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 $A subseteq B$
$begingroup$
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 $A subset B$
proof-verification elementary-set-theory
edited Mar 16 at 11:16
José Carlos Santos
170k23132238
asked Mar 16 at 10:17
seeppseepp
132
$endgroup$
add a comment |
$begingroup$
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 $A subset B$
proof-verification elementary-set-theory
edited Mar 16 at 11:16
José Carlos Santos
170k23132238
asked Mar 16 at 10:17
seeppseepp
132
$endgroup$
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
add a comment |
$begingroup$
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 $A subset B$
proof-verification elementary-set-theory
edited Mar 16 at 11:16
José Carlos Santos
170k23132238
asked Mar 16 at 10:17
seeppseepp
132
$endgroup$
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
proof-verification elementary-set-theory
edited Mar 16 at 11:16
José Carlos Santos
170k23132238
asked Mar 16 at 10:17
seeppseepp
132
edited Mar 16 at 11:16
José Carlos Santos
170k23132238
asked Mar 16 at 10:17
seeppseepp
132
edited Mar 16 at 11:16
José Carlos Santos
170k23132238
edited Mar 16 at 11:16
José Carlos Santos
170k23132238
edited Mar 16 at 11:16
José Carlos Santos
170k23132238
170k23132238
asked Mar 16 at 10:17
seeppseepp
132
asked Mar 16 at 10:17
seeppseepp
132
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $xin A$, $Acap Bsubseteq A$.
answered Mar 16 at 10:31
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
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 $xin A$, $Acap Bsubseteq A$.
answered Mar 16 at 10:31
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $xin A$, $Acap Bsubseteq A$.
answered Mar 16 at 10:31
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $xin A$, $Acap Bsubseteq A$.
answered Mar 16 at 10:31
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
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 $xin A$, $Acap Bsubseteq A$.
answered Mar 16 at 10:31
José Carlos SantosJosé Carlos Santos
170k23132238
edited Mar 16 at 10:52
answered Mar 16 at 10:31
José Carlos SantosJosé Carlos Santos
170k23132238
answered Mar 16 at 10:31
José Carlos SantosJosé Carlos Santos
170k23132238
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
add a comment |
$begingroup$
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
add a comment |
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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