Set notation question: If A = {x∈ℕ: x is even } and B = {x∈ℕ: x is odd }...

[Aion]

Set notation question: If A = {x∈ℕ: x is even } and B = {x∈ℕ: x is odd }...

If A = {\(\displaystyle x \in \mathbb{N}:\) \(\displaystyle x\) is even } and B = {\(\displaystyle x \in \mathbb{N}:\) \(\displaystyle x\) is odd }

Is it valid to write this as:

A = {\(\displaystyle x \in \mathbb{N}:\exists n \in \mathbb{N}(x=2n) \)}

B = {\(\displaystyle x \in \mathbb{N}: x \in \mathbb{N}\wedge\exists m\in \mathbb{N}(x=2m+1) \)}

\(\displaystyle A \cup B\) = {\(\displaystyle x \in \mathbb{N}:\)\(\displaystyle (\exists n \in \mathbb{N}(x=2n)) \lor ( x \in \mathbb{N}\wedge\exists m\in \mathbb{N}(x=2m+1)) \)}\(\displaystyle =\mathbb{N}\)

\(\displaystyle A \cap B = \) {\(\displaystyle x \in \mathbb{N}:\)\(\displaystyle (\exists n \in \mathbb{N}(x=2n)) \wedge ( x \in \mathbb{N}\wedge\exists m\in \mathbb{N}(x=2m+1)) \)} = \(\displaystyle \varnothing\)

 

[Dr.Peterson]

If A = {\(\displaystyle x \in \mathbb{N}:\) \(\displaystyle x\) is even } and B = {\(\displaystyle x \in \mathbb{N}:\) \(\displaystyle x\) is odd }

Is it valid to write this as:

A = {\(\displaystyle x \in \mathbb{N}:\exists n \in \mathbb{N}(x=2n) \)}

B = {\(\displaystyle x \in \mathbb{N}: x \in \mathbb{N}\wedge\exists m\in \mathbb{N}(x=2m+1) \)}

\(\displaystyle A \cup B\) = {\(\displaystyle x \in \mathbb{N}:\)\(\displaystyle (\exists n \in \mathbb{N}(x=2n)) \lor ( x \in \mathbb{N}\wedge\exists m\in \mathbb{N}(x=2m+1)) \)}\(\displaystyle =\mathbb{N}\)

\(\displaystyle A \cap B = \) {\(\displaystyle x \in \mathbb{N}:\)\(\displaystyle (\exists n \in \mathbb{N}(x=2n)) \wedge ( x \in \mathbb{N}\wedge\exists m\in \mathbb{N}(x=2m+1)) \)} = \(\displaystyle \varnothing\)

You repeat "x in N" in several places where it is not needed, because that is already known; but otherwise I see nothing wrong. I would say this:

\(\displaystyle A = \{x \in \mathbb{N}:\exists n \in \mathbb{N}(x=2n) \}\)

\(\displaystyle B = \{x \in \mathbb{N}: \exists m\in \mathbb{N}(x=2m+1) \}\)

\(\displaystyle A \cup B = \{x \in \mathbb{N}: \exists n \in \mathbb{N}(x=2n) \lor \exists m\in \mathbb{N}(x=2m+1) \}=\mathbb{N}\)

\(\displaystyle A \cap B = \{x \in \mathbb{N}: \exists n \in \mathbb{N}(x=2n) \wedge \exists m\in \mathbb{N}(x=2m+1)\} = \varnothing\)