Set theory strange union with cross product

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

Sets:

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

$$B = \{6, 8\}$$

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

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.

• For example, $A=\{a,3,e\}$ and $B=\{(1,0),(\alpha,x)\}$, then $$A\cup B=\{ a,3,e,(1,0),(\alpha,x)\}$$ Sep 23 '19 at 2:40
• 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? Sep 23 '19 at 2:42
• Just like that?! Hahaha, I was overcomplicating myself. Thanks!! Sep 23 '19 at 2:42
• @AnthonyTer You mean subsets of $A\times B$; $4096$ is the size of the number of subsets of $A\times B$. 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.

$$|A|=(\log_24096)/3=4$$