If $Asubseteq B$, then $A'subseteq B'$












0












$begingroup$


Proof or counterexample: If $Asubseteq B$, then $A'subseteq B'$.
I have no idea where to start. Only thing I know is the definition of limit points.



$A'$ is the set of all limit points of $A$.






general-topology examples-counterexamples







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What is the meaning of $A'$?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:01










  • $begingroup$
    $A'$ is the set of all limit points of $A$.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:06










  • $begingroup$
    Have you tried the standard pattern of showing inclusions? By that I mean: Start with „Let $x$ be in $A^prime$. This means X. We want to show that $x$ is in $B^prime$. We know that Y holds.“ and see how you can connect the argument by using the definitions.
    $endgroup$
    – Luke
    Dec 14 '18 at 3:07












  • $begingroup$
    The thing is I have a feeling that this is a false statement, but I cannot find counterexample.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:11
















0












$begingroup$


Proof or counterexample: If $Asubseteq B$, then $A'subseteq B'$.
I have no idea where to start. Only thing I know is the definition of limit points.



$A'$ is the set of all limit points of $A$.






general-topology examples-counterexamples







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What is the meaning of $A'$?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:01










  • $begingroup$
    $A'$ is the set of all limit points of $A$.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:06










  • $begingroup$
    Have you tried the standard pattern of showing inclusions? By that I mean: Start with „Let $x$ be in $A^prime$. This means X. We want to show that $x$ is in $B^prime$. We know that Y holds.“ and see how you can connect the argument by using the definitions.
    $endgroup$
    – Luke
    Dec 14 '18 at 3:07












  • $begingroup$
    The thing is I have a feeling that this is a false statement, but I cannot find counterexample.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:11














0












0








0





$begingroup$


Proof or counterexample: If $Asubseteq B$, then $A'subseteq B'$.
I have no idea where to start. Only thing I know is the definition of limit points.



$A'$ is the set of all limit points of $A$.






general-topology examples-counterexamples







share|cite|improve this question











$endgroup$




Proof or counterexample: If $Asubseteq B$, then $A'subseteq B'$.
I have no idea where to start. Only thing I know is the definition of limit points.



$A'$ is the set of all limit points of $A$.






general-topology examples-counterexamples




general-topology examples-counterexamples






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 3:16









Batominovski

33.1k33293




33.1k33293










asked Dec 14 '18 at 2:58









Adam YoungAdam Young

64




64








  • 2




    $begingroup$
    What is the meaning of $A'$?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:01










  • $begingroup$
    $A'$ is the set of all limit points of $A$.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:06










  • $begingroup$
    Have you tried the standard pattern of showing inclusions? By that I mean: Start with „Let $x$ be in $A^prime$. This means X. We want to show that $x$ is in $B^prime$. We know that Y holds.“ and see how you can connect the argument by using the definitions.
    $endgroup$
    – Luke
    Dec 14 '18 at 3:07












  • $begingroup$
    The thing is I have a feeling that this is a false statement, but I cannot find counterexample.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:11














  • 2




    $begingroup$
    What is the meaning of $A'$?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:01










  • $begingroup$
    $A'$ is the set of all limit points of $A$.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:06










  • $begingroup$
    Have you tried the standard pattern of showing inclusions? By that I mean: Start with „Let $x$ be in $A^prime$. This means X. We want to show that $x$ is in $B^prime$. We know that Y holds.“ and see how you can connect the argument by using the definitions.
    $endgroup$
    – Luke
    Dec 14 '18 at 3:07












  • $begingroup$
    The thing is I have a feeling that this is a false statement, but I cannot find counterexample.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:11








2




2




$begingroup$
What is the meaning of $A'$?
$endgroup$
– Batominovski
Dec 14 '18 at 3:01




$begingroup$
What is the meaning of $A'$?
$endgroup$
– Batominovski
Dec 14 '18 at 3:01












$begingroup$
$A'$ is the set of all limit points of $A$.
$endgroup$
– Adam Young
Dec 14 '18 at 3:06




$begingroup$
$A'$ is the set of all limit points of $A$.
$endgroup$
– Adam Young
Dec 14 '18 at 3:06












$begingroup$
Have you tried the standard pattern of showing inclusions? By that I mean: Start with „Let $x$ be in $A^prime$. This means X. We want to show that $x$ is in $B^prime$. We know that Y holds.“ and see how you can connect the argument by using the definitions.
$endgroup$
– Luke
Dec 14 '18 at 3:07






$begingroup$
Have you tried the standard pattern of showing inclusions? By that I mean: Start with „Let $x$ be in $A^prime$. This means X. We want to show that $x$ is in $B^prime$. We know that Y holds.“ and see how you can connect the argument by using the definitions.
$endgroup$
– Luke
Dec 14 '18 at 3:07














$begingroup$
The thing is I have a feeling that this is a false statement, but I cannot find counterexample.
$endgroup$
– Adam Young
Dec 14 '18 at 3:11




$begingroup$
The thing is I have a feeling that this is a false statement, but I cannot find counterexample.
$endgroup$
– Adam Young
Dec 14 '18 at 3:11










2 Answers
2






active

oldest

votes


















0












$begingroup$

If $xin A'$, then for every open neighborhood $U$ of $x$, there exists point $y_U in U$ such that $y_Uin Asetminus{x}$. As $Asubseteq B$, $y_Uin B$. What conclusion can you make?




Therefore, for every open neighborhood $U$ of $x$, the point $y_Uin U$ is a point distinct from $x$ that is also in $B$. By the definition of limit points, $x$ must be a limit point of $B$, whence $A'subseteq B'$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    I understand what you write here, but I do not see any good for this.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:43










  • $begingroup$
    Can you guess at least whether I'm proving the positive, or providing a counterexample?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:46










  • $begingroup$
    a counterexample?
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:03










  • $begingroup$
    This indicates that you don't really understand the concept of limit points (and probably methods of proof as well). I am proving the positive. That is, I want to show $A'subseteq B'$. If I am to prove this, what must I do?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 4:05












  • $begingroup$
    you need to show a arbitrary point that in $A'$ is also in $B'$
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:15



















1












$begingroup$

Hint: argue by the contrapositive: that is, show that if $x not in B'$, then $x not in A'$.




Take $x not in B'$. Therefore, there is an open neighbourhood $U$ of $x$ with $U setminus {x} cap B = emptyset$. Hence $U setminus {x} cap A subset U setminus {x} cap B = emptyset$, and so $U setminus {x} cap A = emptyset$, or equivalently $x not in A'$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you, I have already understand this concept.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:43











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

If $xin A'$, then for every open neighborhood $U$ of $x$, there exists point $y_U in U$ such that $y_Uin Asetminus{x}$. As $Asubseteq B$, $y_Uin B$. What conclusion can you make?




Therefore, for every open neighborhood $U$ of $x$, the point $y_Uin U$ is a point distinct from $x$ that is also in $B$. By the definition of limit points, $x$ must be a limit point of $B$, whence $A'subseteq B'$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    I understand what you write here, but I do not see any good for this.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:43










  • $begingroup$
    Can you guess at least whether I'm proving the positive, or providing a counterexample?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:46










  • $begingroup$
    a counterexample?
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:03










  • $begingroup$
    This indicates that you don't really understand the concept of limit points (and probably methods of proof as well). I am proving the positive. That is, I want to show $A'subseteq B'$. If I am to prove this, what must I do?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 4:05












  • $begingroup$
    you need to show a arbitrary point that in $A'$ is also in $B'$
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:15
















0












$begingroup$

If $xin A'$, then for every open neighborhood $U$ of $x$, there exists point $y_U in U$ such that $y_Uin Asetminus{x}$. As $Asubseteq B$, $y_Uin B$. What conclusion can you make?




Therefore, for every open neighborhood $U$ of $x$, the point $y_Uin U$ is a point distinct from $x$ that is also in $B$. By the definition of limit points, $x$ must be a limit point of $B$, whence $A'subseteq B'$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    I understand what you write here, but I do not see any good for this.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:43










  • $begingroup$
    Can you guess at least whether I'm proving the positive, or providing a counterexample?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:46










  • $begingroup$
    a counterexample?
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:03










  • $begingroup$
    This indicates that you don't really understand the concept of limit points (and probably methods of proof as well). I am proving the positive. That is, I want to show $A'subseteq B'$. If I am to prove this, what must I do?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 4:05












  • $begingroup$
    you need to show a arbitrary point that in $A'$ is also in $B'$
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:15














0












0








0





$begingroup$

If $xin A'$, then for every open neighborhood $U$ of $x$, there exists point $y_U in U$ such that $y_Uin Asetminus{x}$. As $Asubseteq B$, $y_Uin B$. What conclusion can you make?




Therefore, for every open neighborhood $U$ of $x$, the point $y_Uin U$ is a point distinct from $x$ that is also in $B$. By the definition of limit points, $x$ must be a limit point of $B$, whence $A'subseteq B'$.







share|cite|improve this answer











$endgroup$



If $xin A'$, then for every open neighborhood $U$ of $x$, there exists point $y_U in U$ such that $y_Uin Asetminus{x}$. As $Asubseteq B$, $y_Uin B$. What conclusion can you make?




Therefore, for every open neighborhood $U$ of $x$, the point $y_Uin U$ is a point distinct from $x$ that is also in $B$. By the definition of limit points, $x$ must be a limit point of $B$, whence $A'subseteq B'$.








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 4:27

























answered Dec 14 '18 at 3:23









BatominovskiBatominovski

33.1k33293




33.1k33293












  • $begingroup$
    I understand what you write here, but I do not see any good for this.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:43










  • $begingroup$
    Can you guess at least whether I'm proving the positive, or providing a counterexample?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:46










  • $begingroup$
    a counterexample?
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:03










  • $begingroup$
    This indicates that you don't really understand the concept of limit points (and probably methods of proof as well). I am proving the positive. That is, I want to show $A'subseteq B'$. If I am to prove this, what must I do?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 4:05












  • $begingroup$
    you need to show a arbitrary point that in $A'$ is also in $B'$
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:15


















  • $begingroup$
    I understand what you write here, but I do not see any good for this.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 3:43










  • $begingroup$
    Can you guess at least whether I'm proving the positive, or providing a counterexample?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:46










  • $begingroup$
    a counterexample?
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:03










  • $begingroup$
    This indicates that you don't really understand the concept of limit points (and probably methods of proof as well). I am proving the positive. That is, I want to show $A'subseteq B'$. If I am to prove this, what must I do?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 4:05












  • $begingroup$
    you need to show a arbitrary point that in $A'$ is also in $B'$
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:15
















$begingroup$
I understand what you write here, but I do not see any good for this.
$endgroup$
– Adam Young
Dec 14 '18 at 3:43




$begingroup$
I understand what you write here, but I do not see any good for this.
$endgroup$
– Adam Young
Dec 14 '18 at 3:43












$begingroup$
Can you guess at least whether I'm proving the positive, or providing a counterexample?
$endgroup$
– Batominovski
Dec 14 '18 at 3:46




$begingroup$
Can you guess at least whether I'm proving the positive, or providing a counterexample?
$endgroup$
– Batominovski
Dec 14 '18 at 3:46












$begingroup$
a counterexample?
$endgroup$
– Adam Young
Dec 14 '18 at 4:03




$begingroup$
a counterexample?
$endgroup$
– Adam Young
Dec 14 '18 at 4:03












$begingroup$
This indicates that you don't really understand the concept of limit points (and probably methods of proof as well). I am proving the positive. That is, I want to show $A'subseteq B'$. If I am to prove this, what must I do?
$endgroup$
– Batominovski
Dec 14 '18 at 4:05






$begingroup$
This indicates that you don't really understand the concept of limit points (and probably methods of proof as well). I am proving the positive. That is, I want to show $A'subseteq B'$. If I am to prove this, what must I do?
$endgroup$
– Batominovski
Dec 14 '18 at 4:05














$begingroup$
you need to show a arbitrary point that in $A'$ is also in $B'$
$endgroup$
– Adam Young
Dec 14 '18 at 4:15




$begingroup$
you need to show a arbitrary point that in $A'$ is also in $B'$
$endgroup$
– Adam Young
Dec 14 '18 at 4:15











1












$begingroup$

Hint: argue by the contrapositive: that is, show that if $x not in B'$, then $x not in A'$.




Take $x not in B'$. Therefore, there is an open neighbourhood $U$ of $x$ with $U setminus {x} cap B = emptyset$. Hence $U setminus {x} cap A subset U setminus {x} cap B = emptyset$, and so $U setminus {x} cap A = emptyset$, or equivalently $x not in A'$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you, I have already understand this concept.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:43
















1












$begingroup$

Hint: argue by the contrapositive: that is, show that if $x not in B'$, then $x not in A'$.




Take $x not in B'$. Therefore, there is an open neighbourhood $U$ of $x$ with $U setminus {x} cap B = emptyset$. Hence $U setminus {x} cap A subset U setminus {x} cap B = emptyset$, and so $U setminus {x} cap A = emptyset$, or equivalently $x not in A'$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you, I have already understand this concept.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:43














1












1








1





$begingroup$

Hint: argue by the contrapositive: that is, show that if $x not in B'$, then $x not in A'$.




Take $x not in B'$. Therefore, there is an open neighbourhood $U$ of $x$ with $U setminus {x} cap B = emptyset$. Hence $U setminus {x} cap A subset U setminus {x} cap B = emptyset$, and so $U setminus {x} cap A = emptyset$, or equivalently $x not in A'$.







share|cite|improve this answer











$endgroup$



Hint: argue by the contrapositive: that is, show that if $x not in B'$, then $x not in A'$.




Take $x not in B'$. Therefore, there is an open neighbourhood $U$ of $x$ with $U setminus {x} cap B = emptyset$. Hence $U setminus {x} cap A subset U setminus {x} cap B = emptyset$, and so $U setminus {x} cap A = emptyset$, or equivalently $x not in A'$.








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 5:03

























answered Dec 14 '18 at 4:35









Guido A.Guido A.

7,7231730




7,7231730












  • $begingroup$
    thank you, I have already understand this concept.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:43


















  • $begingroup$
    thank you, I have already understand this concept.
    $endgroup$
    – Adam Young
    Dec 14 '18 at 4:43
















$begingroup$
thank you, I have already understand this concept.
$endgroup$
– Adam Young
Dec 14 '18 at 4:43




$begingroup$
thank you, I have already understand this concept.
$endgroup$
– Adam Young
Dec 14 '18 at 4:43


















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