Question

Cardinality properties:
Suppose you have three sets A, B, and C, satisfying the following conditions:
\( \# (A\cap B)=11 \)
\( \# (A\cap C)=12\)
\( \# (A\cap B \cap C)=5\)
What is the minimun cardinality of set A?

Answer 1

I thought that it would be (11 - 5) + (12 - 5) = 13

Answer 2

I'm not sure if I'm right and I don't know if the fact that B a C are disjoint set could afect the cardinality.

Answer 3

And also I don't know if I can figure out the maximun cardinality with this information. Or even the cardinality of set A.

Answer 4

Can you see the math symbols? I think because I can't =/

Answer 5

well, if B and C are disjoint, wouldn't it make sense that the cardinality of A is at least \[(A \cap B) + (A \cap C)\]? Because the minimum cardinality of B has to be 11, which means that A is at least 11, but there are also 12 elements that are not in B but have to be in A as well.

Answer 6

If they aren't disjoint, your solution of 13 sounds right.

Answer 7

@aacehm You mean \( #(A \cap B) + #(A \cap C) \)

Answer 8

Yeah, but the number signs aren't working.

Answer 9

sorry

Answer 10

Is this equality right?
\( \#A\cap(B\cap C) = \#A + \# (B\cap C) \)

Answer 11

This is so confusing =/

Answer 12

That equality doesn't sound right.\[A = \left\{5, 6, 7..100\right\}\]
\[B = \left\{1, 2, 3\right\}\]
\[C = \left\{3,4,5\right\}\]

Then the left side of your equation would 1, and the right side would be 97

Answer 13

is there information given about which sets are disjoint?

Answer 14

Actually, from statement 3, you can infer that B and C aren't disjoint, because if they were, then it would equal 23, not 5

Answer 15

The minimum cardinality of A is
\( \#(A\cap B)\cup (A \cap C)=\#(A\cap B) +\#(A \cap C) - \#(A\cap B \cap C)=11+12-5=18 \)

Answer 16

Thank you, this is what i got:
\( A\cap(B\cup C)\subseteq A\)
Then
\(\#A\geq \#[ A\cap(B\cup C)]\)
\(\#[ A\cap(B\cup C)]=\#[(A\cap B)\cup(A\cap C)]\)
\(\#[ A\cap(B\cup C)]=\#(A\cap B)+\#(A\cap C)-\#(A\cap\ B\cap C)\)
\(\#[ A\cap(B\cup C)]=11+12-5=18\)

Answer 17

Then as you said @mathmate the miminum cardinality of A is 18

Answer 18

Yep, that's very convincing! Good job!