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Is this set open or closed (or both?)

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6

$begingroup$

I'm trying to figure out whether or not the following set is open or closed.
$$D=(x,y,z)inmathbb R^3mid xgt0,ygt0,z=0$$

I've tried imagining it and to me, it seems like an open set, but maybe it is both open and closed. How would I determine that?

real-analysis general-topology

share|cite|improve this question

edited Jun 14 at 9:43

Thomas Shelby

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asked Jun 14 at 9:36

Counter BoostingCounter Boosting

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$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What do you know about open sets - what properties must they have? Likewise closed sets? "It seems like ..." doesn't refer to properties or definitions - intuition can be a guide to how you might go about the task. What do you know about closed sets and limit points, for example? Take a point in $D$ - what can you say about points in a neighbourhood of the point? (or you may have other definitions or concepts to hand)
  $endgroup$
  – Mark Bennet
  Jun 14 at 9:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was thinking about the set of all the points in the first quadrant of the X,Y plane that's why I thought it was open, but I forgot that it's a 3 dimensional set...
  $endgroup$
  – Counter Boosting
  Jun 14 at 9:48
  

  
  
  
  

add a comment
 | 

6

$begingroup$

I'm trying to figure out whether or not the following set is open or closed.
$$D=(x,y,z)inmathbb R^3mid xgt0,ygt0,z=0$$

I've tried imagining it and to me, it seems like an open set, but maybe it is both open and closed. How would I determine that?

real-analysis general-topology

share|cite|improve this question

edited Jun 14 at 9:43

Thomas Shelby

7,08944 gold badges1212 silver badges3333 bronze badges

asked Jun 14 at 9:36

Counter BoostingCounter Boosting

4244 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What do you know about open sets - what properties must they have? Likewise closed sets? "It seems like ..." doesn't refer to properties or definitions - intuition can be a guide to how you might go about the task. What do you know about closed sets and limit points, for example? Take a point in $D$ - what can you say about points in a neighbourhood of the point? (or you may have other definitions or concepts to hand)
  $endgroup$
  – Mark Bennet
  Jun 14 at 9:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was thinking about the set of all the points in the first quadrant of the X,Y plane that's why I thought it was open, but I forgot that it's a 3 dimensional set...
  $endgroup$
  – Counter Boosting
  Jun 14 at 9:48
  

  
  
  
  

add a comment
 | 

$begingroup$

I'm trying to figure out whether or not the following set is open or closed.
$$D=(x,y,z)inmathbb R^3mid xgt0,ygt0,z=0$$

I've tried imagining it and to me, it seems like an open set, but maybe it is both open and closed. How would I determine that?

real-analysis general-topology

share|cite|improve this question

edited Jun 14 at 9:43

Thomas Shelby

7,08944 gold badges1212 silver badges3333 bronze badges

asked Jun 14 at 9:36

Counter BoostingCounter Boosting

4244 bronze badges

$endgroup$

share|cite|improve this question

edited Jun 14 at 9:43

Thomas Shelby

7,08944 gold badges1212 silver badges3333 bronze badges

asked Jun 14 at 9:36

Counter BoostingCounter Boosting

4244 bronze badges

share|cite|improve this question

edited Jun 14 at 9:43

Thomas Shelby

7,08944 gold badges1212 silver badges3333 bronze badges

asked Jun 14 at 9:36

Counter BoostingCounter Boosting

4244 bronze badges

edited Jun 14 at 9:43

Thomas Shelby

7,08944 gold badges1212 silver badges3333 bronze badges

edited Jun 14 at 9:43

Thomas Shelby

7,08944 gold badges1212 silver badges3333 bronze badges

asked Jun 14 at 9:36

Counter BoostingCounter Boosting

4244 bronze badges

asked Jun 14 at 9:36

Counter BoostingCounter Boosting

4244 bronze badges

3 Answers
3

active

oldest

votes

14

$begingroup$

It is not open because it contains $(1,1,0)$ and every neighborhood if this point contains points with $z neq 0$. It is not closed because $(frac 1 n, frac 1 n,0)$ is a sequence in this set which converges to a point outside the set.

share|cite|improve this answer

answered Jun 14 at 9:39

Kabo MurphyKabo Murphy

129k77 gold badges4646 silver badges9999 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe a bit nitpicking. One has to ask: according to what topology. Of course, if your assume the natural topology of R3 then your answer is perferct. If one takes the induced topology on the subset D (and then D as a subset of that topological space) then D is open and closed.
  $endgroup$
  – lalala
  Jun 14 at 19:40
  

  
  
  
  

add a comment
 | 

9

$begingroup$

No, it is not an open set. For instance, $(1,1,0)in D$, but no open ball centered at $(1,1,0)$ is contained in $D$.

On the other hand, $left(left(frac1n,frac1n,0right)right)_ninmathbb N$ is a sequence of elements of $D$ which converges to $(0,0,0)$. But $(0,0,0)$ does not belong to $D$. What can you deduce from this?

share|cite|improve this answer

edited Jun 19 at 7:50

answered Jun 14 at 9:39

José Carlos SantosJosé Carlos Santos

224k2727 gold badges172172 silver badges300300 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Amazing! Our answers are identical.
  $endgroup$
  – Kabo Murphy
  Jun 14 at 9:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Great minds think alike! $ddotsmile$
  $endgroup$
  – José Carlos Santos
  Jun 14 at 9:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So the set is neither open or closed, thank you, both answers are great so I don't know who to award it to! :D
  $endgroup$
  – Counter Boosting
  Jun 14 at 9:46
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @CounterBoosting I suggest that you take into account the fact that the other answer appeared before mine.
  $endgroup$
  – José Carlos Santos
  Jun 14 at 9:49
  

  
  
  
  

add a comment
 | 

0

$begingroup$

Theorem: Let x$in mathbbR^n$ where the topology induced by the standard metric is assumed. $xrightarrow a $ $iff$ $x^i rightarrow a^i$ for each $1leq ileq n$

Take the sequence $(frac1n,frac1n, 0)$

Theorem 2: Let X be a metric space. $A subset X$ is closed iff the limit of every convergent sequence in A is in A.

Observe that the components of the above sequence all converge to 0 and therefore the sequence converges to $(0,0,0)$ which is not in the set. To show that the set is open, you must find a point in the set such that no matter what radius your ball has, it is never contained in the set. The previous answers wrote $(1,1,0)$ you should prove that what they wrote indeed shows that the set is not open.

share|cite|improve this answer

answered Jul 10 at 12:16

topologicalmagiciantopologicalmagician

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add a comment
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14

$begingroup$

It is not open because it contains $(1,1,0)$ and every neighborhood if this point contains points with $z neq 0$. It is not closed because $(frac 1 n, frac 1 n,0)$ is a sequence in this set which converges to a point outside the set.

share|cite|improve this answer

answered Jun 14 at 9:39

Kabo MurphyKabo Murphy

129k77 gold badges4646 silver badges9999 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe a bit nitpicking. One has to ask: according to what topology. Of course, if your assume the natural topology of R3 then your answer is perferct. If one takes the induced topology on the subset D (and then D as a subset of that topological space) then D is open and closed.
  $endgroup$
  – lalala
  Jun 14 at 19:40
  

  
  
  
  

add a comment
 | 

14

$begingroup$

It is not open because it contains $(1,1,0)$ and every neighborhood if this point contains points with $z neq 0$. It is not closed because $(frac 1 n, frac 1 n,0)$ is a sequence in this set which converges to a point outside the set.

share|cite|improve this answer

answered Jun 14 at 9:39

Kabo MurphyKabo Murphy

129k77 gold badges4646 silver badges9999 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe a bit nitpicking. One has to ask: according to what topology. Of course, if your assume the natural topology of R3 then your answer is perferct. If one takes the induced topology on the subset D (and then D as a subset of that topological space) then D is open and closed.
  $endgroup$
  – lalala
  Jun 14 at 19:40
  

  
  
  
  

add a comment
 | 

$begingroup$

It is not open because it contains $(1,1,0)$ and every neighborhood if this point contains points with $z neq 0$. It is not closed because $(frac 1 n, frac 1 n,0)$ is a sequence in this set which converges to a point outside the set.

share|cite|improve this answer

answered Jun 14 at 9:39

Kabo MurphyKabo Murphy

129k77 gold badges4646 silver badges9999 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 14 at 9:39

Kabo MurphyKabo Murphy

129k77 gold badges4646 silver badges9999 bronze badges

answered Jun 14 at 9:39

Kabo MurphyKabo Murphy

129k77 gold badges4646 silver badges9999 bronze badges

answered Jun 14 at 9:39

Kabo MurphyKabo Murphy

129k77 gold badges4646 silver badges9999 bronze badges

9

$begingroup$

No, it is not an open set. For instance, $(1,1,0)in D$, but no open ball centered at $(1,1,0)$ is contained in $D$.

On the other hand, $left(left(frac1n,frac1n,0right)right)_ninmathbb N$ is a sequence of elements of $D$ which converges to $(0,0,0)$. But $(0,0,0)$ does not belong to $D$. What can you deduce from this?

share|cite|improve this answer

edited Jun 19 at 7:50

answered Jun 14 at 9:39

José Carlos SantosJosé Carlos Santos

224k2727 gold badges172172 silver badges300300 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Amazing! Our answers are identical.
  $endgroup$
  – Kabo Murphy
  Jun 14 at 9:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Great minds think alike! $ddotsmile$
  $endgroup$
  – José Carlos Santos
  Jun 14 at 9:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So the set is neither open or closed, thank you, both answers are great so I don't know who to award it to! :D
  $endgroup$
  – Counter Boosting
  Jun 14 at 9:46
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @CounterBoosting I suggest that you take into account the fact that the other answer appeared before mine.
  $endgroup$
  – José Carlos Santos
  Jun 14 at 9:49
  

  
  
  
  

add a comment
 | 

9

$begingroup$

No, it is not an open set. For instance, $(1,1,0)in D$, but no open ball centered at $(1,1,0)$ is contained in $D$.

On the other hand, $left(left(frac1n,frac1n,0right)right)_ninmathbb N$ is a sequence of elements of $D$ which converges to $(0,0,0)$. But $(0,0,0)$ does not belong to $D$. What can you deduce from this?

share|cite|improve this answer

edited Jun 19 at 7:50

answered Jun 14 at 9:39

José Carlos SantosJosé Carlos Santos

224k2727 gold badges172172 silver badges300300 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Amazing! Our answers are identical.
  $endgroup$
  – Kabo Murphy
  Jun 14 at 9:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Great minds think alike! $ddotsmile$
  $endgroup$
  – José Carlos Santos
  Jun 14 at 9:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So the set is neither open or closed, thank you, both answers are great so I don't know who to award it to! :D
  $endgroup$
  – Counter Boosting
  Jun 14 at 9:46
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @CounterBoosting I suggest that you take into account the fact that the other answer appeared before mine.
  $endgroup$
  – José Carlos Santos
  Jun 14 at 9:49
  

  
  
  
  

add a comment
 | 

$begingroup$

No, it is not an open set. For instance, $(1,1,0)in D$, but no open ball centered at $(1,1,0)$ is contained in $D$.

On the other hand, $left(left(frac1n,frac1n,0right)right)_ninmathbb N$ is a sequence of elements of $D$ which converges to $(0,0,0)$. But $(0,0,0)$ does not belong to $D$. What can you deduce from this?

share|cite|improve this answer

edited Jun 19 at 7:50

answered Jun 14 at 9:39

José Carlos SantosJosé Carlos Santos

224k2727 gold badges172172 silver badges300300 bronze badges

$endgroup$

share|cite|improve this answer

edited Jun 19 at 7:50

answered Jun 14 at 9:39

José Carlos SantosJosé Carlos Santos

224k2727 gold badges172172 silver badges300300 bronze badges

answered Jun 14 at 9:39

José Carlos SantosJosé Carlos Santos

224k2727 gold badges172172 silver badges300300 bronze badges

answered Jun 14 at 9:39

José Carlos SantosJosé Carlos Santos

224k2727 gold badges172172 silver badges300300 bronze badges

0

$begingroup$

Theorem: Let x$in mathbbR^n$ where the topology induced by the standard metric is assumed. $xrightarrow a $ $iff$ $x^i rightarrow a^i$ for each $1leq ileq n$

Take the sequence $(frac1n,frac1n, 0)$

Theorem 2: Let X be a metric space. $A subset X$ is closed iff the limit of every convergent sequence in A is in A.

Observe that the components of the above sequence all converge to 0 and therefore the sequence converges to $(0,0,0)$ which is not in the set. To show that the set is open, you must find a point in the set such that no matter what radius your ball has, it is never contained in the set. The previous answers wrote $(1,1,0)$ you should prove that what they wrote indeed shows that the set is not open.

share|cite|improve this answer

answered Jul 10 at 12:16

topologicalmagiciantopologicalmagician

90611 silver badge1111 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

Theorem: Let x$in mathbbR^n$ where the topology induced by the standard metric is assumed. $xrightarrow a $ $iff$ $x^i rightarrow a^i$ for each $1leq ileq n$

Take the sequence $(frac1n,frac1n, 0)$

Theorem 2: Let X be a metric space. $A subset X$ is closed iff the limit of every convergent sequence in A is in A.

Observe that the components of the above sequence all converge to 0 and therefore the sequence converges to $(0,0,0)$ which is not in the set. To show that the set is open, you must find a point in the set such that no matter what radius your ball has, it is never contained in the set. The previous answers wrote $(1,1,0)$ you should prove that what they wrote indeed shows that the set is not open.

share|cite|improve this answer

answered Jul 10 at 12:16

topologicalmagiciantopologicalmagician

90611 silver badge1111 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Theorem: Let x$in mathbbR^n$ where the topology induced by the standard metric is assumed. $xrightarrow a $ $iff$ $x^i rightarrow a^i$ for each $1leq ileq n$

Take the sequence $(frac1n,frac1n, 0)$

Theorem 2: Let X be a metric space. $A subset X$ is closed iff the limit of every convergent sequence in A is in A.

Observe that the components of the above sequence all converge to 0 and therefore the sequence converges to $(0,0,0)$ which is not in the set. To show that the set is open, you must find a point in the set such that no matter what radius your ball has, it is never contained in the set. The previous answers wrote $(1,1,0)$ you should prove that what they wrote indeed shows that the set is not open.

share|cite|improve this answer

answered Jul 10 at 12:16

topologicalmagiciantopologicalmagician

90611 silver badge1111 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 10 at 12:16

topologicalmagiciantopologicalmagician

90611 silver badge1111 bronze badges

answered Jul 10 at 12:16

topologicalmagiciantopologicalmagician

90611 silver badge1111 bronze badges

answered Jul 10 at 12:16

topologicalmagiciantopologicalmagician

90611 silver badge1111 bronze badges