Logical connectives in set-builder notation

Question

I'm trying to understand what bi-implication and implication mean when they're used inside sets like in

1. $$A = \{x\in \mathbb{N} \mid (x-1)(x-3)\neq 0 \iff x+1=x\}$$
2. $$B = \{x\mid x\in \mathbb{N} \land (x-1)(x-3)\neq 0 \iff x+1=x\}$$
3. $$C = \{x\in \mathbb{N} \mid x \geq 3 \iff x < 0 \}$$

What is the difference between sets $A$ and $B$ ? They both contain natural numbers and I don't see how displaying an intersection makes a difference.

If someone knows a source on the internet to better explain this or similar concepts, do share a link.

Answer 1

1. $$A = \{x\in \mathbb{N} \mid (x-1)(x-3)\neq 0 \iff x+1=x\}$$

This is the set of naturals for which that biimplication is true. This means that as we inspect each natural number, that number belongs in $A$ precisely when it satisfies both the left and right sides of the biimplication or it satisfies neither the left nor right sides of the biimplication. So, $A=\{1,3\}.$

2. $$B = \{x\mid x\in \mathbb{N} \land (x-1)(x-3)\neq 0 \iff x+1=x\}$$

Note that set $B$ is $$B = \{x\mid \Big(x\in \mathbb{N} \land (x-1)(x-3)\neq 0 \Big)\iff x+1=x\}.$$ This time, we are considering the entire universe of discourse rather than necessarily just the naturals. By the above boldfaced reasoning (but this time replacing "natural numbers" with "objects of the universe $U$"), $B=(U{\setminus}\mathbb N)\cup\{1,3\}.$ For example, $2.5\in B$ while $5\not\in B.$

However, what you probably intended is $$B_2 = \{x\mid x\in \mathbb{N} \land \color\red{\Big(}(x-1)(x-3)\neq 0 \iff x+1=x\color\red{\Big)}\},$$ where the parentheses are essential. This specification is an abbreviation of the very first one above; so, sets $A$ and $B_2$ are the same.

3. $$C = \{x\in \mathbb{N} \mid x \geq 3 \iff x < 0 \}$$

By the above boldfaced reasoning, $C=\{0,1,2\}.$

Answer 2

For 1), $x + 1 = x$ is not satisfied by any natural number, so you can substitute FALSE. For $(x-1))(x-3) \neq 0 \iff \mathrm{FALSE}$ to be satisfied, you need to make sure $(x-1)(x-3) \neq 0$ is not satisfied. That is, it must be the case that $(x-1)(x-3) = 0$. So $A = \{1,3\}$.

3. is quite similar. As for 2), I don't understand what $\cap$ means in this context. Neither “$x \in \mathbb{N}$” nor “$(x-1))(x-3) \neq 0 \iff x+1=x$” are sets.