Proving theorem on indexing sets.
$begingroup$
Let $Lambda$ be an arbitrary indexing set and let {$B_alpha$}$_{alphain Lambda}$ be a collection of sets indexed by $Lambda$. Let $C$ be any set. Then the following identities hold:
- $Ccapbigcup_{alphainLambda}B_alpha = bigcup_{alphainLambda}(Ccap B_{alpha})$
I haven't got much into it at all, but would the notion of DeMorgan's Laws apply to this theorem at all? Just poking ideas at it.
My attempt at reading this, which is probably incorrect: "The intersection of a set $C$ and the union over $alphainLambda$ of the $B_{alpha}$'s is equal to the union over $alphainLambda$ on the set $C$'s intersection over $B_{alpha}$'s.
Apologies if I completely butchered it - this is the first time seeing these notations presented in this form. Also, we are supposed to prove why that theorem (1) holds true.
real-analysis elementary-set-theory proof-explanation
asked Jan 8 at 2:33
RyanRyan
1156
$endgroup$
add a comment |
$begingroup$
Let $Lambda$ be an arbitrary indexing set and let {$B_alpha$}$_{alphain Lambda}$ be a collection of sets indexed by $Lambda$. Let $C$ be any set. Then the following identities hold:
- $Ccapbigcup_{alphainLambda}B_alpha = bigcup_{alphainLambda}(Ccap B_{alpha})$
I haven't got much into it at all, but would the notion of DeMorgan's Laws apply to this theorem at all? Just poking ideas at it.
My attempt at reading this, which is probably incorrect: "The intersection of a set $C$ and the union over $alphainLambda$ of the $B_{alpha}$'s is equal to the union over $alphainLambda$ on the set $C$'s intersection over $B_{alpha}$'s.
Apologies if I completely butchered it - this is the first time seeing these notations presented in this form. Also, we are supposed to prove why that theorem (1) holds true.
real-analysis elementary-set-theory proof-explanation
asked Jan 8 at 2:33
RyanRyan
1156
$endgroup$
4
$begingroup$
That's not De Morgan's laws (no negations anywhere); it is just distributivity.
$endgroup$
– Henning Makholm
Jan 8 at 2:41
$begingroup$
So, in order to prove this identity holds true, how can I begin?
$endgroup$
– Ryan
Jan 8 at 5:59
add a comment |
$begingroup$
Let $Lambda$ be an arbitrary indexing set and let {$B_alpha$}$_{alphain Lambda}$ be a collection of sets indexed by $Lambda$. Let $C$ be any set. Then the following identities hold:
- $Ccapbigcup_{alphainLambda}B_alpha = bigcup_{alphainLambda}(Ccap B_{alpha})$
I haven't got much into it at all, but would the notion of DeMorgan's Laws apply to this theorem at all? Just poking ideas at it.
My attempt at reading this, which is probably incorrect: "The intersection of a set $C$ and the union over $alphainLambda$ of the $B_{alpha}$'s is equal to the union over $alphainLambda$ on the set $C$'s intersection over $B_{alpha}$'s.
Apologies if I completely butchered it - this is the first time seeing these notations presented in this form. Also, we are supposed to prove why that theorem (1) holds true.
real-analysis elementary-set-theory proof-explanation
asked Jan 8 at 2:33
RyanRyan
1156
$endgroup$
Let $Lambda$ be an arbitrary indexing set and let {$B_alpha$}$_{alphain Lambda}$ be a collection of sets indexed by $Lambda$. Let $C$ be any set. Then the following identities hold:
- $Ccapbigcup_{alphainLambda}B_alpha = bigcup_{alphainLambda}(Ccap B_{alpha})$
I haven't got much into it at all, but would the notion of DeMorgan's Laws apply to this theorem at all? Just poking ideas at it.
My attempt at reading this, which is probably incorrect: "The intersection of a set $C$ and the union over $alphainLambda$ of the $B_{alpha}$'s is equal to the union over $alphainLambda$ on the set $C$'s intersection over $B_{alpha}$'s.
Apologies if I completely butchered it - this is the first time seeing these notations presented in this form. Also, we are supposed to prove why that theorem (1) holds true.
real-analysis elementary-set-theory proof-explanation
real-analysis elementary-set-theory proof-explanation
asked Jan 8 at 2:33
RyanRyan
1156
asked Jan 8 at 2:33
RyanRyan
1156
asked Jan 8 at 2:33
RyanRyan
1156
asked Jan 8 at 2:33
RyanRyan
1156
asked Jan 8 at 2:33
RyanRyan
1156
1156
4
$begingroup$
That's not De Morgan's laws (no negations anywhere); it is just distributivity.
$endgroup$
– Henning Makholm
Jan 8 at 2:41
$begingroup$
So, in order to prove this identity holds true, how can I begin?
$endgroup$
– Ryan
Jan 8 at 5:59
add a comment |
4
$begingroup$
That's not De Morgan's laws (no negations anywhere); it is just distributivity.
$endgroup$
– Henning Makholm
Jan 8 at 2:41
$begingroup$
So, in order to prove this identity holds true, how can I begin?
$endgroup$
– Ryan
Jan 8 at 5:59
4
4
$begingroup$
That's not De Morgan's laws (no negations anywhere); it is just distributivity.
$endgroup$
– Henning Makholm
Jan 8 at 2:41
$begingroup$
That's not De Morgan's laws (no negations anywhere); it is just distributivity.
$endgroup$
– Henning Makholm
Jan 8 at 2:41
$begingroup$
So, in order to prove this identity holds true, how can I begin?
$endgroup$
– Ryan
Jan 8 at 5:59
$begingroup$
So, in order to prove this identity holds true, how can I begin?
$endgroup$
– Ryan
Jan 8 at 5:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The general case is probably a bit distracting. Try a simple example first. Prove that
$C cap (B_1 cup B_2) = (C cap B_1) cup (C cap B_2)$
That is, show that each set is a subset of the other. If you can get this, then the ideas involved in the proof to your general case is essentially identical.
answered Jan 9 at 5:05
MetricMetric
91839
$endgroup$
$begingroup$
Thanks! Appreciate your feedback, given the errors in my original question.
$endgroup$
– Ryan
Jan 9 at 5:06
$begingroup$
You're welcome. I'm glad I could help!
$endgroup$
– Metric
Jan 9 at 5:06
add a comment |
$begingroup$
Tough it out point by point.
x in C $cap$ $cup${ B$_a$ : a in A }
iff x in C and exists a in A with x in B$_a$
iff ....
answered Jan 8 at 4:06
William ElliotWilliam Elliot
7,7322720
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The general case is probably a bit distracting. Try a simple example first. Prove that
$C cap (B_1 cup B_2) = (C cap B_1) cup (C cap B_2)$
That is, show that each set is a subset of the other. If you can get this, then the ideas involved in the proof to your general case is essentially identical.
answered Jan 9 at 5:05
MetricMetric
91839
$endgroup$
$begingroup$
Thanks! Appreciate your feedback, given the errors in my original question.
$endgroup$
– Ryan
Jan 9 at 5:06
$begingroup$
You're welcome. I'm glad I could help!
$endgroup$
– Metric
Jan 9 at 5:06
add a comment |
$begingroup$
The general case is probably a bit distracting. Try a simple example first. Prove that
$C cap (B_1 cup B_2) = (C cap B_1) cup (C cap B_2)$
That is, show that each set is a subset of the other. If you can get this, then the ideas involved in the proof to your general case is essentially identical.
answered Jan 9 at 5:05
MetricMetric
91839
$endgroup$
$begingroup$
Thanks! Appreciate your feedback, given the errors in my original question.
$endgroup$
– Ryan
Jan 9 at 5:06
$begingroup$
You're welcome. I'm glad I could help!
$endgroup$
– Metric
Jan 9 at 5:06
add a comment |
$begingroup$
The general case is probably a bit distracting. Try a simple example first. Prove that
$C cap (B_1 cup B_2) = (C cap B_1) cup (C cap B_2)$
That is, show that each set is a subset of the other. If you can get this, then the ideas involved in the proof to your general case is essentially identical.
answered Jan 9 at 5:05
MetricMetric
91839
$endgroup$
The general case is probably a bit distracting. Try a simple example first. Prove that
$C cap (B_1 cup B_2) = (C cap B_1) cup (C cap B_2)$
That is, show that each set is a subset of the other. If you can get this, then the ideas involved in the proof to your general case is essentially identical.
answered Jan 9 at 5:05
MetricMetric
91839
answered Jan 9 at 5:05
MetricMetric
91839
answered Jan 9 at 5:05
MetricMetric
91839
answered Jan 9 at 5:05
MetricMetric
91839
91839
$begingroup$
Thanks! Appreciate your feedback, given the errors in my original question.
$endgroup$
– Ryan
Jan 9 at 5:06
$begingroup$
You're welcome. I'm glad I could help!
$endgroup$
– Metric
Jan 9 at 5:06
add a comment |
$begingroup$
Thanks! Appreciate your feedback, given the errors in my original question.
$endgroup$
– Ryan
Jan 9 at 5:06
$begingroup$
You're welcome. I'm glad I could help!
$endgroup$
– Metric
Jan 9 at 5:06
$begingroup$
Thanks! Appreciate your feedback, given the errors in my original question.
$endgroup$
– Ryan
Jan 9 at 5:06
$begingroup$
Thanks! Appreciate your feedback, given the errors in my original question.
$endgroup$
– Ryan
Jan 9 at 5:06
$begingroup$
You're welcome. I'm glad I could help!
$endgroup$
– Metric
Jan 9 at 5:06
$begingroup$
You're welcome. I'm glad I could help!
$endgroup$
– Metric
Jan 9 at 5:06
add a comment |
$begingroup$
Tough it out point by point.
x in C $cap$ $cup${ B$_a$ : a in A }
iff x in C and exists a in A with x in B$_a$
iff ....
answered Jan 8 at 4:06
William ElliotWilliam Elliot
7,7322720
$endgroup$
add a comment |
$begingroup$
Tough it out point by point.
x in C $cap$ $cup${ B$_a$ : a in A }
iff x in C and exists a in A with x in B$_a$
iff ....
answered Jan 8 at 4:06
William ElliotWilliam Elliot
7,7322720
$endgroup$
add a comment |
$begingroup$
Tough it out point by point.
x in C $cap$ $cup${ B$_a$ : a in A }
iff x in C and exists a in A with x in B$_a$
iff ....
answered Jan 8 at 4:06
William ElliotWilliam Elliot
7,7322720
$endgroup$
Tough it out point by point.
x in C $cap$ $cup${ B$_a$ : a in A }
iff x in C and exists a in A with x in B$_a$
iff ....
answered Jan 8 at 4:06
William ElliotWilliam Elliot
7,7322720
answered Jan 8 at 4:06
William ElliotWilliam Elliot
7,7322720
answered Jan 8 at 4:06
William ElliotWilliam Elliot
7,7322720
answered Jan 8 at 4:06
William ElliotWilliam Elliot
7,7322720
7,7322720
add a comment |
add a comment |
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4
$begingroup$
That's not De Morgan's laws (no negations anywhere); it is just distributivity.
$endgroup$
– Henning Makholm
Jan 8 at 2:41
$begingroup$
So, in order to prove this identity holds true, how can I begin?
$endgroup$
– Ryan
Jan 8 at 5:59