Elementary Set Theory : Are my proofs correct?
$begingroup$
Prove that the arbitrary intersection indexed over the power set is the empty set and that the arbitrary union indexed over the power set is the entire set itself.
$a) bigcap_{Ain P(X)} A = varnothing $
$b) bigcup_{Ain P(X)} A = X $
Where $P(X)$ is the power set of $X$
Proof:-
a) Let $x in bigcap_{A in P(X)}A$
Then $forall Asubseteq Xquad (orquad forall A in P(X)$)
$x in A $
Therefore $bigcap_{A in P(X)}A subseteq A quad (forall Ain P(X)$)
Now $varnothing in P(X)$
Thus $bigcap_{A in P(X)}A subseteq varnothing$
We know that the empty set is a subset of every set, in particular:
$varnothing subseteq bigcap_{A in P(X)}A$
Therefore we've demonstrated set inclusion in both directions and by the definition of set equality, it follows that,
$bigcap_{A in P(X)}A = varnothing$.
b) Let $x in bigcup_{A in P(X)}A$
Then
$exists A in P(X) $ such that $xin A$
and since $A subseteq X $ it follows that $x in A$
Therefore $bigcup_{A in P(X)}A subseteq X$
Conversely, let $x in X$
Then $exists A in P(X)$ such that $ x in A $
And, so $x inbigcup_{A in P(X)}A$
Therefore we've demonstrated set inclusion in both directions and by the definition of set equality, it follows that,
$bigcup_{A in P(X)}A=X$.
Quod Erat Demonstrandum.
proof-verification elementary-set-theory
edited Jan 20 at 19:05
Andrés E. Caicedo
65.5k8158249
asked Jan 20 at 15:05
ambrish abhijatyaambrish abhijatya
33
$endgroup$
add a comment |
$begingroup$
Prove that the arbitrary intersection indexed over the power set is the empty set and that the arbitrary union indexed over the power set is the entire set itself.
$a) bigcap_{Ain P(X)} A = varnothing $
$b) bigcup_{Ain P(X)} A = X $
Where $P(X)$ is the power set of $X$
Proof:-
a) Let $x in bigcap_{A in P(X)}A$
Then $forall Asubseteq Xquad (orquad forall A in P(X)$)
$x in A $
Therefore $bigcap_{A in P(X)}A subseteq A quad (forall Ain P(X)$)
Now $varnothing in P(X)$
Thus $bigcap_{A in P(X)}A subseteq varnothing$
We know that the empty set is a subset of every set, in particular:
$varnothing subseteq bigcap_{A in P(X)}A$
Therefore we've demonstrated set inclusion in both directions and by the definition of set equality, it follows that,
$bigcap_{A in P(X)}A = varnothing$.
b) Let $x in bigcup_{A in P(X)}A$
Then
$exists A in P(X) $ such that $xin A$
and since $A subseteq X $ it follows that $x in A$
Therefore $bigcup_{A in P(X)}A subseteq X$
Conversely, let $x in X$
Then $exists A in P(X)$ such that $ x in A $
And, so $x inbigcup_{A in P(X)}A$
Therefore we've demonstrated set inclusion in both directions and by the definition of set equality, it follows that,
$bigcup_{A in P(X)}A=X$.
Quod Erat Demonstrandum.
proof-verification elementary-set-theory
edited Jan 20 at 19:05
Andrés E. Caicedo
65.5k8158249
asked Jan 20 at 15:05
ambrish abhijatyaambrish abhijatya
33
$endgroup$
add a comment |
$begingroup$
Prove that the arbitrary intersection indexed over the power set is the empty set and that the arbitrary union indexed over the power set is the entire set itself.
$a) bigcap_{Ain P(X)} A = varnothing $
$b) bigcup_{Ain P(X)} A = X $
Where $P(X)$ is the power set of $X$
Proof:-
a) Let $x in bigcap_{A in P(X)}A$
Then $forall Asubseteq Xquad (orquad forall A in P(X)$)
$x in A $
Therefore $bigcap_{A in P(X)}A subseteq A quad (forall Ain P(X)$)
Now $varnothing in P(X)$
Thus $bigcap_{A in P(X)}A subseteq varnothing$
We know that the empty set is a subset of every set, in particular:
$varnothing subseteq bigcap_{A in P(X)}A$
Therefore we've demonstrated set inclusion in both directions and by the definition of set equality, it follows that,
$bigcap_{A in P(X)}A = varnothing$.
b) Let $x in bigcup_{A in P(X)}A$
Then
$exists A in P(X) $ such that $xin A$
and since $A subseteq X $ it follows that $x in A$
Therefore $bigcup_{A in P(X)}A subseteq X$
Conversely, let $x in X$
Then $exists A in P(X)$ such that $ x in A $
And, so $x inbigcup_{A in P(X)}A$
Therefore we've demonstrated set inclusion in both directions and by the definition of set equality, it follows that,
$bigcup_{A in P(X)}A=X$.
Quod Erat Demonstrandum.
proof-verification elementary-set-theory
edited Jan 20 at 19:05
Andrés E. Caicedo
65.5k8158249
asked Jan 20 at 15:05
ambrish abhijatyaambrish abhijatya
33
$endgroup$
Prove that the arbitrary intersection indexed over the power set is the empty set and that the arbitrary union indexed over the power set is the entire set itself.
$a) bigcap_{Ain P(X)} A = varnothing $
$b) bigcup_{Ain P(X)} A = X $
Where $P(X)$ is the power set of $X$
Proof:-
a) Let $x in bigcap_{A in P(X)}A$
Then $forall Asubseteq Xquad (orquad forall A in P(X)$)
$x in A $
Therefore $bigcap_{A in P(X)}A subseteq A quad (forall Ain P(X)$)
Now $varnothing in P(X)$
Thus $bigcap_{A in P(X)}A subseteq varnothing$
We know that the empty set is a subset of every set, in particular:
$varnothing subseteq bigcap_{A in P(X)}A$
Therefore we've demonstrated set inclusion in both directions and by the definition of set equality, it follows that,
$bigcap_{A in P(X)}A = varnothing$.
b) Let $x in bigcup_{A in P(X)}A$
Then
$exists A in P(X) $ such that $xin A$
and since $A subseteq X $ it follows that $x in A$
Therefore $bigcup_{A in P(X)}A subseteq X$
Conversely, let $x in X$
Then $exists A in P(X)$ such that $ x in A $
And, so $x inbigcup_{A in P(X)}A$
Therefore we've demonstrated set inclusion in both directions and by the definition of set equality, it follows that,
$bigcup_{A in P(X)}A=X$.
Quod Erat Demonstrandum.
proof-verification elementary-set-theory
proof-verification elementary-set-theory
edited Jan 20 at 19:05
Andrés E. Caicedo
65.5k8158249
asked Jan 20 at 15:05
ambrish abhijatyaambrish abhijatya
33
edited Jan 20 at 19:05
Andrés E. Caicedo
65.5k8158249
asked Jan 20 at 15:05
ambrish abhijatyaambrish abhijatya
33
edited Jan 20 at 19:05
Andrés E. Caicedo
65.5k8158249
edited Jan 20 at 19:05
Andrés E. Caicedo
65.5k8158249
edited Jan 20 at 19:05
Andrés E. Caicedo
65.5k8158249
65.5k8158249
asked Jan 20 at 15:05
ambrish abhijatyaambrish abhijatya
33
asked Jan 20 at 15:05
ambrish abhijatyaambrish abhijatya
33
asked Jan 20 at 15:05
ambrish abhijatyaambrish abhijatya
33
33
add a comment |
add a comment |
1 Answer
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$begingroup$
a) You stated correctly that an element $x$ of the intersection must be an element of every $Asubseteq X$. For $A$ we can take $varnothing$ so the assumption that $x$ is an element of the intersection leads directly to the conclusion that $xinvarnothing$. This is absurd so you are allowed to reject the assumption. This for every $xin X$ and the final conclusion is then that the intersection is empty.
b)
"...and since $A subseteq X $ it follows that $x in A$"
I suspect a typo here and that you meant to end this with "...that $xin X$"
(You stated correctly that - if $x$ is an element of the union - we have $xin A$ for some $Asubseteq X$. From this you can conlude directly that $xin X$.)
answered Jan 20 at 15:46
drhabdrhab
102k545136
$endgroup$
$begingroup$
I was trying to emphasize that any member of the power set of a set is the subset of that set. Is that redundant?
$endgroup$
– ambrish abhijatya
Jan 20 at 17:06
$begingroup$
That is not redundant.
$endgroup$
– drhab
Jan 20 at 17:55
add a comment |
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1 Answer
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$begingroup$
a) You stated correctly that an element $x$ of the intersection must be an element of every $Asubseteq X$. For $A$ we can take $varnothing$ so the assumption that $x$ is an element of the intersection leads directly to the conclusion that $xinvarnothing$. This is absurd so you are allowed to reject the assumption. This for every $xin X$ and the final conclusion is then that the intersection is empty.
b)
"...and since $A subseteq X $ it follows that $x in A$"
I suspect a typo here and that you meant to end this with "...that $xin X$"
(You stated correctly that - if $x$ is an element of the union - we have $xin A$ for some $Asubseteq X$. From this you can conlude directly that $xin X$.)
answered Jan 20 at 15:46
drhabdrhab
102k545136
$endgroup$
$begingroup$
I was trying to emphasize that any member of the power set of a set is the subset of that set. Is that redundant?
$endgroup$
– ambrish abhijatya
Jan 20 at 17:06
$begingroup$
That is not redundant.
$endgroup$
– drhab
Jan 20 at 17:55
add a comment |
$begingroup$
a) You stated correctly that an element $x$ of the intersection must be an element of every $Asubseteq X$. For $A$ we can take $varnothing$ so the assumption that $x$ is an element of the intersection leads directly to the conclusion that $xinvarnothing$. This is absurd so you are allowed to reject the assumption. This for every $xin X$ and the final conclusion is then that the intersection is empty.
b)
"...and since $A subseteq X $ it follows that $x in A$"
I suspect a typo here and that you meant to end this with "...that $xin X$"
(You stated correctly that - if $x$ is an element of the union - we have $xin A$ for some $Asubseteq X$. From this you can conlude directly that $xin X$.)
answered Jan 20 at 15:46
drhabdrhab
102k545136
$endgroup$
$begingroup$
I was trying to emphasize that any member of the power set of a set is the subset of that set. Is that redundant?
$endgroup$
– ambrish abhijatya
Jan 20 at 17:06
$begingroup$
That is not redundant.
$endgroup$
– drhab
Jan 20 at 17:55
add a comment |
$begingroup$
a) You stated correctly that an element $x$ of the intersection must be an element of every $Asubseteq X$. For $A$ we can take $varnothing$ so the assumption that $x$ is an element of the intersection leads directly to the conclusion that $xinvarnothing$. This is absurd so you are allowed to reject the assumption. This for every $xin X$ and the final conclusion is then that the intersection is empty.
b)
"...and since $A subseteq X $ it follows that $x in A$"
I suspect a typo here and that you meant to end this with "...that $xin X$"
(You stated correctly that - if $x$ is an element of the union - we have $xin A$ for some $Asubseteq X$. From this you can conlude directly that $xin X$.)
answered Jan 20 at 15:46
drhabdrhab
102k545136
$endgroup$
a) You stated correctly that an element $x$ of the intersection must be an element of every $Asubseteq X$. For $A$ we can take $varnothing$ so the assumption that $x$ is an element of the intersection leads directly to the conclusion that $xinvarnothing$. This is absurd so you are allowed to reject the assumption. This for every $xin X$ and the final conclusion is then that the intersection is empty.
b)
"...and since $A subseteq X $ it follows that $x in A$"
I suspect a typo here and that you meant to end this with "...that $xin X$"
(You stated correctly that - if $x$ is an element of the union - we have $xin A$ for some $Asubseteq X$. From this you can conlude directly that $xin X$.)
answered Jan 20 at 15:46
drhabdrhab
102k545136
answered Jan 20 at 15:46
drhabdrhab
102k545136
answered Jan 20 at 15:46
drhabdrhab
102k545136
answered Jan 20 at 15:46
drhabdrhab
102k545136
102k545136
$begingroup$
I was trying to emphasize that any member of the power set of a set is the subset of that set. Is that redundant?
$endgroup$
– ambrish abhijatya
Jan 20 at 17:06
$begingroup$
That is not redundant.
$endgroup$
– drhab
Jan 20 at 17:55
add a comment |
$begingroup$
I was trying to emphasize that any member of the power set of a set is the subset of that set. Is that redundant?
$endgroup$
– ambrish abhijatya
Jan 20 at 17:06
$begingroup$
That is not redundant.
$endgroup$
– drhab
Jan 20 at 17:55
$begingroup$
I was trying to emphasize that any member of the power set of a set is the subset of that set. Is that redundant?
$endgroup$
– ambrish abhijatya
Jan 20 at 17:06
$begingroup$
I was trying to emphasize that any member of the power set of a set is the subset of that set. Is that redundant?
$endgroup$
– ambrish abhijatya
Jan 20 at 17:06
$begingroup$
That is not redundant.
$endgroup$
– drhab
Jan 20 at 17:55
$begingroup$
That is not redundant.
$endgroup$
– drhab
Jan 20 at 17:55
add a comment |
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