I am trying to write out the domain of the function $\tan x$ in set-builder notation. I have seen this written out as $$\left\{x \in \mathbb{R} \mid (\exists n \in \mathbb{Z})\left[x \neq \frac{\pi}{2} + \pi n\right]\right\},$$ with the existential quantifier. My question is that since the domain excludes an infinite number of asymptotes, would it not be more correct to use the universal quantifier to write the domain like so? $$\left\{x \in \mathbb{R} \mid (\forall n \in \mathbb{Z})\left[x \neq \frac{\pi}{2} + \pi n\right]\right\}$$
2 Answers
You are correct, you should have a universal quantifier. The first set you are describing is just $\mathbb R$.
(Indeed, given any $x$, I can find an $n$ such that $x\neq\frac\pi2 +\pi n$. Just take $n=x$ and it is trivially true.)
Although personally I would just write the domain more concisely as $$\mathbb R\smallsetminus \{(2n+1)\tfrac\pi2 : n\in\mathbb Z\},$$ or just as $$\mathbb R\smallsetminus (\pi\mathbb Z+\tfrac\pi2)$$ if you're comfortable with Minkowski set operations.
Note. If you want to use an existential quantifier, you would have to rephrase the condition as a negation, i.e., as $\lnot\big[(\exists n\in\mathbb Z)(x=\tfrac\pi2+\pi n)\big]$.
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1Perhaps $\Bbb R\smallsetminus\bigl(\frac\pi2+\Bbb Z\pi\bigr)$ is a bit simpler. – Bernard Nov 10 '20 at 22:26
Yes, the second domain is correct. Note that the first domain includes $\pi/2$ because for $n=1,n\pi+\pi/2=3\pi/2\ne\pi/2$.
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