Closed set in topological space generated by sets of the form [a, b).Exhaustion of open sets by closed...
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Closed set in topological space generated by sets of the form [a, b).
Exhaustion of open sets by closed setszero dimensional topological spaceTopological proof that the interval $[a,b)subset mathbb{R}$ is not closedTopology on a finite set with closed singletons is discreteHow to find closed sets in a given topological subspaceGiven two topologies on a set , prove that the set is finite.Topological embedding on adjunction spaceBasis of topological space and topological subspacePreimage under $f$ of any closed set is closed.A set is compact in complement topology iff closed in standard topology
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Let $tau$ be the topology on $mathbb{R}$ generated by $B = {[a, b)subset mathbb{R}: -infty<a<b<infty}$. Then the set ${xin mathbb{R}: 4sin^2xleq 1}cup big{frac{pi}{2}big}$ is closed in $(mathbb{R}, tau)$.
My effort:
We know that $4sin^2xleq 1$ whenever $-frac{1}{2}leq sin x leq frac{1}{2}$, i.e. $x in big[-frac{pi}{6}, frac{pi}{6}big]cup big[-frac{11pi}{6}, frac{13pi}{6}big]cupcdots$. How to proceed further?
real-analysis general-topology
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add a comment |
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Let $tau$ be the topology on $mathbb{R}$ generated by $B = {[a, b)subset mathbb{R}: -infty<a<b<infty}$. Then the set ${xin mathbb{R}: 4sin^2xleq 1}cup big{frac{pi}{2}big}$ is closed in $(mathbb{R}, tau)$.
My effort:
We know that $4sin^2xleq 1$ whenever $-frac{1}{2}leq sin x leq frac{1}{2}$, i.e. $x in big[-frac{pi}{6}, frac{pi}{6}big]cup big[-frac{11pi}{6}, frac{13pi}{6}big]cupcdots$. How to proceed further?
real-analysis general-topology
$endgroup$
add a comment |
$begingroup$
Let $tau$ be the topology on $mathbb{R}$ generated by $B = {[a, b)subset mathbb{R}: -infty<a<b<infty}$. Then the set ${xin mathbb{R}: 4sin^2xleq 1}cup big{frac{pi}{2}big}$ is closed in $(mathbb{R}, tau)$.
My effort:
We know that $4sin^2xleq 1$ whenever $-frac{1}{2}leq sin x leq frac{1}{2}$, i.e. $x in big[-frac{pi}{6}, frac{pi}{6}big]cup big[-frac{11pi}{6}, frac{13pi}{6}big]cupcdots$. How to proceed further?
real-analysis general-topology
$endgroup$
Let $tau$ be the topology on $mathbb{R}$ generated by $B = {[a, b)subset mathbb{R}: -infty<a<b<infty}$. Then the set ${xin mathbb{R}: 4sin^2xleq 1}cup big{frac{pi}{2}big}$ is closed in $(mathbb{R}, tau)$.
My effort:
We know that $4sin^2xleq 1$ whenever $-frac{1}{2}leq sin x leq frac{1}{2}$, i.e. $x in big[-frac{pi}{6}, frac{pi}{6}big]cup big[-frac{11pi}{6}, frac{13pi}{6}big]cupcdots$. How to proceed further?
real-analysis general-topology
real-analysis general-topology
asked 6 hours ago
Mittal GMittal G
1,250516
1,250516
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2 Answers
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Since $left{fracpi2right}$ is closed in $(mathbb{R},tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac{11pi}{6}, frac{13pi}{6}right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbb{R},tau)$.
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I know it! And…?
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– José Carlos Santos
6 hours ago
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Sorry! I got it! $(a,b)$ is open in the above topology also.
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– Ajay Kumar Nair
6 hours ago
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Indeed. That's important.
$endgroup$
– José Carlos Santos
6 hours ago
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:
$$(a, b) = cup_{n in Bbb N} [a-frac{1}{n}, b).$$
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.
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $left{fracpi2right}$ is closed in $(mathbb{R},tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac{11pi}{6}, frac{13pi}{6}right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbb{R},tau)$.
$endgroup$
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
6 hours ago
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
6 hours ago
add a comment |
$begingroup$
Since $left{fracpi2right}$ is closed in $(mathbb{R},tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac{11pi}{6}, frac{13pi}{6}right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbb{R},tau)$.
$endgroup$
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
6 hours ago
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
6 hours ago
add a comment |
$begingroup$
Since $left{fracpi2right}$ is closed in $(mathbb{R},tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac{11pi}{6}, frac{13pi}{6}right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbb{R},tau)$.
$endgroup$
Since $left{fracpi2right}$ is closed in $(mathbb{R},tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac{11pi}{6}, frac{13pi}{6}right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbb{R},tau)$.
answered 6 hours ago
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
6 hours ago
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
6 hours ago
add a comment |
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
6 hours ago
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
I know it! And…?
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
6 hours ago
$begingroup$
Sorry! I got it! $(a,b)$ is open in the above topology also.
$endgroup$
– Ajay Kumar Nair
6 hours ago
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
Indeed. That's important.
$endgroup$
– José Carlos Santos
6 hours ago
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:
$$(a, b) = cup_{n in Bbb N} [a-frac{1}{n}, b).$$
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.
$endgroup$
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:
$$(a, b) = cup_{n in Bbb N} [a-frac{1}{n}, b).$$
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.
$endgroup$
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:
$$(a, b) = cup_{n in Bbb N} [a-frac{1}{n}, b).$$
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.
$endgroup$
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:
$$(a, b) = cup_{n in Bbb N} [a-frac{1}{n}, b).$$
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 6 hours ago
Robert ShoreRobert Shore
1,55615
1,55615
add a comment |
add a comment |
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