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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

2

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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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asked 6 hours ago

Mittal GMittal G

1,250516

$endgroup$

add a comment | 

2

$begingroup$

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

share|cite|improve this question

asked 6 hours ago

Mittal GMittal G

1,250516

$endgroup$

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

share|cite|improve this question

asked 6 hours ago

Mittal GMittal G

1,250516

$endgroup$

share|cite|improve this question

asked 6 hours ago

Mittal GMittal G

1,250516

share|cite|improve this question

asked 6 hours ago

Mittal GMittal G

1,250516

asked 6 hours ago

Mittal GMittal G

1,250516

asked 6 hours ago

Mittal GMittal G

1,250516

2 Answers
2

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oldest

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6

$begingroup$

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 $(mathbb{R},tau)$.

share|cite|improve this answer

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

164k22131234

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

2

$begingroup$

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.

share|cite|improve this answer

answered 6 hours ago

Robert ShoreRobert Shore

1,55615

$endgroup$

add a comment | 

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6

$begingroup$

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 $(mathbb{R},tau)$.

share|cite|improve this answer

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

164k22131234

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

6

$begingroup$

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 $(mathbb{R},tau)$.

share|cite|improve this answer

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

164k22131234

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $(mathbb{R},tau)$.

share|cite|improve this answer

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

164k22131234

$endgroup$

share|cite|improve this answer

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

164k22131234

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

164k22131234

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

164k22131234

2

$begingroup$

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.

share|cite|improve this answer

answered 6 hours ago

Robert ShoreRobert Shore

1,55615

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered 6 hours ago

Robert ShoreRobert Shore

1,55615

$endgroup$

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.

share|cite|improve this answer

answered 6 hours ago

Robert ShoreRobert Shore

1,55615

$endgroup$

share|cite|improve this answer

answered 6 hours ago

Robert ShoreRobert Shore

1,55615

answered 6 hours ago

Robert ShoreRobert Shore

1,55615

answered 6 hours ago

Robert ShoreRobert Shore

1,55615