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Open set in a subspace topology

0

$begingroup$

Let $C=1/n_n=1^infty$ be a subspace of $mathbbR$, on $mathbbR$ we have the standard topology.

Question. Why the one-point set $1/n$ are all open in $C$ with the subspace topology?

Thanks!

general-topology

share|cite|improve this question

asked Apr 2 at 17:15

Jack J.Jack J.

4082519

$endgroup$

add a comment | 

0

$begingroup$

Let $C=1/n_n=1^infty$ be a subspace of $mathbbR$, on $mathbbR$ we have the standard topology.

Question. Why the one-point set $1/n$ are all open in $C$ with the subspace topology?

Thanks!

general-topology

share|cite|improve this question

asked Apr 2 at 17:15

Jack J.Jack J.

4082519

$endgroup$

add a comment | 

$begingroup$

Let $C=1/n_n=1^infty$ be a subspace of $mathbbR$, on $mathbbR$ we have the standard topology.

Question. Why the one-point set $1/n$ are all open in $C$ with the subspace topology?

Thanks!

general-topology

share|cite|improve this question

asked Apr 2 at 17:15

Jack J.Jack J.

4082519

$endgroup$

share|cite|improve this question

asked Apr 2 at 17:15

Jack J.Jack J.

4082519

share|cite|improve this question

asked Apr 2 at 17:15

Jack J.Jack J.

4082519

asked Apr 2 at 17:15

Jack J.Jack J.

4082519

asked Apr 2 at 17:15

Jack J.Jack J.

4082519

3 Answers
3

active

oldest

votes

2

$begingroup$

$O$ is open in $C$

$ iff$ there is an open $U$ $in mathbbR$ s.t. $O = U cap C$.

Choose an interval $(1/n-epsilon , 1/n +epsilon)$, open in $mathbbR$ with $epsilon$ small enough s.t.

$ C cap (1/n-epsilon, 1/n+epsilon)=$$1/n$.

$...1/(n+1), 1/n, 1/(n-1),...$

$1/n-1/(n+1)= dfrac1n(n+1)$;

$1/(n-1)-1/n= dfrac1n(n-1) >$

$dfrac1n(n+1)$.

Choose for example $epsilon =dfrac12n(n+1)$.

share|cite|improve this answer

answered Apr 2 at 17:59

Peter SzilasPeter Szilas

12k2822

$endgroup$

add a comment | 

3

$begingroup$

Because the intervall $$left(frac1n- frac12n(n+1), frac1n+ frac12n(n+1)right)$$

is open, contains $frac1n$, but does not contain any other point of $C$.

share|cite|improve this answer

answered Apr 2 at 17:18

TheSilverDoeTheSilverDoe

5,594316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer.Shouldn't we write $1/n$ as $Ccap V$, where $V$ is an open set of $mathbbR$?
  $endgroup$
  – Jack J.
  Apr 2 at 17:21
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Actually, what I say in my answer is equivalent to say that $lbrace 1/n rbrace = C cap V$, where $V$ is the open intervall I constructed.
  $endgroup$
  – TheSilverDoe
  Apr 2 at 17:27
  

  
  
  
  

add a comment | 

2

$begingroup$

A subset $V$ of $C$ is open in the subspace topology of $mathbf R$ if $V = Ccap U$ for some open set $U$ in $mathbf R$. So we are saying that the one-point sets $V=1,1/2,1/3,dots$ each can be written as $Ccap U_1$, $Ccap U_2$, $Ccap U_3$, and so on, for open sets $U_1,U_2,U_3,ldotssubsetmathbf R$. Can you draw a picture to figure out what the sets $U_1,U_2,U_3,dots$ could be?

share|cite|improve this answer

answered Apr 2 at 17:21

Alex OrtizAlex Ortiz

11.6k21442

$endgroup$

add a comment | 

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2

$begingroup$

$O$ is open in $C$

$ iff$ there is an open $U$ $in mathbbR$ s.t. $O = U cap C$.

Choose an interval $(1/n-epsilon , 1/n +epsilon)$, open in $mathbbR$ with $epsilon$ small enough s.t.

$ C cap (1/n-epsilon, 1/n+epsilon)=$$1/n$.

$...1/(n+1), 1/n, 1/(n-1),...$

$1/n-1/(n+1)= dfrac1n(n+1)$;

$1/(n-1)-1/n= dfrac1n(n-1) >$

$dfrac1n(n+1)$.

Choose for example $epsilon =dfrac12n(n+1)$.

share|cite|improve this answer

answered Apr 2 at 17:59

Peter SzilasPeter Szilas

12k2822

$endgroup$

add a comment | 

2

$begingroup$

$O$ is open in $C$

$ iff$ there is an open $U$ $in mathbbR$ s.t. $O = U cap C$.

Choose an interval $(1/n-epsilon , 1/n +epsilon)$, open in $mathbbR$ with $epsilon$ small enough s.t.

$ C cap (1/n-epsilon, 1/n+epsilon)=$$1/n$.

$...1/(n+1), 1/n, 1/(n-1),...$

$1/n-1/(n+1)= dfrac1n(n+1)$;

$1/(n-1)-1/n= dfrac1n(n-1) >$

$dfrac1n(n+1)$.

Choose for example $epsilon =dfrac12n(n+1)$.

share|cite|improve this answer

answered Apr 2 at 17:59

Peter SzilasPeter Szilas

12k2822

$endgroup$

add a comment | 

$begingroup$

$O$ is open in $C$

$ iff$ there is an open $U$ $in mathbbR$ s.t. $O = U cap C$.

Choose an interval $(1/n-epsilon , 1/n +epsilon)$, open in $mathbbR$ with $epsilon$ small enough s.t.

$ C cap (1/n-epsilon, 1/n+epsilon)=$$1/n$.

$...1/(n+1), 1/n, 1/(n-1),...$

$1/n-1/(n+1)= dfrac1n(n+1)$;

$1/(n-1)-1/n= dfrac1n(n-1) >$

$dfrac1n(n+1)$.

Choose for example $epsilon =dfrac12n(n+1)$.

share|cite|improve this answer

answered Apr 2 at 17:59

Peter SzilasPeter Szilas

12k2822

$endgroup$

share|cite|improve this answer

answered Apr 2 at 17:59

Peter SzilasPeter Szilas

12k2822

answered Apr 2 at 17:59

Peter SzilasPeter Szilas

12k2822

answered Apr 2 at 17:59

Peter SzilasPeter Szilas

12k2822

3

$begingroup$

Because the intervall $$left(frac1n- frac12n(n+1), frac1n+ frac12n(n+1)right)$$

is open, contains $frac1n$, but does not contain any other point of $C$.

share|cite|improve this answer

answered Apr 2 at 17:18

TheSilverDoeTheSilverDoe

5,594316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer.Shouldn't we write $1/n$ as $Ccap V$, where $V$ is an open set of $mathbbR$?
  $endgroup$
  – Jack J.
  Apr 2 at 17:21
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Actually, what I say in my answer is equivalent to say that $lbrace 1/n rbrace = C cap V$, where $V$ is the open intervall I constructed.
  $endgroup$
  – TheSilverDoe
  Apr 2 at 17:27
  

  
  
  
  

add a comment | 

3

$begingroup$

Because the intervall $$left(frac1n- frac12n(n+1), frac1n+ frac12n(n+1)right)$$

is open, contains $frac1n$, but does not contain any other point of $C$.

share|cite|improve this answer

answered Apr 2 at 17:18

TheSilverDoeTheSilverDoe

5,594316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer.Shouldn't we write $1/n$ as $Ccap V$, where $V$ is an open set of $mathbbR$?
  $endgroup$
  – Jack J.
  Apr 2 at 17:21
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Actually, what I say in my answer is equivalent to say that $lbrace 1/n rbrace = C cap V$, where $V$ is the open intervall I constructed.
  $endgroup$
  – TheSilverDoe
  Apr 2 at 17:27
  

  
  
  
  

add a comment | 

$begingroup$

Because the intervall $$left(frac1n- frac12n(n+1), frac1n+ frac12n(n+1)right)$$

is open, contains $frac1n$, but does not contain any other point of $C$.

share|cite|improve this answer

answered Apr 2 at 17:18

TheSilverDoeTheSilverDoe

5,594316

$endgroup$

share|cite|improve this answer

answered Apr 2 at 17:18

TheSilverDoeTheSilverDoe

5,594316

answered Apr 2 at 17:18

TheSilverDoeTheSilverDoe

5,594316

answered Apr 2 at 17:18

TheSilverDoeTheSilverDoe

5,594316

2

$begingroup$

A subset $V$ of $C$ is open in the subspace topology of $mathbf R$ if $V = Ccap U$ for some open set $U$ in $mathbf R$. So we are saying that the one-point sets $V=1,1/2,1/3,dots$ each can be written as $Ccap U_1$, $Ccap U_2$, $Ccap U_3$, and so on, for open sets $U_1,U_2,U_3,ldotssubsetmathbf R$. Can you draw a picture to figure out what the sets $U_1,U_2,U_3,dots$ could be?

share|cite|improve this answer

answered Apr 2 at 17:21

Alex OrtizAlex Ortiz

11.6k21442

$endgroup$

add a comment | 

2

$begingroup$

A subset $V$ of $C$ is open in the subspace topology of $mathbf R$ if $V = Ccap U$ for some open set $U$ in $mathbf R$. So we are saying that the one-point sets $V=1,1/2,1/3,dots$ each can be written as $Ccap U_1$, $Ccap U_2$, $Ccap U_3$, and so on, for open sets $U_1,U_2,U_3,ldotssubsetmathbf R$. Can you draw a picture to figure out what the sets $U_1,U_2,U_3,dots$ could be?

share|cite|improve this answer

answered Apr 2 at 17:21

Alex OrtizAlex Ortiz

11.6k21442

$endgroup$

add a comment | 

$begingroup$

A subset $V$ of $C$ is open in the subspace topology of $mathbf R$ if $V = Ccap U$ for some open set $U$ in $mathbf R$. So we are saying that the one-point sets $V=1,1/2,1/3,dots$ each can be written as $Ccap U_1$, $Ccap U_2$, $Ccap U_3$, and so on, for open sets $U_1,U_2,U_3,ldotssubsetmathbf R$. Can you draw a picture to figure out what the sets $U_1,U_2,U_3,dots$ could be?

share|cite|improve this answer

answered Apr 2 at 17:21

Alex OrtizAlex Ortiz

11.6k21442

$endgroup$

share|cite|improve this answer

answered Apr 2 at 17:21

Alex OrtizAlex Ortiz

11.6k21442

answered Apr 2 at 17:21

Alex OrtizAlex Ortiz

11.6k21442

answered Apr 2 at 17:21

Alex OrtizAlex Ortiz

11.6k21442