$(A times B)^c = (A^c times Y) cup (X times B^c)$
$begingroup$
Let $X$ and $Y$ be universal sets and $A subseteq X$ and $B subseteq Y$.
Then, I want to prove that $(A times B)^c=(A^c times Y) cup (X times B^c)$.
I think that i have to suppose that $(x,y) in (A times B)^c$, then $(x,y) in (X times Y) setminus (Atimes B) Rightarrow (x,y) in (A^c times Y) cup (X times B^c)$, but is this correct? And how should I prove that?
Any help is appreciated...
elementary-set-theory
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be universal sets and $A subseteq X$ and $B subseteq Y$.
Then, I want to prove that $(A times B)^c=(A^c times Y) cup (X times B^c)$.
I think that i have to suppose that $(x,y) in (A times B)^c$, then $(x,y) in (X times Y) setminus (Atimes B) Rightarrow (x,y) in (A^c times Y) cup (X times B^c)$, but is this correct? And how should I prove that?
Any help is appreciated...
elementary-set-theory
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be universal sets and $A subseteq X$ and $B subseteq Y$.
Then, I want to prove that $(A times B)^c=(A^c times Y) cup (X times B^c)$.
I think that i have to suppose that $(x,y) in (A times B)^c$, then $(x,y) in (X times Y) setminus (Atimes B) Rightarrow (x,y) in (A^c times Y) cup (X times B^c)$, but is this correct? And how should I prove that?
Any help is appreciated...
elementary-set-theory
$endgroup$
Let $X$ and $Y$ be universal sets and $A subseteq X$ and $B subseteq Y$.
Then, I want to prove that $(A times B)^c=(A^c times Y) cup (X times B^c)$.
I think that i have to suppose that $(x,y) in (A times B)^c$, then $(x,y) in (X times Y) setminus (Atimes B) Rightarrow (x,y) in (A^c times Y) cup (X times B^c)$, but is this correct? And how should I prove that?
Any help is appreciated...
elementary-set-theory
elementary-set-theory
asked Jan 2 at 10:46
t.ysnt.ysn
1397
1397
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $(x,y)in(Atimes B)^c$. Then $(x,y)notin Atimes B$ which implies that $xnotin A$ or $ynotin B$. Hence, $(x,y)in A^ctimes Y$ or $(x,y)in Xtimes B^c$.
For the other direction. Let $(x,y)in (A^c times Y) cup (X times B^c)$. This implies that $(x,y)in A^c times Y$ or $(x,y)in X times B^c$, namely $xnotin A$ or $ynotin B$. Thus, $(x,y)notin Atimes B$.
To see this result and get some intuition check for example: $A=B=[0,1]$ and $X=Y=mathbb{R}$.
$endgroup$
$begingroup$
Thanks a lot. :)
$endgroup$
– t.ysn
Jan 2 at 11:07
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Let $(x,y)in(Atimes B)^c$. Then $(x,y)notin Atimes B$ which implies that $xnotin A$ or $ynotin B$. Hence, $(x,y)in A^ctimes Y$ or $(x,y)in Xtimes B^c$.
For the other direction. Let $(x,y)in (A^c times Y) cup (X times B^c)$. This implies that $(x,y)in A^c times Y$ or $(x,y)in X times B^c$, namely $xnotin A$ or $ynotin B$. Thus, $(x,y)notin Atimes B$.
To see this result and get some intuition check for example: $A=B=[0,1]$ and $X=Y=mathbb{R}$.
$endgroup$
$begingroup$
Thanks a lot. :)
$endgroup$
– t.ysn
Jan 2 at 11:07
add a comment |
$begingroup$
Let $(x,y)in(Atimes B)^c$. Then $(x,y)notin Atimes B$ which implies that $xnotin A$ or $ynotin B$. Hence, $(x,y)in A^ctimes Y$ or $(x,y)in Xtimes B^c$.
For the other direction. Let $(x,y)in (A^c times Y) cup (X times B^c)$. This implies that $(x,y)in A^c times Y$ or $(x,y)in X times B^c$, namely $xnotin A$ or $ynotin B$. Thus, $(x,y)notin Atimes B$.
To see this result and get some intuition check for example: $A=B=[0,1]$ and $X=Y=mathbb{R}$.
$endgroup$
$begingroup$
Thanks a lot. :)
$endgroup$
– t.ysn
Jan 2 at 11:07
add a comment |
$begingroup$
Let $(x,y)in(Atimes B)^c$. Then $(x,y)notin Atimes B$ which implies that $xnotin A$ or $ynotin B$. Hence, $(x,y)in A^ctimes Y$ or $(x,y)in Xtimes B^c$.
For the other direction. Let $(x,y)in (A^c times Y) cup (X times B^c)$. This implies that $(x,y)in A^c times Y$ or $(x,y)in X times B^c$, namely $xnotin A$ or $ynotin B$. Thus, $(x,y)notin Atimes B$.
To see this result and get some intuition check for example: $A=B=[0,1]$ and $X=Y=mathbb{R}$.
$endgroup$
Let $(x,y)in(Atimes B)^c$. Then $(x,y)notin Atimes B$ which implies that $xnotin A$ or $ynotin B$. Hence, $(x,y)in A^ctimes Y$ or $(x,y)in Xtimes B^c$.
For the other direction. Let $(x,y)in (A^c times Y) cup (X times B^c)$. This implies that $(x,y)in A^c times Y$ or $(x,y)in X times B^c$, namely $xnotin A$ or $ynotin B$. Thus, $(x,y)notin Atimes B$.
To see this result and get some intuition check for example: $A=B=[0,1]$ and $X=Y=mathbb{R}$.
answered Jan 2 at 10:53
Test123Test123
2,782828
2,782828
$begingroup$
Thanks a lot. :)
$endgroup$
– t.ysn
Jan 2 at 11:07
add a comment |
$begingroup$
Thanks a lot. :)
$endgroup$
– t.ysn
Jan 2 at 11:07
$begingroup$
Thanks a lot. :)
$endgroup$
– t.ysn
Jan 2 at 11:07
$begingroup$
Thanks a lot. :)
$endgroup$
– t.ysn
Jan 2 at 11:07
add a comment |
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