Mixing
Is there any notation for testing if A is in B?
$begingroup$
I want to do a union over sets $A_i$, but only if $A_i in B_i$. I don't know any way to write this concisely. I was trying to write something with unions and intersections but I can never get it to exclude $A_i$ if it's not entirely contained in the corresponding $B_i$
elementary-set-theory logic
asked Jan 23 at 21:59
swedishfishedswedishfished
887
$endgroup$
add a comment |
$begingroup$
I want to do a union over sets $A_i$, but only if $A_i in B_i$. I don't know any way to write this concisely. I was trying to write something with unions and intersections but I can never get it to exclude $A_i$ if it's not entirely contained in the corresponding $B_i$
elementary-set-theory logic
asked Jan 23 at 21:59
swedishfishedswedishfished
887
$endgroup$
$begingroup$
Do you mean that $B$ is a set of sets, and you want to union only the sets that are in it? (e.g. $B = {{1}, {2, 4, 6}, {-1, -2, -3}}$). Or is B a set containing similar elements to A, and you want to union the ones where A is a subset of B? Because that's a different notation.
$endgroup$
– ConMan
Jan 23 at 22:23
add a comment |
$begingroup$
I want to do a union over sets $A_i$, but only if $A_i in B_i$. I don't know any way to write this concisely. I was trying to write something with unions and intersections but I can never get it to exclude $A_i$ if it's not entirely contained in the corresponding $B_i$
elementary-set-theory logic
asked Jan 23 at 21:59
swedishfishedswedishfished
887
$endgroup$
I want to do a union over sets $A_i$, but only if $A_i in B_i$. I don't know any way to write this concisely. I was trying to write something with unions and intersections but I can never get it to exclude $A_i$ if it's not entirely contained in the corresponding $B_i$
elementary-set-theory logic
elementary-set-theory logic
asked Jan 23 at 21:59
swedishfishedswedishfished
887
asked Jan 23 at 21:59
swedishfishedswedishfished
887
asked Jan 23 at 21:59
swedishfishedswedishfished
887
asked Jan 23 at 21:59
swedishfishedswedishfished
887
asked Jan 23 at 21:59
swedishfishedswedishfished
887
887
$begingroup$
Do you mean that $B$ is a set of sets, and you want to union only the sets that are in it? (e.g. $B = {{1}, {2, 4, 6}, {-1, -2, -3}}$). Or is B a set containing similar elements to A, and you want to union the ones where A is a subset of B? Because that's a different notation.
$endgroup$
– ConMan
Jan 23 at 22:23
add a comment |
$begingroup$
Do you mean that $B$ is a set of sets, and you want to union only the sets that are in it? (e.g. $B = {{1}, {2, 4, 6}, {-1, -2, -3}}$). Or is B a set containing similar elements to A, and you want to union the ones where A is a subset of B? Because that's a different notation.
$endgroup$
– ConMan
Jan 23 at 22:23
$begingroup$
Do you mean that $B$ is a set of sets, and you want to union only the sets that are in it? (e.g. $B = {{1}, {2, 4, 6}, {-1, -2, -3}}$). Or is B a set containing similar elements to A, and you want to union the ones where A is a subset of B? Because that's a different notation.
$endgroup$
– ConMan
Jan 23 at 22:23
$begingroup$
Do you mean that $B$ is a set of sets, and you want to union only the sets that are in it? (e.g. $B = {{1}, {2, 4, 6}, {-1, -2, -3}}$). Or is B a set containing similar elements to A, and you want to union the ones where A is a subset of B? Because that's a different notation.
$endgroup$
– ConMan
Jan 23 at 22:23
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Perhaps something like: $$bigcuplimits_{i:A_i in B_i}A_i$$
answered Jan 23 at 22:02
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
You've already written it. One way of writing the overall expression would be: $$bigcup{A_imid iin I land A_iin B_i}$$
answered Jan 23 at 22:03
Derek ElkinsDerek Elkins
17.2k11437
$endgroup$
add a comment |
$begingroup$
You could define indicator functions on sets:
$$
Acdot 1_{text{statement}} =
begin{cases}
A & 1_{text{statement}}=1\
emptyset & 1_{text{statement}}=0
end{cases}
$$
And then Union over all the sets times the indicator functions.
answered Jan 23 at 23:09
Felix B.Felix B.
674317
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Perhaps something like: $$bigcuplimits_{i:A_i in B_i}A_i$$
answered Jan 23 at 22:02
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
Perhaps something like: $$bigcuplimits_{i:A_i in B_i}A_i$$
answered Jan 23 at 22:02
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
Perhaps something like: $$bigcuplimits_{i:A_i in B_i}A_i$$
answered Jan 23 at 22:02
HenryHenry
101k481168
$endgroup$
Perhaps something like: $$bigcuplimits_{i:A_i in B_i}A_i$$
answered Jan 23 at 22:02
HenryHenry
101k481168
answered Jan 23 at 22:02
HenryHenry
101k481168
answered Jan 23 at 22:02
HenryHenry
101k481168
answered Jan 23 at 22:02
HenryHenry
101k481168
101k481168
add a comment |
add a comment |
$begingroup$
You've already written it. One way of writing the overall expression would be: $$bigcup{A_imid iin I land A_iin B_i}$$
answered Jan 23 at 22:03
Derek ElkinsDerek Elkins
17.2k11437
$endgroup$
add a comment |
$begingroup$
You've already written it. One way of writing the overall expression would be: $$bigcup{A_imid iin I land A_iin B_i}$$
answered Jan 23 at 22:03
Derek ElkinsDerek Elkins
17.2k11437
$endgroup$
add a comment |
$begingroup$
You've already written it. One way of writing the overall expression would be: $$bigcup{A_imid iin I land A_iin B_i}$$
answered Jan 23 at 22:03
Derek ElkinsDerek Elkins
17.2k11437
$endgroup$
You've already written it. One way of writing the overall expression would be: $$bigcup{A_imid iin I land A_iin B_i}$$
answered Jan 23 at 22:03
Derek ElkinsDerek Elkins
17.2k11437
answered Jan 23 at 22:03
Derek ElkinsDerek Elkins
17.2k11437
answered Jan 23 at 22:03
Derek ElkinsDerek Elkins
17.2k11437
answered Jan 23 at 22:03
Derek ElkinsDerek Elkins
17.2k11437
17.2k11437
add a comment |
add a comment |
$begingroup$
You could define indicator functions on sets:
$$
Acdot 1_{text{statement}} =
begin{cases}
A & 1_{text{statement}}=1\
emptyset & 1_{text{statement}}=0
end{cases}
$$
And then Union over all the sets times the indicator functions.
answered Jan 23 at 23:09
Felix B.Felix B.
674317
$endgroup$
add a comment |
$begingroup$
You could define indicator functions on sets:
$$
Acdot 1_{text{statement}} =
begin{cases}
A & 1_{text{statement}}=1\
emptyset & 1_{text{statement}}=0
end{cases}
$$
And then Union over all the sets times the indicator functions.
answered Jan 23 at 23:09
Felix B.Felix B.
674317
$endgroup$
add a comment |
$begingroup$
You could define indicator functions on sets:
$$
Acdot 1_{text{statement}} =
begin{cases}
A & 1_{text{statement}}=1\
emptyset & 1_{text{statement}}=0
end{cases}
$$
And then Union over all the sets times the indicator functions.
answered Jan 23 at 23:09
Felix B.Felix B.
674317
$endgroup$
You could define indicator functions on sets:
$$
Acdot 1_{text{statement}} =
begin{cases}
A & 1_{text{statement}}=1\
emptyset & 1_{text{statement}}=0
end{cases}
$$
And then Union over all the sets times the indicator functions.
answered Jan 23 at 23:09
Felix B.Felix B.
674317
answered Jan 23 at 23:09
Felix B.Felix B.
674317
answered Jan 23 at 23:09
Felix B.Felix B.
674317
answered Jan 23 at 23:09
Felix B.Felix B.
674317
674317
add a comment |
add a comment |
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$begingroup$
Do you mean that $B$ is a set of sets, and you want to union only the sets that are in it? (e.g. $B = {{1}, {2, 4, 6}, {-1, -2, -3}}$). Or is B a set containing similar elements to A, and you want to union the ones where A is a subset of B? Because that's a different notation.
$endgroup$
– ConMan
Jan 23 at 22:23