Prove $A cap B subseteq A $Prove: If $A subseteq C$ and $B subseteq D$, then $A cap B subseteq C cap D$Let $A, B, C$ be sets. Prove that if $B subseteq C$, then $(Acap B)subseteq (Acap C)$Prove $A subseteq (B cap C) iff (A subseteq B)$ and $(A subseteq C)$Please check this proof: $A cap B subseteq C wedge A^c cap B subseteq C Rightarrow B subseteq C$Prove that $ (A cup B) cap C subseteq A cup (B cap C)$Let $ADelta Csubseteq ADelta B$. Prove $Acap B subseteq C$. (Proof.v)Prove that $inf(A, B) = A cap B$having some trouble with proof that if $A subseteq B$ then $(A cap C) subseteq (B cap C)$Prove that if $ Acup B subseteq C cup D, A cap B =$ ∅ $land C subseteq A implies B subseteq D.$Suppose $A cap C subseteq B cap C$ and $A cup Csubseteq B cup C$. Prove that $A subseteq B$

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Prove $A cap B subseteq A $


Prove: If $A subseteq C$ and $B subseteq D$, then $A cap B subseteq C cap D$Let $A, B, C$ be sets. Prove that if $B subseteq C$, then $(Acap B)subseteq (Acap C)$Prove $A subseteq (B cap C) iff (A subseteq B)$ and $(A subseteq C)$Please check this proof: $A cap B subseteq C wedge A^c cap B subseteq C Rightarrow B subseteq C$Prove that $ (A cup B) cap C subseteq A cup (B cap C)$Let $ADelta Csubseteq ADelta B$. Prove $Acap B subseteq C$. (Proof.v)Prove that $inf(A, B) = A cap B$having some trouble with proof that if $A subseteq B$ then $(A cap C) subseteq (B cap C)$Prove that if $ Acup B subseteq C cup D, A cap B =$ ∅ $land C subseteq A implies B subseteq D.$Suppose $A cap C subseteq B cap C$ and $A cup Csubseteq B cup C$. Prove that $A subseteq B$













0












$begingroup$


I am learning from Apostle's Calculus Vol.1 and could really do with some verification.



Question 12(sect. 12.5): prove $A cap B subseteq A $



Proof: Let $x in A cap B$ if $x in A$ and $x in B$. Thus, $A cap B subseteq A $ if $A subseteq B$ or $A subset B$










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    I'm lost after your "Thus".
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 10:26










  • $begingroup$
    More simply : $x in A cap B text iff x in A text and x in B$. Thus, $(x in A cap B) to (x in B)$ and this is simply the def of $subseteq$.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 16 at 10:26
















0












$begingroup$


I am learning from Apostle's Calculus Vol.1 and could really do with some verification.



Question 12(sect. 12.5): prove $A cap B subseteq A $



Proof: Let $x in A cap B$ if $x in A$ and $x in B$. Thus, $A cap B subseteq A $ if $A subseteq B$ or $A subset B$










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    I'm lost after your "Thus".
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 10:26










  • $begingroup$
    More simply : $x in A cap B text iff x in A text and x in B$. Thus, $(x in A cap B) to (x in B)$ and this is simply the def of $subseteq$.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 16 at 10:26














0












0








0





$begingroup$


I am learning from Apostle's Calculus Vol.1 and could really do with some verification.



Question 12(sect. 12.5): prove $A cap B subseteq A $



Proof: Let $x in A cap B$ if $x in A$ and $x in B$. Thus, $A cap B subseteq A $ if $A subseteq B$ or $A subset B$










share|cite|improve this question











$endgroup$




I am learning from Apostle's Calculus Vol.1 and could really do with some verification.



Question 12(sect. 12.5): prove $A cap B subseteq A $



Proof: Let $x in A cap B$ if $x in A$ and $x in B$. Thus, $A cap B subseteq A $ if $A subseteq B$ or $A subset B$







proof-verification elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 16 at 11:16









José Carlos Santos

170k23132238




170k23132238










asked Mar 16 at 10:17









seeppseepp

132




132







  • 1




    $begingroup$
    I'm lost after your "Thus".
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 10:26










  • $begingroup$
    More simply : $x in A cap B text iff x in A text and x in B$. Thus, $(x in A cap B) to (x in B)$ and this is simply the def of $subseteq$.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 16 at 10:26













  • 1




    $begingroup$
    I'm lost after your "Thus".
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 10:26










  • $begingroup$
    More simply : $x in A cap B text iff x in A text and x in B$. Thus, $(x in A cap B) to (x in B)$ and this is simply the def of $subseteq$.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 16 at 10:26








1




1




$begingroup$
I'm lost after your "Thus".
$endgroup$
– Lord Shark the Unknown
Mar 16 at 10:26




$begingroup$
I'm lost after your "Thus".
$endgroup$
– Lord Shark the Unknown
Mar 16 at 10:26












$begingroup$
More simply : $x in A cap B text iff x in A text and x in B$. Thus, $(x in A cap B) to (x in B)$ and this is simply the def of $subseteq$.
$endgroup$
– Mauro ALLEGRANZA
Mar 16 at 10:26





$begingroup$
More simply : $x in A cap B text iff x in A text and x in B$. Thus, $(x in A cap B) to (x in B)$ and this is simply the def of $subseteq$.
$endgroup$
– Mauro ALLEGRANZA
Mar 16 at 10:26











1 Answer
1






active

oldest

votes


















2












$begingroup$

Your first sentence introduces an element $x$, which is not mentioned in the second one. It doesn't make sense.



Simply say that if $xin Acap B$, then, by the definition of $Acap B$, $xin A$. Since thus occurs for each $xin A$, $Acap Bsubseteq A$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Reading the question I get the impression that $subset$ and $subseteq$ might be distinct notions for the OP. So to avoid confusion I think it is better to write $Acap Bsubseteq A$.
    $endgroup$
    – drhab
    Mar 16 at 10:46










  • $begingroup$
    @drhab Your are right.
    $endgroup$
    – seepp
    Mar 16 at 10:52










  • $begingroup$
    Thank you for your answer, it has clarified some things for me.
    $endgroup$
    – seepp
    Mar 16 at 10:56










  • $begingroup$
    It should be "Since this occurs for each $xin Acap B$".
    $endgroup$
    – Henning Makholm
    Mar 16 at 11:00










Your Answer





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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

Your first sentence introduces an element $x$, which is not mentioned in the second one. It doesn't make sense.



Simply say that if $xin Acap B$, then, by the definition of $Acap B$, $xin A$. Since thus occurs for each $xin A$, $Acap Bsubseteq A$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Reading the question I get the impression that $subset$ and $subseteq$ might be distinct notions for the OP. So to avoid confusion I think it is better to write $Acap Bsubseteq A$.
    $endgroup$
    – drhab
    Mar 16 at 10:46










  • $begingroup$
    @drhab Your are right.
    $endgroup$
    – seepp
    Mar 16 at 10:52










  • $begingroup$
    Thank you for your answer, it has clarified some things for me.
    $endgroup$
    – seepp
    Mar 16 at 10:56










  • $begingroup$
    It should be "Since this occurs for each $xin Acap B$".
    $endgroup$
    – Henning Makholm
    Mar 16 at 11:00















2












$begingroup$

Your first sentence introduces an element $x$, which is not mentioned in the second one. It doesn't make sense.



Simply say that if $xin Acap B$, then, by the definition of $Acap B$, $xin A$. Since thus occurs for each $xin A$, $Acap Bsubseteq A$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Reading the question I get the impression that $subset$ and $subseteq$ might be distinct notions for the OP. So to avoid confusion I think it is better to write $Acap Bsubseteq A$.
    $endgroup$
    – drhab
    Mar 16 at 10:46










  • $begingroup$
    @drhab Your are right.
    $endgroup$
    – seepp
    Mar 16 at 10:52










  • $begingroup$
    Thank you for your answer, it has clarified some things for me.
    $endgroup$
    – seepp
    Mar 16 at 10:56










  • $begingroup$
    It should be "Since this occurs for each $xin Acap B$".
    $endgroup$
    – Henning Makholm
    Mar 16 at 11:00













2












2








2





$begingroup$

Your first sentence introduces an element $x$, which is not mentioned in the second one. It doesn't make sense.



Simply say that if $xin Acap B$, then, by the definition of $Acap B$, $xin A$. Since thus occurs for each $xin A$, $Acap Bsubseteq A$.






share|cite|improve this answer











$endgroup$



Your first sentence introduces an element $x$, which is not mentioned in the second one. It doesn't make sense.



Simply say that if $xin Acap B$, then, by the definition of $Acap B$, $xin A$. Since thus occurs for each $xin A$, $Acap Bsubseteq A$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 16 at 10:52

























answered Mar 16 at 10:31









José Carlos SantosJosé Carlos Santos

170k23132238




170k23132238











  • $begingroup$
    Reading the question I get the impression that $subset$ and $subseteq$ might be distinct notions for the OP. So to avoid confusion I think it is better to write $Acap Bsubseteq A$.
    $endgroup$
    – drhab
    Mar 16 at 10:46










  • $begingroup$
    @drhab Your are right.
    $endgroup$
    – seepp
    Mar 16 at 10:52










  • $begingroup$
    Thank you for your answer, it has clarified some things for me.
    $endgroup$
    – seepp
    Mar 16 at 10:56










  • $begingroup$
    It should be "Since this occurs for each $xin Acap B$".
    $endgroup$
    – Henning Makholm
    Mar 16 at 11:00
















  • $begingroup$
    Reading the question I get the impression that $subset$ and $subseteq$ might be distinct notions for the OP. So to avoid confusion I think it is better to write $Acap Bsubseteq A$.
    $endgroup$
    – drhab
    Mar 16 at 10:46










  • $begingroup$
    @drhab Your are right.
    $endgroup$
    – seepp
    Mar 16 at 10:52










  • $begingroup$
    Thank you for your answer, it has clarified some things for me.
    $endgroup$
    – seepp
    Mar 16 at 10:56










  • $begingroup$
    It should be "Since this occurs for each $xin Acap B$".
    $endgroup$
    – Henning Makholm
    Mar 16 at 11:00















$begingroup$
Reading the question I get the impression that $subset$ and $subseteq$ might be distinct notions for the OP. So to avoid confusion I think it is better to write $Acap Bsubseteq A$.
$endgroup$
– drhab
Mar 16 at 10:46




$begingroup$
Reading the question I get the impression that $subset$ and $subseteq$ might be distinct notions for the OP. So to avoid confusion I think it is better to write $Acap Bsubseteq A$.
$endgroup$
– drhab
Mar 16 at 10:46












$begingroup$
@drhab Your are right.
$endgroup$
– seepp
Mar 16 at 10:52




$begingroup$
@drhab Your are right.
$endgroup$
– seepp
Mar 16 at 10:52












$begingroup$
Thank you for your answer, it has clarified some things for me.
$endgroup$
– seepp
Mar 16 at 10:56




$begingroup$
Thank you for your answer, it has clarified some things for me.
$endgroup$
– seepp
Mar 16 at 10:56












$begingroup$
It should be "Since this occurs for each $xin Acap B$".
$endgroup$
– Henning Makholm
Mar 16 at 11:00




$begingroup$
It should be "Since this occurs for each $xin Acap B$".
$endgroup$
– Henning Makholm
Mar 16 at 11:00

















draft saved

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