The Cardinality of set
$begingroup$
about the cardinality of these sets?
(a) $(A cup B)$
(b) $(A cap B)$
(c) $A times B$.
(d) $A smallsetminus B$.
(e) $P(A cup B)$
I know if I want to find the union I use Cardinal Number formula:
$$n(Acup B) = n(A) + n(B) - n(Acap B)$$
and for intersection
$n(Acap B) = n(A) + n(B) - n(Acup B)$
but I don't have all information only $A$ and $B$ . how to get cardinality in this way ?
elementary-set-theory
asked Jan 18 at 22:01
JaredJared
356
$endgroup$
add a comment |
$begingroup$
about the cardinality of these sets?
(a) $(A cup B)$
(b) $(A cap B)$
(c) $A times B$.
(d) $A smallsetminus B$.
(e) $P(A cup B)$
I know if I want to find the union I use Cardinal Number formula:
$$n(Acup B) = n(A) + n(B) - n(Acap B)$$
and for intersection
$n(Acap B) = n(A) + n(B) - n(Acup B)$
but I don't have all information only $A$ and $B$ . how to get cardinality in this way ?
elementary-set-theory
asked Jan 18 at 22:01
JaredJared
356
$endgroup$
3
$begingroup$
"if not, explain why not; in that case, what can be said about the cardinality of these sets?" You can not find $|Acup B|$ when only given $|A|$ and $|B|$.
$endgroup$
– SmileyCraft
Jan 18 at 22:04
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 18 at 22:23
add a comment |
$begingroup$
about the cardinality of these sets?
(a) $(A cup B)$
(b) $(A cap B)$
(c) $A times B$.
(d) $A smallsetminus B$.
(e) $P(A cup B)$
I know if I want to find the union I use Cardinal Number formula:
$$n(Acup B) = n(A) + n(B) - n(Acap B)$$
and for intersection
$n(Acap B) = n(A) + n(B) - n(Acup B)$
but I don't have all information only $A$ and $B$ . how to get cardinality in this way ?
elementary-set-theory
asked Jan 18 at 22:01
JaredJared
356
$endgroup$
about the cardinality of these sets?
(a) $(A cup B)$
(b) $(A cap B)$
(c) $A times B$.
(d) $A smallsetminus B$.
(e) $P(A cup B)$
I know if I want to find the union I use Cardinal Number formula:
$$n(Acup B) = n(A) + n(B) - n(Acap B)$$
and for intersection
$n(Acap B) = n(A) + n(B) - n(Acup B)$
but I don't have all information only $A$ and $B$ . how to get cardinality in this way ?
elementary-set-theory
elementary-set-theory
asked Jan 18 at 22:01
JaredJared
356
asked Jan 18 at 22:01
JaredJared
356
edited Jan 21 at 4:15
Jared
asked Jan 18 at 22:01
JaredJared
356
asked Jan 18 at 22:01
JaredJared
356
asked Jan 18 at 22:01
JaredJared
356
356
3
$begingroup$
"if not, explain why not; in that case, what can be said about the cardinality of these sets?" You can not find $|Acup B|$ when only given $|A|$ and $|B|$.
$endgroup$
– SmileyCraft
Jan 18 at 22:04
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 18 at 22:23
add a comment |
3
$begingroup$
"if not, explain why not; in that case, what can be said about the cardinality of these sets?" You can not find $|Acup B|$ when only given $|A|$ and $|B|$.
$endgroup$
– SmileyCraft
Jan 18 at 22:04
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 18 at 22:23
3
3
$begingroup$
"if not, explain why not; in that case, what can be said about the cardinality of these sets?" You can not find $|Acup B|$ when only given $|A|$ and $|B|$.
$endgroup$
– SmileyCraft
Jan 18 at 22:04
$begingroup$
"if not, explain why not; in that case, what can be said about the cardinality of these sets?" You can not find $|Acup B|$ when only given $|A|$ and $|B|$.
$endgroup$
– SmileyCraft
Jan 18 at 22:04
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 18 at 22:23
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 18 at 22:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider two scenarios:
$A = {1,2,3,4,5}$ and $B = {6,7,8,9,10,11,12}$
$A = {1,2,3,4,5}$ and $B={1,2,3,4,5,6,7}$
It should be clear that you get different answers for the cardinalities for each except for the cartesian product where you do get the same answer for each. You are however able to find bounds for the cardinalities of each expression. You should be able to argue, using exactly the inclusion-exclusion principle that you mention in your post ($|Acup B|=|A|+|B|-|Acap B|$) that the various cardinalities are either minimized or maximized when there is as much overlap as possible or no overlap at all.
answered Jan 18 at 22:08
JMoravitzJMoravitz
49.4k44091
$endgroup$
$begingroup$
Would you please explain more? So I need to argue can't get the cardinality for all the them?
$endgroup$
– Jared
Jan 18 at 22:22
$begingroup$
@Jared If we were to suppose that $|Acup B|$ is possible to determine from the information $|A|$ and $|B|$ alone, then it would have been the case that regardless what we let $A$ or $B$ actually equal that $|Acup B|$ would have been the same. Since we have two examples here and in each of those examples $|Acup B|$ is equal to something different and not the same, that directly implies that $|A|$ and $|B|$ alone are not enough information to uniquely determine $|Acup B|$. No further argument or proof is necessary.
$endgroup$
– JMoravitz
Jan 18 at 22:31
$begingroup$
@Jared As alluded to above, we can however find bounds for the sizes of the various sets. For example $max(|A|,|B|)leq |Acup B| = |A|+|B|-|Acap B| leq |A|+|B|$ so we learn that $7leq |Acup B|leq 12$. Further, the examples I outline above actually achieve those bounds so we know the bounds are strict. You can come up with similar bounds for the other expressions, and you can argue as alluded to earlier that the examples where $|Acap B|$ is either minimized or maximized achieve the extremes of the bounds.
$endgroup$
– JMoravitz
Jan 18 at 22:34
$begingroup$
Thanks , but for A×B and A- B Difference that can be determine?
$endgroup$
– Jared
Jan 18 at 22:40
$begingroup$
@Jared $|Atimes B|= |A|times |B|$ in every case. $|Asetminus B|$ however depends on how much overlap there is, again shown by the examples I provided above. Note that $|A|=|Acap B|+|Asetminus B|$
$endgroup$
– JMoravitz
Jan 18 at 22:41
|
show 3 more comments
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Consider two scenarios:
$A = {1,2,3,4,5}$ and $B = {6,7,8,9,10,11,12}$
$A = {1,2,3,4,5}$ and $B={1,2,3,4,5,6,7}$
It should be clear that you get different answers for the cardinalities for each except for the cartesian product where you do get the same answer for each. You are however able to find bounds for the cardinalities of each expression. You should be able to argue, using exactly the inclusion-exclusion principle that you mention in your post ($|Acup B|=|A|+|B|-|Acap B|$) that the various cardinalities are either minimized or maximized when there is as much overlap as possible or no overlap at all.
answered Jan 18 at 22:08
JMoravitzJMoravitz
49.4k44091
$endgroup$
$begingroup$
Would you please explain more? So I need to argue can't get the cardinality for all the them?
$endgroup$
– Jared
Jan 18 at 22:22
$begingroup$
@Jared If we were to suppose that $|Acup B|$ is possible to determine from the information $|A|$ and $|B|$ alone, then it would have been the case that regardless what we let $A$ or $B$ actually equal that $|Acup B|$ would have been the same. Since we have two examples here and in each of those examples $|Acup B|$ is equal to something different and not the same, that directly implies that $|A|$ and $|B|$ alone are not enough information to uniquely determine $|Acup B|$. No further argument or proof is necessary.
$endgroup$
– JMoravitz
Jan 18 at 22:31
$begingroup$
@Jared As alluded to above, we can however find bounds for the sizes of the various sets. For example $max(|A|,|B|)leq |Acup B| = |A|+|B|-|Acap B| leq |A|+|B|$ so we learn that $7leq |Acup B|leq 12$. Further, the examples I outline above actually achieve those bounds so we know the bounds are strict. You can come up with similar bounds for the other expressions, and you can argue as alluded to earlier that the examples where $|Acap B|$ is either minimized or maximized achieve the extremes of the bounds.
$endgroup$
– JMoravitz
Jan 18 at 22:34
$begingroup$
Thanks , but for A×B and A- B Difference that can be determine?
$endgroup$
– Jared
Jan 18 at 22:40
$begingroup$
@Jared $|Atimes B|= |A|times |B|$ in every case. $|Asetminus B|$ however depends on how much overlap there is, again shown by the examples I provided above. Note that $|A|=|Acap B|+|Asetminus B|$
$endgroup$
– JMoravitz
Jan 18 at 22:41
|
show 3 more comments
$begingroup$
Consider two scenarios:
$A = {1,2,3,4,5}$ and $B = {6,7,8,9,10,11,12}$
$A = {1,2,3,4,5}$ and $B={1,2,3,4,5,6,7}$
It should be clear that you get different answers for the cardinalities for each except for the cartesian product where you do get the same answer for each. You are however able to find bounds for the cardinalities of each expression. You should be able to argue, using exactly the inclusion-exclusion principle that you mention in your post ($|Acup B|=|A|+|B|-|Acap B|$) that the various cardinalities are either minimized or maximized when there is as much overlap as possible or no overlap at all.
answered Jan 18 at 22:08
JMoravitzJMoravitz
49.4k44091
$endgroup$
$begingroup$
Would you please explain more? So I need to argue can't get the cardinality for all the them?
$endgroup$
– Jared
Jan 18 at 22:22
$begingroup$
@Jared If we were to suppose that $|Acup B|$ is possible to determine from the information $|A|$ and $|B|$ alone, then it would have been the case that regardless what we let $A$ or $B$ actually equal that $|Acup B|$ would have been the same. Since we have two examples here and in each of those examples $|Acup B|$ is equal to something different and not the same, that directly implies that $|A|$ and $|B|$ alone are not enough information to uniquely determine $|Acup B|$. No further argument or proof is necessary.
$endgroup$
– JMoravitz
Jan 18 at 22:31
$begingroup$
@Jared As alluded to above, we can however find bounds for the sizes of the various sets. For example $max(|A|,|B|)leq |Acup B| = |A|+|B|-|Acap B| leq |A|+|B|$ so we learn that $7leq |Acup B|leq 12$. Further, the examples I outline above actually achieve those bounds so we know the bounds are strict. You can come up with similar bounds for the other expressions, and you can argue as alluded to earlier that the examples where $|Acap B|$ is either minimized or maximized achieve the extremes of the bounds.
$endgroup$
– JMoravitz
Jan 18 at 22:34
$begingroup$
Thanks , but for A×B and A- B Difference that can be determine?
$endgroup$
– Jared
Jan 18 at 22:40
$begingroup$
@Jared $|Atimes B|= |A|times |B|$ in every case. $|Asetminus B|$ however depends on how much overlap there is, again shown by the examples I provided above. Note that $|A|=|Acap B|+|Asetminus B|$
$endgroup$
– JMoravitz
Jan 18 at 22:41
|
show 3 more comments
$begingroup$
Consider two scenarios:
$A = {1,2,3,4,5}$ and $B = {6,7,8,9,10,11,12}$
$A = {1,2,3,4,5}$ and $B={1,2,3,4,5,6,7}$
It should be clear that you get different answers for the cardinalities for each except for the cartesian product where you do get the same answer for each. You are however able to find bounds for the cardinalities of each expression. You should be able to argue, using exactly the inclusion-exclusion principle that you mention in your post ($|Acup B|=|A|+|B|-|Acap B|$) that the various cardinalities are either minimized or maximized when there is as much overlap as possible or no overlap at all.
answered Jan 18 at 22:08
JMoravitzJMoravitz
49.4k44091
$endgroup$
Consider two scenarios:
$A = {1,2,3,4,5}$ and $B = {6,7,8,9,10,11,12}$
$A = {1,2,3,4,5}$ and $B={1,2,3,4,5,6,7}$
It should be clear that you get different answers for the cardinalities for each except for the cartesian product where you do get the same answer for each. You are however able to find bounds for the cardinalities of each expression. You should be able to argue, using exactly the inclusion-exclusion principle that you mention in your post ($|Acup B|=|A|+|B|-|Acap B|$) that the various cardinalities are either minimized or maximized when there is as much overlap as possible or no overlap at all.
answered Jan 18 at 22:08
JMoravitzJMoravitz
49.4k44091
answered Jan 18 at 22:08
JMoravitzJMoravitz
49.4k44091
answered Jan 18 at 22:08
JMoravitzJMoravitz
49.4k44091
answered Jan 18 at 22:08
JMoravitzJMoravitz
49.4k44091
49.4k44091
$begingroup$
Would you please explain more? So I need to argue can't get the cardinality for all the them?
$endgroup$
– Jared
Jan 18 at 22:22
$begingroup$
@Jared If we were to suppose that $|Acup B|$ is possible to determine from the information $|A|$ and $|B|$ alone, then it would have been the case that regardless what we let $A$ or $B$ actually equal that $|Acup B|$ would have been the same. Since we have two examples here and in each of those examples $|Acup B|$ is equal to something different and not the same, that directly implies that $|A|$ and $|B|$ alone are not enough information to uniquely determine $|Acup B|$. No further argument or proof is necessary.
$endgroup$
– JMoravitz
Jan 18 at 22:31
$begingroup$
@Jared As alluded to above, we can however find bounds for the sizes of the various sets. For example $max(|A|,|B|)leq |Acup B| = |A|+|B|-|Acap B| leq |A|+|B|$ so we learn that $7leq |Acup B|leq 12$. Further, the examples I outline above actually achieve those bounds so we know the bounds are strict. You can come up with similar bounds for the other expressions, and you can argue as alluded to earlier that the examples where $|Acap B|$ is either minimized or maximized achieve the extremes of the bounds.
$endgroup$
– JMoravitz
Jan 18 at 22:34
$begingroup$
Thanks , but for A×B and A- B Difference that can be determine?
$endgroup$
– Jared
Jan 18 at 22:40
$begingroup$
@Jared $|Atimes B|= |A|times |B|$ in every case. $|Asetminus B|$ however depends on how much overlap there is, again shown by the examples I provided above. Note that $|A|=|Acap B|+|Asetminus B|$
$endgroup$
– JMoravitz
Jan 18 at 22:41
|
show 3 more comments
$begingroup$
Would you please explain more? So I need to argue can't get the cardinality for all the them?
$endgroup$
– Jared
Jan 18 at 22:22
$begingroup$
@Jared If we were to suppose that $|Acup B|$ is possible to determine from the information $|A|$ and $|B|$ alone, then it would have been the case that regardless what we let $A$ or $B$ actually equal that $|Acup B|$ would have been the same. Since we have two examples here and in each of those examples $|Acup B|$ is equal to something different and not the same, that directly implies that $|A|$ and $|B|$ alone are not enough information to uniquely determine $|Acup B|$. No further argument or proof is necessary.
$endgroup$
– JMoravitz
Jan 18 at 22:31
$begingroup$
@Jared As alluded to above, we can however find bounds for the sizes of the various sets. For example $max(|A|,|B|)leq |Acup B| = |A|+|B|-|Acap B| leq |A|+|B|$ so we learn that $7leq |Acup B|leq 12$. Further, the examples I outline above actually achieve those bounds so we know the bounds are strict. You can come up with similar bounds for the other expressions, and you can argue as alluded to earlier that the examples where $|Acap B|$ is either minimized or maximized achieve the extremes of the bounds.
$endgroup$
– JMoravitz
Jan 18 at 22:34
$begingroup$
Thanks , but for A×B and A- B Difference that can be determine?
$endgroup$
– Jared
Jan 18 at 22:40
$begingroup$
@Jared $|Atimes B|= |A|times |B|$ in every case. $|Asetminus B|$ however depends on how much overlap there is, again shown by the examples I provided above. Note that $|A|=|Acap B|+|Asetminus B|$
$endgroup$
– JMoravitz
Jan 18 at 22:41
$begingroup$
Would you please explain more? So I need to argue can't get the cardinality for all the them?
$endgroup$
– Jared
Jan 18 at 22:22
$begingroup$
Would you please explain more? So I need to argue can't get the cardinality for all the them?
$endgroup$
– Jared
Jan 18 at 22:22
$begingroup$
@Jared If we were to suppose that $|Acup B|$ is possible to determine from the information $|A|$ and $|B|$ alone, then it would have been the case that regardless what we let $A$ or $B$ actually equal that $|Acup B|$ would have been the same. Since we have two examples here and in each of those examples $|Acup B|$ is equal to something different and not the same, that directly implies that $|A|$ and $|B|$ alone are not enough information to uniquely determine $|Acup B|$. No further argument or proof is necessary.
$endgroup$
– JMoravitz
Jan 18 at 22:31
$begingroup$
@Jared If we were to suppose that $|Acup B|$ is possible to determine from the information $|A|$ and $|B|$ alone, then it would have been the case that regardless what we let $A$ or $B$ actually equal that $|Acup B|$ would have been the same. Since we have two examples here and in each of those examples $|Acup B|$ is equal to something different and not the same, that directly implies that $|A|$ and $|B|$ alone are not enough information to uniquely determine $|Acup B|$. No further argument or proof is necessary.
$endgroup$
– JMoravitz
Jan 18 at 22:31
$begingroup$
@Jared As alluded to above, we can however find bounds for the sizes of the various sets. For example $max(|A|,|B|)leq |Acup B| = |A|+|B|-|Acap B| leq |A|+|B|$ so we learn that $7leq |Acup B|leq 12$. Further, the examples I outline above actually achieve those bounds so we know the bounds are strict. You can come up with similar bounds for the other expressions, and you can argue as alluded to earlier that the examples where $|Acap B|$ is either minimized or maximized achieve the extremes of the bounds.
$endgroup$
– JMoravitz
Jan 18 at 22:34
$begingroup$
@Jared As alluded to above, we can however find bounds for the sizes of the various sets. For example $max(|A|,|B|)leq |Acup B| = |A|+|B|-|Acap B| leq |A|+|B|$ so we learn that $7leq |Acup B|leq 12$. Further, the examples I outline above actually achieve those bounds so we know the bounds are strict. You can come up with similar bounds for the other expressions, and you can argue as alluded to earlier that the examples where $|Acap B|$ is either minimized or maximized achieve the extremes of the bounds.
$endgroup$
– JMoravitz
Jan 18 at 22:34
$begingroup$
Thanks , but for A×B and A- B Difference that can be determine?
$endgroup$
– Jared
Jan 18 at 22:40
$begingroup$
Thanks , but for A×B and A- B Difference that can be determine?
$endgroup$
– Jared
Jan 18 at 22:40
$begingroup$
@Jared $|Atimes B|= |A|times |B|$ in every case. $|Asetminus B|$ however depends on how much overlap there is, again shown by the examples I provided above. Note that $|A|=|Acap B|+|Asetminus B|$
$endgroup$
– JMoravitz
Jan 18 at 22:41
$begingroup$
@Jared $|Atimes B|= |A|times |B|$ in every case. $|Asetminus B|$ however depends on how much overlap there is, again shown by the examples I provided above. Note that $|A|=|Acap B|+|Asetminus B|$
$endgroup$
– JMoravitz
Jan 18 at 22:41
|
show 3 more comments
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"if not, explain why not; in that case, what can be said about the cardinality of these sets?" You can not find $|Acup B|$ when only given $|A|$ and $|B|$.
$endgroup$
– SmileyCraft
Jan 18 at 22:04
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 18 at 22:23