I don´t know what property of sets is being used in this exercise:


$A = \{4, 5, 6, 7\}$

$B = \{6, 8\}$

$C = \{4, 7, 9\}$

Determine: $A \cup (B \times C)$:

I just did $B \times C$ and but I don't understand how can you make union between a set and another set but with ordered pairs.

Also, I have a set $A$ and another one $B, \vert B \vert = 3.$ I need to obtain $\vert A \vert$ based of the fact that there are $4096$ relations between $A$ and $B$. I did $\vert A \times B \vert = \vert A \vert \vert B \vert = 4096$ but when I get $\vert A \vert$ it makes no sense because $\vert A \vert$ would be $1365.33.$

I hope someone can help me, I would appreciate it a lot. Thanks.

  • 3
    $\begingroup$ For example, $A=\{a,3,e\}$ and $B=\{(1,0),(\alpha,x)\}$, then $$A\cup B=\{ a,3,e,(1,0),(\alpha,x)\}$$ $\endgroup$
    – azif00
    Sep 23 '19 at 2:40
  • $\begingroup$ To help with the relation question, all relations between $A$ and $B$ are actually subsets of $|A \times B|$. How many subsets are there of a given set? $\endgroup$ Sep 23 '19 at 2:42
  • $\begingroup$ Just like that?! Hahaha, I was overcomplicating myself. Thanks!! $\endgroup$ Sep 23 '19 at 2:42
  • $\begingroup$ @AnthonyTer You mean subsets of $A\times B$; $4096$ is the size of the number of subsets of $A\times B$. $\endgroup$
    – Vsotvep
    Sep 23 '19 at 4:40

If $A = \{apples, oranges, bananas\}$ and $B= \{3,7,9\}$ then $A\cup B = \{apples, oranges, bananas,3,7,9\}$.

There is no restriction that if $A$ is a set of fruit that the only things you can do with it is in the realm of fruit nor that if $B$ is in the realm of numbers that the only things we can do with it are in the realms of numbers.

$A = \{4,5,6,7\}$ in the realm of single numbers.

And $B\times C =\{(6,4),(6,7),(6,9),(8,4),(8,7),(8,9)\}$ in the real of ordered pairs

And so $A \cup (B\times C) =\{4,5,6,7,(6,4),(6,7),(6,9),(8,4),(8,7),(8,9)\}$ which is not restricted to either the ream of number nor the realm of ordered pairs.


As for relations. Note: one relation is subset of $A\times B$. A relation is not an element of $A\times B$.

Example if $A = \mathbb N$ and $B=\mathbb N$ and the relation is "the first number divides the second" then "the first number divides the second" $\ne (a,b)$ for any one pair $(a,b)$.

Instead "the first number divides the second" $= \{(1,2),(5,15), (7,35),,.......\}$ which is a whole subset of ordered pairs.

So if $C= \{1,5\}$ and $D=\{2,3,5\}$

Then what are the possible relations.

First is: $\emptyset$. That is no number is related to any other.

The second is $\{(1,2)\}$. That is $1$ is related to $2$ but nothing else are.

Another is $\{(1,2),(1,5), (5,3),(5,2)\}$. Which is a arbitrary $1$ is relateed to $2$ and $5$ and $5$ is related to $3$ and $2$. Why? Who cares?

The last is $\{(1,2),(1,3),(1,5),(5,2),(5,3),(5,5)\} = C\times D$. Everything is related to everything!

So how many total relations are there?

Well every subset of $C\times D$ can be a relation. so if $\mathscr P(C\times D)=\{$ the subsets of $C\times D\}$ then the number of relations is $|\mathscr P(C\times D)|$.

Which is a different concept then $|C\times D|$.


So you figure that if $|A| = k$ and $|B| =3$ then $|A\times B| = 3k$. That's fine.

But $4096 = |\mathscr P(A\times B)|$.

So you need to figure out if $|A\times B| = 3k$ then what does $|\mathscr P(A\times B)| = f(3k)$ is.

Can you?

Hint: If I eyeball factor $4096$ I get that $4|4096$ so $4096 = 4*1024$ and if I eyeball factor $1024$ I see it is divisible by $2$ so $4096 = 8*512$ and if I eyeball factor I get .... hey, wait a minute!


You are on the right track, Andres. Given $A=\{4,5,6,7\},$ $B=\{6,8\},$ $C=\{4,7,9\},$

$B \times C = \{ (6,4), (6,7), (6,9), (8,4), (8,7), (8,9) \}$.

The fact that the members of $B \times C$ are ordered pairs does not affect the union operation. We still end up with another set whose members belong to $A$ or $B \times C$ or both.

Hence, $A \cup (B \times C) =\{4,5,6,7,(6,4), (6,7), (6,9), (8,4), (8,7), (8,9)\}$

Note that when you draw a comparison between $6\in A$ and $(6,7)\in B \times C$, you are not drawing a comparison with anything inside the ordered pair. We are drawing a comparison between MEMBERS of $A$, which are the integers $4,5,6,7$, and the MEMBERS of $B\times C$, which are the ordered pairs themselves. The integer $6$ is not the same as the ordered pair $(6,7)$ - even though the ordered pair contains $6$. The two are fundamentally different "things," so we include $6$ in our union and we also include $(6,7)$ in our union.


As for your question re the relation... we are given sets $A$ and $B$ s.t. $|B|=3$ and we know there are $4096$ relations between $A$ and $B$. There are a few things you need to be aware of in order to solve this:

(1) Definition of a relation, a subset, a Cartesian product, and a powerset.

(2) How to determine the cardinality of a Cartesian product and a powerset.

Recall that a relation from $A$ to $B$ is a subset of $A \times B$. We are told there are $4096$ possible relations; in other words, there are $4096$ possible subsets of $A \times B$.

Recall that a powerset of some set $S$, denoted as $\mathfrak{P}$$(S)$, is the set of all subsets of $S$ and has a cardinality of $2^n$ where $|S|=n$. Hence, if there are $4096$ possible subsets of $A \times B$, then that means the set of all subsets of $A \times B$, or the $\mathfrak{P}$$(A \times B)$, has a cardinality of $4096.$ In other words, $|\mathfrak{P}$$(A \times B)|=4096.$

And, if $|A \times B|=n$, then $|\mathfrak{P}$$(A \times B)|=4096=2^n$. We now have an equation we can work with.

Now, also recall that $|A \times B|=|A| \cdot |B|$. We know $|B|=3$, so $|A \times B|=|A| \cdot 3$. In addition, we have already established that $|A \times B|=n$, so $n=|A| \cdot 3$. By substitution with $n$, we have

$2^n=2^{|A|\cdot 3}=4096$. Now we can solve for $|A|$ to obtain the answer to the problem.



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