Show that $C- (A ∪ B) = (C - A)cap (C - B)$ [closed]

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For sets $A, B, C$: $$C- (A ∪ B) = (C - A)cap (C - B).$$




I'm not sure how to formulate a correct proof for this statement. I assume that any element $x$ in set $C$ can't be in either $A$ or $B$, however I'm wrong about these things around 90% of the time.






elementary-set-theory







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closed as off-topic by Jendrik Stelzner, user21820, Arnaud D., Did, user99914 Sep 12 at 1:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, user21820, Did, Community
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Use De Morgan's laws: en.wikipedia.org/wiki/De_Morgan%27s_laws
    – Robert Z
    Sep 7 at 11:27










  • Possible duplicate of How to prove a set equality?
    – Arnaud D.
    Sep 11 at 12:14














up vote
0
down vote

favorite













For sets $A, B, C$: $$C- (A ∪ B) = (C - A)cap (C - B).$$




I'm not sure how to formulate a correct proof for this statement. I assume that any element $x$ in set $C$ can't be in either $A$ or $B$, however I'm wrong about these things around 90% of the time.






elementary-set-theory







share|cite|improve this question















closed as off-topic by Jendrik Stelzner, user21820, Arnaud D., Did, user99914 Sep 12 at 1:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, user21820, Did, Community
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Use De Morgan's laws: en.wikipedia.org/wiki/De_Morgan%27s_laws
    – Robert Z
    Sep 7 at 11:27










  • Possible duplicate of How to prove a set equality?
    – Arnaud D.
    Sep 11 at 12:14












up vote
0
down vote

favorite









up vote
0
down vote

favorite












For sets $A, B, C$: $$C- (A ∪ B) = (C - A)cap (C - B).$$




I'm not sure how to formulate a correct proof for this statement. I assume that any element $x$ in set $C$ can't be in either $A$ or $B$, however I'm wrong about these things around 90% of the time.






elementary-set-theory







share|cite|improve this question
















For sets $A, B, C$: $$C- (A ∪ B) = (C - A)cap (C - B).$$




I'm not sure how to formulate a correct proof for this statement. I assume that any element $x$ in set $C$ can't be in either $A$ or $B$, however I'm wrong about these things around 90% of the time.






elementary-set-theory




elementary-set-theory






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share|cite|improve this question













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edited Sep 7 at 11:30









Robert Z

85.8k1056124




85.8k1056124










asked Sep 7 at 11:25









Peter Celinski

423




423




closed as off-topic by Jendrik Stelzner, user21820, Arnaud D., Did, user99914 Sep 12 at 1:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, user21820, Did, Community
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Jendrik Stelzner, user21820, Arnaud D., Did, user99914 Sep 12 at 1:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, user21820, Did, Community
If this question can be reworded to fit the rules in the help center, please edit the question.











  • Use De Morgan's laws: en.wikipedia.org/wiki/De_Morgan%27s_laws
    – Robert Z
    Sep 7 at 11:27










  • Possible duplicate of How to prove a set equality?
    – Arnaud D.
    Sep 11 at 12:14
















  • Use De Morgan's laws: en.wikipedia.org/wiki/De_Morgan%27s_laws
    – Robert Z
    Sep 7 at 11:27










  • Possible duplicate of How to prove a set equality?
    – Arnaud D.
    Sep 11 at 12:14















Use De Morgan's laws: en.wikipedia.org/wiki/De_Morgan%27s_laws
– Robert Z
Sep 7 at 11:27




Use De Morgan's laws: en.wikipedia.org/wiki/De_Morgan%27s_laws
– Robert Z
Sep 7 at 11:27












Possible duplicate of How to prove a set equality?
– Arnaud D.
Sep 11 at 12:14




Possible duplicate of How to prove a set equality?
– Arnaud D.
Sep 11 at 12:14










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










Let $X$ be a set, $A,B,C$ be subsets of $X$.



$C-(A cup B) = C cap (A cup B)^c = $



$C cap (A^ccap B^c) =$



$ (Ccap A^c)cap (C cap B^c)=$



$(C-A)cap (C-B)$.



Used:



de Morgan: $(Acup B)^c= A^c cap B^c$;



$C-D= C cap D^c$, where $D=A,B$.






share|cite|improve this answer






















  • What do you mean with universal set? Surely, something like $Xsupseteq Acup Bcup C$ where $X$ really is a set, however I really dislike that terminology as universal set comes with this intuitive connection to the set of all sets(also called the universal set sometimes) with is (besides being not a set but a class) a topic to be carefully dealt with (like sethood in general).
    – zzuussee
    Sep 7 at 12:09










  • Ok, not universal .Superset.
    – Peter Szilas
    Sep 7 at 12:34










  • I mean, your line of reasoning is a very nice way to show the equality just through set algebra (+1), this phrase just makes me nervous whenever I read it.
    – zzuussee
    Sep 7 at 12:42






  • 1




    This of course assumes that you have already proved the relations with complement. And the complement is probably defined via set difference …
    – celtschk
    Sep 7 at 13:08






  • 1




    @zzuussee: Note that only certain foundational systems have a problem with the universal set. Others do not, so it could be misleading to say that the universal set is actually not a set but a class. And in ZFC one has to go a convoluted route, namely to let $X = bigcup bigcup A,B,C$, which is just an artifact of ZFC.
    – user21820
    Sep 10 at 5:45


















up vote
2
down vote













You could approach this via double inclusion, i.e. showing that



$$C-(Acup B)subseteq(C-A)cap (C-B)text and (C-A)cap (C-B)subseteq C-(Acup B)$$



I'll just show the former one and leave the rest to you:



Let $xin C-(Acup B)$, i.e. $xin C$ and $xnotin(Acup B)$, i.e. $xnotin A$ and $xnotin B$. Thus, by the first, $xin C-A$, by the second, $xin C-B$. Thus $xin (C-A)cap (C-B)$






share|cite|improve this answer




















  • Thanks a bunch, this cleared it right up
    – Peter Celinski
    Sep 7 at 11:51










  • @PeterCelinski You're very welcome.
    – zzuussee
    Sep 7 at 11:51

















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Let $X$ be a set, $A,B,C$ be subsets of $X$.



$C-(A cup B) = C cap (A cup B)^c = $



$C cap (A^ccap B^c) =$



$ (Ccap A^c)cap (C cap B^c)=$



$(C-A)cap (C-B)$.



Used:



de Morgan: $(Acup B)^c= A^c cap B^c$;



$C-D= C cap D^c$, where $D=A,B$.






share|cite|improve this answer






















  • What do you mean with universal set? Surely, something like $Xsupseteq Acup Bcup C$ where $X$ really is a set, however I really dislike that terminology as universal set comes with this intuitive connection to the set of all sets(also called the universal set sometimes) with is (besides being not a set but a class) a topic to be carefully dealt with (like sethood in general).
    – zzuussee
    Sep 7 at 12:09










  • Ok, not universal .Superset.
    – Peter Szilas
    Sep 7 at 12:34










  • I mean, your line of reasoning is a very nice way to show the equality just through set algebra (+1), this phrase just makes me nervous whenever I read it.
    – zzuussee
    Sep 7 at 12:42






  • 1




    This of course assumes that you have already proved the relations with complement. And the complement is probably defined via set difference …
    – celtschk
    Sep 7 at 13:08






  • 1




    @zzuussee: Note that only certain foundational systems have a problem with the universal set. Others do not, so it could be misleading to say that the universal set is actually not a set but a class. And in ZFC one has to go a convoluted route, namely to let $X = bigcup bigcup A,B,C$, which is just an artifact of ZFC.
    – user21820
    Sep 10 at 5:45















up vote
2
down vote



accepted










Let $X$ be a set, $A,B,C$ be subsets of $X$.



$C-(A cup B) = C cap (A cup B)^c = $



$C cap (A^ccap B^c) =$



$ (Ccap A^c)cap (C cap B^c)=$



$(C-A)cap (C-B)$.



Used:



de Morgan: $(Acup B)^c= A^c cap B^c$;



$C-D= C cap D^c$, where $D=A,B$.






share|cite|improve this answer






















  • What do you mean with universal set? Surely, something like $Xsupseteq Acup Bcup C$ where $X$ really is a set, however I really dislike that terminology as universal set comes with this intuitive connection to the set of all sets(also called the universal set sometimes) with is (besides being not a set but a class) a topic to be carefully dealt with (like sethood in general).
    – zzuussee
    Sep 7 at 12:09










  • Ok, not universal .Superset.
    – Peter Szilas
    Sep 7 at 12:34










  • I mean, your line of reasoning is a very nice way to show the equality just through set algebra (+1), this phrase just makes me nervous whenever I read it.
    – zzuussee
    Sep 7 at 12:42






  • 1




    This of course assumes that you have already proved the relations with complement. And the complement is probably defined via set difference …
    – celtschk
    Sep 7 at 13:08






  • 1




    @zzuussee: Note that only certain foundational systems have a problem with the universal set. Others do not, so it could be misleading to say that the universal set is actually not a set but a class. And in ZFC one has to go a convoluted route, namely to let $X = bigcup bigcup A,B,C$, which is just an artifact of ZFC.
    – user21820
    Sep 10 at 5:45













up vote
2
down vote



accepted







up vote
2
down vote



accepted






Let $X$ be a set, $A,B,C$ be subsets of $X$.



$C-(A cup B) = C cap (A cup B)^c = $



$C cap (A^ccap B^c) =$



$ (Ccap A^c)cap (C cap B^c)=$



$(C-A)cap (C-B)$.



Used:



de Morgan: $(Acup B)^c= A^c cap B^c$;



$C-D= C cap D^c$, where $D=A,B$.






share|cite|improve this answer














Let $X$ be a set, $A,B,C$ be subsets of $X$.



$C-(A cup B) = C cap (A cup B)^c = $



$C cap (A^ccap B^c) =$



$ (Ccap A^c)cap (C cap B^c)=$



$(C-A)cap (C-B)$.



Used:



de Morgan: $(Acup B)^c= A^c cap B^c$;



$C-D= C cap D^c$, where $D=A,B$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 7 at 12:35

























answered Sep 7 at 11:58









Peter Szilas

8,4252617




8,4252617











  • What do you mean with universal set? Surely, something like $Xsupseteq Acup Bcup C$ where $X$ really is a set, however I really dislike that terminology as universal set comes with this intuitive connection to the set of all sets(also called the universal set sometimes) with is (besides being not a set but a class) a topic to be carefully dealt with (like sethood in general).
    – zzuussee
    Sep 7 at 12:09










  • Ok, not universal .Superset.
    – Peter Szilas
    Sep 7 at 12:34










  • I mean, your line of reasoning is a very nice way to show the equality just through set algebra (+1), this phrase just makes me nervous whenever I read it.
    – zzuussee
    Sep 7 at 12:42






  • 1




    This of course assumes that you have already proved the relations with complement. And the complement is probably defined via set difference …
    – celtschk
    Sep 7 at 13:08






  • 1




    @zzuussee: Note that only certain foundational systems have a problem with the universal set. Others do not, so it could be misleading to say that the universal set is actually not a set but a class. And in ZFC one has to go a convoluted route, namely to let $X = bigcup bigcup A,B,C$, which is just an artifact of ZFC.
    – user21820
    Sep 10 at 5:45

















  • What do you mean with universal set? Surely, something like $Xsupseteq Acup Bcup C$ where $X$ really is a set, however I really dislike that terminology as universal set comes with this intuitive connection to the set of all sets(also called the universal set sometimes) with is (besides being not a set but a class) a topic to be carefully dealt with (like sethood in general).
    – zzuussee
    Sep 7 at 12:09










  • Ok, not universal .Superset.
    – Peter Szilas
    Sep 7 at 12:34










  • I mean, your line of reasoning is a very nice way to show the equality just through set algebra (+1), this phrase just makes me nervous whenever I read it.
    – zzuussee
    Sep 7 at 12:42






  • 1




    This of course assumes that you have already proved the relations with complement. And the complement is probably defined via set difference …
    – celtschk
    Sep 7 at 13:08






  • 1




    @zzuussee: Note that only certain foundational systems have a problem with the universal set. Others do not, so it could be misleading to say that the universal set is actually not a set but a class. And in ZFC one has to go a convoluted route, namely to let $X = bigcup bigcup A,B,C$, which is just an artifact of ZFC.
    – user21820
    Sep 10 at 5:45
















What do you mean with universal set? Surely, something like $Xsupseteq Acup Bcup C$ where $X$ really is a set, however I really dislike that terminology as universal set comes with this intuitive connection to the set of all sets(also called the universal set sometimes) with is (besides being not a set but a class) a topic to be carefully dealt with (like sethood in general).
– zzuussee
Sep 7 at 12:09




What do you mean with universal set? Surely, something like $Xsupseteq Acup Bcup C$ where $X$ really is a set, however I really dislike that terminology as universal set comes with this intuitive connection to the set of all sets(also called the universal set sometimes) with is (besides being not a set but a class) a topic to be carefully dealt with (like sethood in general).
– zzuussee
Sep 7 at 12:09












Ok, not universal .Superset.
– Peter Szilas
Sep 7 at 12:34




Ok, not universal .Superset.
– Peter Szilas
Sep 7 at 12:34












I mean, your line of reasoning is a very nice way to show the equality just through set algebra (+1), this phrase just makes me nervous whenever I read it.
– zzuussee
Sep 7 at 12:42




I mean, your line of reasoning is a very nice way to show the equality just through set algebra (+1), this phrase just makes me nervous whenever I read it.
– zzuussee
Sep 7 at 12:42




1




1




This of course assumes that you have already proved the relations with complement. And the complement is probably defined via set difference …
– celtschk
Sep 7 at 13:08




This of course assumes that you have already proved the relations with complement. And the complement is probably defined via set difference …
– celtschk
Sep 7 at 13:08




1




1




@zzuussee: Note that only certain foundational systems have a problem with the universal set. Others do not, so it could be misleading to say that the universal set is actually not a set but a class. And in ZFC one has to go a convoluted route, namely to let $X = bigcup bigcup A,B,C$, which is just an artifact of ZFC.
– user21820
Sep 10 at 5:45





@zzuussee: Note that only certain foundational systems have a problem with the universal set. Others do not, so it could be misleading to say that the universal set is actually not a set but a class. And in ZFC one has to go a convoluted route, namely to let $X = bigcup bigcup A,B,C$, which is just an artifact of ZFC.
– user21820
Sep 10 at 5:45











up vote
2
down vote













You could approach this via double inclusion, i.e. showing that



$$C-(Acup B)subseteq(C-A)cap (C-B)text and (C-A)cap (C-B)subseteq C-(Acup B)$$



I'll just show the former one and leave the rest to you:



Let $xin C-(Acup B)$, i.e. $xin C$ and $xnotin(Acup B)$, i.e. $xnotin A$ and $xnotin B$. Thus, by the first, $xin C-A$, by the second, $xin C-B$. Thus $xin (C-A)cap (C-B)$






share|cite|improve this answer




















  • Thanks a bunch, this cleared it right up
    – Peter Celinski
    Sep 7 at 11:51










  • @PeterCelinski You're very welcome.
    – zzuussee
    Sep 7 at 11:51














up vote
2
down vote













You could approach this via double inclusion, i.e. showing that



$$C-(Acup B)subseteq(C-A)cap (C-B)text and (C-A)cap (C-B)subseteq C-(Acup B)$$



I'll just show the former one and leave the rest to you:



Let $xin C-(Acup B)$, i.e. $xin C$ and $xnotin(Acup B)$, i.e. $xnotin A$ and $xnotin B$. Thus, by the first, $xin C-A$, by the second, $xin C-B$. Thus $xin (C-A)cap (C-B)$






share|cite|improve this answer




















  • Thanks a bunch, this cleared it right up
    – Peter Celinski
    Sep 7 at 11:51










  • @PeterCelinski You're very welcome.
    – zzuussee
    Sep 7 at 11:51












up vote
2
down vote










up vote
2
down vote









You could approach this via double inclusion, i.e. showing that



$$C-(Acup B)subseteq(C-A)cap (C-B)text and (C-A)cap (C-B)subseteq C-(Acup B)$$



I'll just show the former one and leave the rest to you:



Let $xin C-(Acup B)$, i.e. $xin C$ and $xnotin(Acup B)$, i.e. $xnotin A$ and $xnotin B$. Thus, by the first, $xin C-A$, by the second, $xin C-B$. Thus $xin (C-A)cap (C-B)$






share|cite|improve this answer












You could approach this via double inclusion, i.e. showing that



$$C-(Acup B)subseteq(C-A)cap (C-B)text and (C-A)cap (C-B)subseteq C-(Acup B)$$



I'll just show the former one and leave the rest to you:



Let $xin C-(Acup B)$, i.e. $xin C$ and $xnotin(Acup B)$, i.e. $xnotin A$ and $xnotin B$. Thus, by the first, $xin C-A$, by the second, $xin C-B$. Thus $xin (C-A)cap (C-B)$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 7 at 11:35









zzuussee

2,446725




2,446725











  • Thanks a bunch, this cleared it right up
    – Peter Celinski
    Sep 7 at 11:51










  • @PeterCelinski You're very welcome.
    – zzuussee
    Sep 7 at 11:51
















  • Thanks a bunch, this cleared it right up
    – Peter Celinski
    Sep 7 at 11:51










  • @PeterCelinski You're very welcome.
    – zzuussee
    Sep 7 at 11:51















Thanks a bunch, this cleared it right up
– Peter Celinski
Sep 7 at 11:51




Thanks a bunch, this cleared it right up
– Peter Celinski
Sep 7 at 11:51












@PeterCelinski You're very welcome.
– zzuussee
Sep 7 at 11:51




@PeterCelinski You're very welcome.
– zzuussee
Sep 7 at 11:51


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