Easy proof that $\{(x,y) \in \mathbb{R}^2 : y=\tan(x) \}$ is a closed set. I'd like to know whether there's an &quot;easy&quot; proof that $$A:= \{(x,y) \in \mathbb{R}^2 : y=\tan(x) \} .$$
I've tried to prove that its complement is open, but given an $(x,y)$ such that $\tan(x) \neq y$, it's a bit of a grind to find (in the general case) an open set that contains $(x,y)$ and does not intersect $A$. Is there a simpler argument using basic euclidean topology?
Some of the comments here gave me an idea: If $U:=\{{(x,y) \in \mathbb{R}^2: cos(x) \neq 0}\}$ and $f:U -&gt; \mathbb{R}$ is such that $f(x,y)=y-tan(x)$, then$A=f^{-1}(\{0\})$, therefore A is closed in U. Does this imply that A is closed in \mathbb{R}^2, though?
If $p\in A^c:=\Bbb R^2\setminus A$, let the perpendicular from $p$ to $A$ have length $r$: then the open neighbourhood of $p$ of radius $r$ is a subset of $A^c$, so $A^c$ is open. You can formalize the result that $r&gt;0$ with $A$'s radius of curvature.
You know that
Introducing $A_k=] -\pi/2+k\pi,\pi/2+k\pi[$
$$ f(x, y) =y-tan(x) $$
$f$ is continous on each set$G_k={ A_k} \times \mathbb{R} $ where sit the zeros.
$$ A=f^{-1}(\{0\}) $$
because $f$ is continous on each$G_k$ where sit the zeros and the reciprocal image of a closed set by a continous function is closed hence
A is closed in $\cup G_k$
In order to get to the slides take $(x_n,y_n)=((\pi/2+1/n) ,\tan(1/n-\pi/2)) $
$$y_n-\tan(x_n) = \dfrac{\sin(\pi)} {\cos(\pi/2 +1/n)\cos(\pi/2-1/n)} =0$$
So you can complete the missing lines...