In the usual topology. In particular, is the graph of the function $f: (-\pi/2, \pi/2) \to \mathbb{R}$, where $f(x) = \tan(x)$, closed?
edit: in the generic case is already clear for me that it is not, for example, the graph of $f(x) = e^{-x}, x>0$. But still, I'm curious about the particular case above. Or something like the graph of $e^x$, on the entire line.
On
OK, I got it for the graph of $tan(x)$. Suppose $(x_n, tan(x_n)) \in Graph(f)$. If $x_n \to \pi/2$ or $x_n \to -\pi/2$, then the sequence $(x_n, f(x_n))$ is not convergent. Therefore, a convergent sequence $(x_n, f(x_n))$ in the curve is such that $(x_n)$ converges to $x \in (-\pi/2, \pi/2)$, so $x \in [-\pi/2 + 1/n, \pi/2 - 1/n]$ for some $n$ and, therefore, $f(x_n) \to f(x)$.
On
The graph of $$f: (-\pi/2, \pi/2) \to \mathbb{R}$$ where $$ f(x) = tan(x)$$ is closed.
As you know the complement of the graph is the union of two open sets, thus the graph is closed.
The fact that a function has an asymptote does not prevent the graph to be closed.
Similarly the graph of $$ f(x)=e^x $$ on the entire real line is closed.
In general, the graph of a continuous function over a closed subset of $\mathbb{R}$ is closed.
To show this, suppose $f : C \to \mathbb{R}$ is continuous, and $(x_n, f(x_n)) \to (x, y)$. Then $x_n \to x$. Since $C$ is closed, we have $x \in C$. Since $f$ is continuous on $C$, we have $f(x_n) \to f(x)$. But $f(x_n) \to y$ too, since $(x_n, f(x_n)) \to (x, y)$. Therefore, by uniqueness of limits, $y = f(x)$, hence $(x, y)$ is in the graph of $f$.
Try thinking of counterexamples where $C$ is not closed, or when $f$ is not continuous.