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

#### Aion

##### New member
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 = {$$\displaystyle x \in \mathbb{N}: x \in \mathbb{N}\wedge\exists m\in \mathbb{N}(x=2m+1)$$}

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

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

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#### Dr.Peterson

##### Elite Member
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:

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

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

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

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