I just started a course of discrete mathematics. We’ve seen basic sets and operations (union, intersection, etc...) with easy examples. But, I now have an homework in which I have to proceed with unions and intersections where the sets contains functions. I'm not so sure of what to do, I'm not asking to make this question, but if someone could help me by providing an example, it would be very nice. Thank you.

As an example, $$ \begin{align} A &= \{2x+3\ |\ x \in \{1,2,3,4\}\} \\ B &= \{2x + 1\ |\ x \in \mathbb{N}\} \\ C &= \{4x + 1\ |\ x \in \mathbb{N}\} \end{align} $$

Give a definition of the union A and B and it's cardinality

Give a definition of the intersection between A and B and it's cardinality

  • $\begingroup$ One way to look at, say, the union $B \cup C$ and the intersection $B \cap C$, would be to note that any element $4x+1 \in C$ can be written as $2 \cdot (2x) +1$ for some $x \in \mathbb{N}$. Since $(2x) \in \mathbb{N}$ as well, this means that $4x+1 \in B$ for all $x \in \mathbb{N}$. That is, $C \subseteq B$ and then, $B \cup C = B$. However, not every element in $B$ is in $C$. Namely, if $x$ is an odd number, we have $x=2k+1$ for some $k \in \mathbb{N}$. Then $2x+1=2(2k+1)+1=(4k+1)+1$, and $4k+1$ is not divisible by 4. But if $x$ is an even number then $2x+1 \in C$. Thus, $B \cap C = C$. $\endgroup$ – B. Núñez Sep 23 '18 at 17:58
  • $\begingroup$ @BastiánNúñez Let's say I did A union B, if I added the functions : (2x + 3) + (2x + 1) and defined that x can be any N, is it the right way to do it? $\endgroup$ – monad Sep 23 '18 at 18:09
  • $\begingroup$ No, since $(2x+3)+(2x+1)=4x+4$, and that is always an even number and the sets $A$, $B$ and $C$ only contain odd numbers, so an element of the form $4x+4$ cannot belong to any of those sets. @Paul's answer shows a better way to do this intuitively. $\endgroup$ – B. Núñez Sep 23 '18 at 20:04

I think the easiest way is to write it out:

$$ \begin{array}{} A &=& \{2x+3\ |\ x \in \{1,2,3,4\}\} &=& \{5, 7, 9, 11\}\\ B &=& \{2x + 1\ |\ x \in \mathbb{N}\} &=& \{3,5,7,9,11,13,15,\dots\}\\ C &=& \{4x + 1\ |\ x \in \mathbb{N}\} &=& \{5,9,13,17,21,\dots\} \end{array} $$

Now you can apply union and intersection similar to easy examples.

If it's not clear I'll expand the answer.

(note: There are two common definitions of $\mathbb{N}$: one where $0 \in \mathbb{N}$ and one where $0 \not\in \mathbb{N}$. I assumed the latter.)

  • 1
    $\begingroup$ Thank you, I realised after giving x a value. $\endgroup$ – monad Sep 28 '18 at 1:35

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