.



Now, $bigcaplimits_n = 1^infty left( a - dfrac1n, a + dfrac1n right) = leftlbrace a rightrbrace$ because no other point can be common to all these infinitely many open intervals (We can also prove this using the Archemedian Property).



Therefore,
$$bigcaplimits_n = 1^infty bigcuplimits_m = n^infty left( a - dfrac1m, a + dfrac1m right) = leftlbrace a rightrbrace$$






share|cite|improve this answer




















  • Thanks for the answer. If I interchange the union and intersection in this question. What would the answer be? The expression would reduce to $bigcuplimits_n=1^infty a$, and finally the answer would be a?
    – mathnewbie
    Aug 25 at 12:24











  • Yes. That will be the case. The expression would become $bigcuplimits_n = 1^infty leftlbrace a rightrbrace = leftlbrace a rightrbrace$.
    – Aniruddha Deshmukh
    Aug 25 at 12:41














up vote
3
down vote













First, observe that $forall m geq n$, $left( a - dfrac1m, a + dfrac1m right) subseteq left( a - dfrac1n, a + dfrac1n right)$. Therefore, $forall n in mathbbN$, $bigcuplimits_m = n^infty left( a - dfrac1m, a + dfrac1m right) = left( a - dfrac1n, a + dfrac1n right)$.



Now, $bigcaplimits_n = 1^infty left( a - dfrac1n, a + dfrac1n right) = leftlbrace a rightrbrace$ because no other point can be common to all these infinitely many open intervals (We can also prove this using the Archemedian Property).



Therefore,
$$bigcaplimits_n = 1^infty bigcuplimits_m = n^infty left( a - dfrac1m, a + dfrac1m right) = leftlbrace a rightrbrace$$






share|cite|improve this answer




















  • Thanks for the answer. If I interchange the union and intersection in this question. What would the answer be? The expression would reduce to $bigcuplimits_n=1^infty a$, and finally the answer would be a?
    – mathnewbie
    Aug 25 at 12:24











  • Yes. That will be the case. The expression would become $bigcuplimits_n = 1^infty leftlbrace a rightrbrace = leftlbrace a rightrbrace$.
    – Aniruddha Deshmukh
    Aug 25 at 12:41












up vote
3
down vote










up vote
3
down vote









First, observe that $forall m geq n$, $left( a - dfrac1m, a + dfrac1m right) subseteq left( a - dfrac1n, a + dfrac1n right)$. Therefore, $forall n in mathbbN$, $bigcuplimits_m = n^infty left( a - dfrac1m, a + dfrac1m right) = left( a - dfrac1n, a + dfrac1n right)$.



Now, $bigcaplimits_n = 1^infty left( a - dfrac1n, a + dfrac1n right) = leftlbrace a rightrbrace$ because no other point can be common to all these infinitely many open intervals (We can also prove this using the Archemedian Property).



Therefore,
$$bigcaplimits_n = 1^infty bigcuplimits_m = n^infty left( a - dfrac1m, a + dfrac1m right) = leftlbrace a rightrbrace$$






share|cite|improve this answer












First, observe that $forall m geq n$, $left( a - dfrac1m, a + dfrac1m right) subseteq left( a - dfrac1n, a + dfrac1n right)$. Therefore, $forall n in mathbbN$, $bigcuplimits_m = n^infty left( a - dfrac1m, a + dfrac1m right) = left( a - dfrac1n, a + dfrac1n right)$.



Now, $bigcaplimits_n = 1^infty left( a - dfrac1n, a + dfrac1n right) = leftlbrace a rightrbrace$ because no other point can be common to all these infinitely many open intervals (We can also prove this using the Archemedian Property).



Therefore,
$$bigcaplimits_n = 1^infty bigcuplimits_m = n^infty left( a - dfrac1m, a + dfrac1m right) = leftlbrace a rightrbrace$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 25 at 5:06









Aniruddha Deshmukh

663417




663417











  • Thanks for the answer. If I interchange the union and intersection in this question. What would the answer be? The expression would reduce to $bigcuplimits_n=1^infty a$, and finally the answer would be a?
    – mathnewbie
    Aug 25 at 12:24











  • Yes. That will be the case. The expression would become $bigcuplimits_n = 1^infty leftlbrace a rightrbrace = leftlbrace a rightrbrace$.
    – Aniruddha Deshmukh
    Aug 25 at 12:41
















  • Thanks for the answer. If I interchange the union and intersection in this question. What would the answer be? The expression would reduce to $bigcuplimits_n=1^infty a$, and finally the answer would be a?
    – mathnewbie
    Aug 25 at 12:24











  • Yes. That will be the case. The expression would become $bigcuplimits_n = 1^infty leftlbrace a rightrbrace = leftlbrace a rightrbrace$.
    – Aniruddha Deshmukh
    Aug 25 at 12:41















Thanks for the answer. If I interchange the union and intersection in this question. What would the answer be? The expression would reduce to $bigcuplimits_n=1^infty a$, and finally the answer would be a?
– mathnewbie
Aug 25 at 12:24





Thanks for the answer. If I interchange the union and intersection in this question. What would the answer be? The expression would reduce to $bigcuplimits_n=1^infty a$, and finally the answer would be a?
– mathnewbie
Aug 25 at 12:24













Yes. That will be the case. The expression would become $bigcuplimits_n = 1^infty leftlbrace a rightrbrace = leftlbrace a rightrbrace$.
– Aniruddha Deshmukh
Aug 25 at 12:41




Yes. That will be the case. The expression would become $bigcuplimits_n = 1^infty leftlbrace a rightrbrace = leftlbrace a rightrbrace$.
– Aniruddha Deshmukh
Aug 25 at 12:41


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