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

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

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

Is it valid to write this as:

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

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

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

AB=\displaystyle A \cap B = {xN:\displaystyle x \in \mathbb{N}:(nN(x=2n))(xNmN(x=2m+1))\displaystyle (\exists n \in \mathbb{N}(x=2n)) \wedge ( x \in \mathbb{N}\wedge\exists m\in \mathbb{N}(x=2m+1)) } = \displaystyle \varnothing
 
Last edited:
If A = {xN:\displaystyle x \in \mathbb{N}: x\displaystyle x is even } and B = {xN:\displaystyle x \in \mathbb{N}: x\displaystyle x is odd }

Is it valid to write this as:

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

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

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

AB=\displaystyle A \cap B = {xN:\displaystyle x \in \mathbb{N}:(nN(x=2n))(xNmN(x=2m+1))\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:

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

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

AB={xN:nN(x=2n)mN(x=2m+1)}=N\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}

AB={xN:nN(x=2n)mN(x=2m+1)}=\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
 
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