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 "easy" 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 $f:U -> \mathbb{R}$ 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?
 A: 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 $r>0$ with $A$'s radius of curvature.
A: If $X$ is any topological space and $A\subseteq X$ and $X=\bigcup_n U_n$, where $U_n$ are open, then $A$ is closed if and only if $A\cap U_n$ is closed for all $n$.
Now, if you consider $X=\mathbf R^2$ and $U_n=(-n,n)^2$, then it is clear that each $A\cap U_n$ is closed (e.g. because $A\cap \overline U_n$ is compact).
A: The graph of a continuous function $f: X \to Y$ is closed in $X \times Y$, assuming $Y$ is hausdoff (you can google this result or even better, prove it). The $\tan$ function is not continuous however but it is continuous on each interval $(-\frac{\pi}{2}+ n \pi , \frac{\pi}{2}+ n \pi )$.
Hence, you only must check that it is closed in those slices that are not covered by those intervals, so only for $x =\frac{\pi}{2}+n \pi$, so just find an open set that does not intersect $\tan$ for each $(\frac{\pi}{2}+n \pi, y)$. This is much more easier to do just using facts about the $\tan$ function.
A: Note that $A= \{(x,y) | x \notin {\pi \over 2} \mathbb{Z}, \ y = \tan x \}$.
Suppose $(x_k, \tan x_k ) \to (x,y)$. In particular, for large $k$ we have
$|y-\tan x_k| \le 1$ and so
$\tan x_k \in [y-1,y+1]$.
Hence $x_k \in \cup_{n \in \mathbb{Z}} I_n$, where $I_n = [\arctan (y-1)+n \pi, \arctan (y+1) +n \pi]$.
Note that if $n \neq n'$ and $a \in I_n, b \in I_{n'}$ then
$|a-b| \ge \pi-\max(|\arctan (y-1)|, |\arctan (y+1)|) > 0$.
Since $x_k \to x$, there must be some $n$ such that $x_k \in I_n$ for a countable number of $k$ and since $I_n$ is closed, we have $x \in I_n$.
Since $\tan$ is continuous on $I_n$, we see that $\tan x_k \to \tan x = y$
and so $(x,y) \in A$.
A: 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...
