$Ysubseteq X$ iff $Xcup Y^c=Omega$, and $Xcap Y=emptyset$ iff $X^ccup Y^c=Omega$
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Let $X$ and $Y$ be subsets of the universe $Omega$. Prove the following:
1) $Ysubseteq X$ iff $Xcup Y^c=Omega$
2) $Xcap Y=emptyset$ iff $X^ccup Y^c=Omega$
Here $^c$ denotes the complement
The statements do logically makes sense, however I'm having trouble proving it formally.
discrete-mathematics
asked Sep 9 at 11:27
Mictej
304
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up vote
2
down vote
favorite
Let $X$ and $Y$ be subsets of the universe $Omega$. Prove the following:
1) $Ysubseteq X$ iff $Xcup Y^c=Omega$
2) $Xcap Y=emptyset$ iff $X^ccup Y^c=Omega$
Here $^c$ denotes the complement
The statements do logically makes sense, however I'm having trouble proving it formally.
discrete-mathematics
asked Sep 9 at 11:27
Mictej
304
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $X$ and $Y$ be subsets of the universe $Omega$. Prove the following:
1) $Ysubseteq X$ iff $Xcup Y^c=Omega$
2) $Xcap Y=emptyset$ iff $X^ccup Y^c=Omega$
Here $^c$ denotes the complement
The statements do logically makes sense, however I'm having trouble proving it formally.
discrete-mathematics
asked Sep 9 at 11:27
Mictej
304
Let $X$ and $Y$ be subsets of the universe $Omega$. Prove the following:
1) $Ysubseteq X$ iff $Xcup Y^c=Omega$
2) $Xcap Y=emptyset$ iff $X^ccup Y^c=Omega$
Here $^c$ denotes the complement
The statements do logically makes sense, however I'm having trouble proving it formally.
discrete-mathematics
discrete-mathematics
asked Sep 9 at 11:27
Mictej
304
asked Sep 9 at 11:27
Mictej
304
edited Sep 9 at 11:35
asked Sep 9 at 11:27
Mictej
304
asked Sep 9 at 11:27
Mictej
304
asked Sep 9 at 11:27
Mictej
304
304
add a comment |Â
add a comment |Â
1 Answer
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Let's prove the first complication.
If $Ysubseteq X$ then we have to prove that $Xcup Y^c=Omega$.
To prove it let's show both inclusions: it is clear that $Xcup Y^csubseteq Omega,$ so we only have to prove the other one. Take $xin Omega:$ if it belongs to $X$ we're done, otherwise it belongs to $X^c,$ that by hypothesis on $Y$ is a subset of $Y^c.$ So it certainly belongs to the union of $X$ and $Y^c.$
The other arrow is similar: if $Xcup Y^c=Omega$ then we have to prove that $Ysubseteq X.$
By absurd, suppose that $exists xin Ysetminus X.$ Then it wouldn't belong to $Xcup Y^c$ because it is nor in $X$ neither in $Y^c.$ So it is an element of $Omega$ which doesn't belong to $Xcup Y^c,$ which goes against the hypotheses.
The second statement is similar to prove, but if you want I'll make it.
answered Sep 9 at 11:40
Riccardo Ceccon
875320
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let's prove the first complication.
If $Ysubseteq X$ then we have to prove that $Xcup Y^c=Omega$.
To prove it let's show both inclusions: it is clear that $Xcup Y^csubseteq Omega,$ so we only have to prove the other one. Take $xin Omega:$ if it belongs to $X$ we're done, otherwise it belongs to $X^c,$ that by hypothesis on $Y$ is a subset of $Y^c.$ So it certainly belongs to the union of $X$ and $Y^c.$
The other arrow is similar: if $Xcup Y^c=Omega$ then we have to prove that $Ysubseteq X.$
By absurd, suppose that $exists xin Ysetminus X.$ Then it wouldn't belong to $Xcup Y^c$ because it is nor in $X$ neither in $Y^c.$ So it is an element of $Omega$ which doesn't belong to $Xcup Y^c,$ which goes against the hypotheses.
The second statement is similar to prove, but if you want I'll make it.
answered Sep 9 at 11:40
Riccardo Ceccon
875320
add a comment |Â
up vote
2
down vote
accepted
Let's prove the first complication.
If $Ysubseteq X$ then we have to prove that $Xcup Y^c=Omega$.
To prove it let's show both inclusions: it is clear that $Xcup Y^csubseteq Omega,$ so we only have to prove the other one. Take $xin Omega:$ if it belongs to $X$ we're done, otherwise it belongs to $X^c,$ that by hypothesis on $Y$ is a subset of $Y^c.$ So it certainly belongs to the union of $X$ and $Y^c.$
The other arrow is similar: if $Xcup Y^c=Omega$ then we have to prove that $Ysubseteq X.$
By absurd, suppose that $exists xin Ysetminus X.$ Then it wouldn't belong to $Xcup Y^c$ because it is nor in $X$ neither in $Y^c.$ So it is an element of $Omega$ which doesn't belong to $Xcup Y^c,$ which goes against the hypotheses.
The second statement is similar to prove, but if you want I'll make it.
answered Sep 9 at 11:40
Riccardo Ceccon
875320
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let's prove the first complication.
If $Ysubseteq X$ then we have to prove that $Xcup Y^c=Omega$.
To prove it let's show both inclusions: it is clear that $Xcup Y^csubseteq Omega,$ so we only have to prove the other one. Take $xin Omega:$ if it belongs to $X$ we're done, otherwise it belongs to $X^c,$ that by hypothesis on $Y$ is a subset of $Y^c.$ So it certainly belongs to the union of $X$ and $Y^c.$
The other arrow is similar: if $Xcup Y^c=Omega$ then we have to prove that $Ysubseteq X.$
By absurd, suppose that $exists xin Ysetminus X.$ Then it wouldn't belong to $Xcup Y^c$ because it is nor in $X$ neither in $Y^c.$ So it is an element of $Omega$ which doesn't belong to $Xcup Y^c,$ which goes against the hypotheses.
The second statement is similar to prove, but if you want I'll make it.
answered Sep 9 at 11:40
Riccardo Ceccon
875320
Let's prove the first complication.
If $Ysubseteq X$ then we have to prove that $Xcup Y^c=Omega$.
To prove it let's show both inclusions: it is clear that $Xcup Y^csubseteq Omega,$ so we only have to prove the other one. Take $xin Omega:$ if it belongs to $X$ we're done, otherwise it belongs to $X^c,$ that by hypothesis on $Y$ is a subset of $Y^c.$ So it certainly belongs to the union of $X$ and $Y^c.$
The other arrow is similar: if $Xcup Y^c=Omega$ then we have to prove that $Ysubseteq X.$
By absurd, suppose that $exists xin Ysetminus X.$ Then it wouldn't belong to $Xcup Y^c$ because it is nor in $X$ neither in $Y^c.$ So it is an element of $Omega$ which doesn't belong to $Xcup Y^c,$ which goes against the hypotheses.
The second statement is similar to prove, but if you want I'll make it.
answered Sep 9 at 11:40
Riccardo Ceccon
875320
answered Sep 9 at 11:40
Riccardo Ceccon
875320
answered Sep 9 at 11:40
Riccardo Ceccon
875320
answered Sep 9 at 11:40
Riccardo Ceccon
875320
875320
add a comment |Â
add a comment |Â
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