Closed set in topological space generated by sets of the form [a, b).
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Let $tau$ be the topology on $mathbbR$ generated by $B = [a, b)subset mathbbR: -infty<a<b<infty$. Then the set $xin mathbbR: 4sin^2xleq 1cup bigfracpi2big$ is closed in $(mathbbR, tau)$.
My effort:
We know that $4sin^2xleq 1$ whenever $-frac12leq sin x leq frac12$, i.e. $x in big[-fracpi6, fracpi6big]cup big[-frac11pi6, frac13pi6big]cupcdots$. How to proceed further?
real-analysis general-topology
asked Feb 25 at 17:47
Mittal GMittal G
1,376516
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add a comment |
$begingroup$
Let $tau$ be the topology on $mathbbR$ generated by $B = [a, b)subset mathbbR: -infty<a<b<infty$. Then the set $xin mathbbR: 4sin^2xleq 1cup bigfracpi2big$ is closed in $(mathbbR, tau)$.
My effort:
We know that $4sin^2xleq 1$ whenever $-frac12leq sin x leq frac12$, i.e. $x in big[-fracpi6, fracpi6big]cup big[-frac11pi6, frac13pi6big]cupcdots$. How to proceed further?
real-analysis general-topology
asked Feb 25 at 17:47
Mittal GMittal G
1,376516
$endgroup$
add a comment |
$begingroup$
Let $tau$ be the topology on $mathbbR$ generated by $B = [a, b)subset mathbbR: -infty<a<b<infty$. Then the set $xin mathbbR: 4sin^2xleq 1cup bigfracpi2big$ is closed in $(mathbbR, tau)$.
My effort:
We know that $4sin^2xleq 1$ whenever $-frac12leq sin x leq frac12$, i.e. $x in big[-fracpi6, fracpi6big]cup big[-frac11pi6, frac13pi6big]cupcdots$. How to proceed further?
real-analysis general-topology
asked Feb 25 at 17:47
Mittal GMittal G
1,376516
$endgroup$
Let $tau$ be the topology on $mathbbR$ generated by $B = [a, b)subset mathbbR: -infty<a<b<infty$. Then the set $xin mathbbR: 4sin^2xleq 1cup bigfracpi2big$ is closed in $(mathbbR, tau)$.
My effort:
We know that $4sin^2xleq 1$ whenever $-frac12leq sin x leq frac12$, i.e. $x in big[-fracpi6, fracpi6big]cup big[-frac11pi6, frac13pi6big]cupcdots$. How to proceed further?
real-analysis general-topology
real-analysis general-topology
asked Feb 25 at 17:47
Mittal GMittal G
1,376516
asked Feb 25 at 17:47
Mittal GMittal G
1,376516
asked Feb 25 at 17:47
Mittal GMittal G
1,376516
asked Feb 25 at 17:47
Mittal GMittal G
1,376516
asked Feb 25 at 17:47
Mittal GMittal G
1,376516
1,376516
add a comment |
add a comment |
2 Answers
2
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Since $leftfracpi2right$ is closed in $(mathbbR,tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac11pi6, frac13pi6right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbbR,tau)$.
answered Feb 25 at 17:55
José Carlos SantosJosé Carlos Santos
169k23132238
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I know it! And…?
$endgroup$
– José Carlos Santos
Feb 25 at 17:59
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
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– Ajay Kumar Nair
Feb 25 at 18:00
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
Feb 25 at 18:02
add a comment |
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One way to show that a set is closed in a topology is to show that its complement is open. In this particular topology, any open interval is open:
$$(a, b) = cup_n in Bbb N [a-frac1n, b).$$
And it's easy to see that your set is the complement of a union of open intervals, which we now know are open in $tau$, so we're done.
answered Feb 25 at 17:59
Robert ShoreRobert Shore
3,540324
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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
$begingroup$
Since $leftfracpi2right$ is closed in $(mathbbR,tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac11pi6, frac13pi6right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbbR,tau)$.
answered Feb 25 at 17:55
José Carlos SantosJosé Carlos Santos
169k23132238
$endgroup$
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
Feb 25 at 17:59
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
Feb 25 at 18:00
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
Feb 25 at 18:02
add a comment |
$begingroup$
Since $leftfracpi2right$ is closed in $(mathbbR,tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac11pi6, frac13pi6right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbbR,tau)$.
answered Feb 25 at 17:55
José Carlos SantosJosé Carlos Santos
169k23132238
$endgroup$
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
Feb 25 at 17:59
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
Feb 25 at 18:00
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
Feb 25 at 18:02
add a comment |
$begingroup$
Since $leftfracpi2right$ is closed in $(mathbbR,tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac11pi6, frac13pi6right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbbR,tau)$.
answered Feb 25 at 17:55
José Carlos SantosJosé Carlos Santos
169k23132238
$endgroup$
Since $leftfracpi2right$ is closed in $(mathbbR,tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac11pi6, frac13pi6right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbbR,tau)$.
answered Feb 25 at 17:55
José Carlos SantosJosé Carlos Santos
169k23132238
answered Feb 25 at 17:55
José Carlos SantosJosé Carlos Santos
169k23132238
answered Feb 25 at 17:55
José Carlos SantosJosé Carlos Santos
169k23132238
answered Feb 25 at 17:55
José Carlos SantosJosé Carlos Santos
169k23132238
169k23132238
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
Feb 25 at 17:59
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
Feb 25 at 18:00
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
Feb 25 at 18:02
add a comment |
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
Feb 25 at 17:59
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
Feb 25 at 18:00
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
Feb 25 at 18:02
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
Feb 25 at 17:59
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
Feb 25 at 17:59
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
Feb 25 at 18:00
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
Feb 25 at 18:00
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
Feb 25 at 18:02
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
Feb 25 at 18:02
add a comment |
$begingroup$
One way to show that a set is closed in a topology is to show that its complement is open. In this particular topology, any open interval is open:
$$(a, b) = cup_n in Bbb N [a-frac1n, b).$$
And it's easy to see that your set is the complement of a union of open intervals, which we now know are open in $tau$, so we're done.
answered Feb 25 at 17:59
Robert ShoreRobert Shore
3,540324
$endgroup$
add a comment |
$begingroup$
One way to show that a set is closed in a topology is to show that its complement is open. In this particular topology, any open interval is open:
$$(a, b) = cup_n in Bbb N [a-frac1n, b).$$
And it's easy to see that your set is the complement of a union of open intervals, which we now know are open in $tau$, so we're done.
answered Feb 25 at 17:59
Robert ShoreRobert Shore
3,540324
$endgroup$
add a comment |
$begingroup$
One way to show that a set is closed in a topology is to show that its complement is open. In this particular topology, any open interval is open:
$$(a, b) = cup_n in Bbb N [a-frac1n, b).$$
And it's easy to see that your set is the complement of a union of open intervals, which we now know are open in $tau$, so we're done.
answered Feb 25 at 17:59
Robert ShoreRobert Shore
3,540324
$endgroup$
One way to show that a set is closed in a topology is to show that its complement is open. In this particular topology, any open interval is open:
$$(a, b) = cup_n in Bbb N [a-frac1n, b).$$
And it's easy to see that your set is the complement of a union of open intervals, which we now know are open in $tau$, so we're done.
answered Feb 25 at 17:59
Robert ShoreRobert Shore
3,540324
answered Feb 25 at 17:59
Robert ShoreRobert Shore
3,540324
answered Feb 25 at 17:59
Robert ShoreRobert Shore
3,540324
answered Feb 25 at 17:59
Robert ShoreRobert Shore
3,540324
3,540324
add a comment |
add a comment |
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