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

33k33293




33k33293










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





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038875%2fif-a-subseteq-b-then-a-subseteq-b%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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

33k33293




33k33293












  • $begingroup$
    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.

7,3901730




7,3901730












  • $begingroup$
    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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038875%2fif-a-subseteq-b-then-a-subseteq-b%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more