Are curves closed in $\mathbb{R}\times \mathbb{R}$ with the standard topology? Given the graph of the curve $y=\frac{1}{x}$, can we determine if the curve is closed or open in $\mathbb{R}^{2}$ with the standard topology? 
 A: The curve is the inverse image of a closed set by a continuous function:
$$g(x,y)=xy,$$
$$\text{graph}=g^{-1}(\{1\}).$$
A: Perhaps the easiest way to show closedness of 
$$G=\{(x,\ 1/x)\mid x\ne 0\}\subset\Bbb R^2$$
is to note that it is the preimage of $\{1\}$ under the map 
$$\Bbb R\times\Bbb R→\Bbb R,\qquad (x,y)\mapsto x\cdot y$$
and this map is continuous and $\{1\}$ is closed.  
Alternatively, note that $G=G(f)$, the graph of the continuous map
$$f:\Bbb R\setminus\{0\}\to\Bbb R,\qquad x\mapsto f(x)=1/x$$
and if $f:X→Y$ is continuous and $Y$ is Hausdorff, then $G(f)$ is closed in $X×Y$. This, however, only gives closedness of $G$ in $\Bbb R\setminus\{0\}×\Bbb R$. You still had to show that no point in $\{0\}×\Bbb R$ is in the closure of $G$.
A: Note that $y = \frac{1}{x}$ has a vertical asymptote at $x = 0$, and really consists of two disjoint curves (in quadrants one and three). Picture the plane $\mathbb R^2$ with these two disjoint curves deleted. This is the complemement of $y = \frac{1}{x}$. Try to prove that this complement is open by showing that each of its points is an interior point. (I would take an $\epsilon$-neighborhood about each point $x$ in the complement, such that $\epsilon$ is less than the shortest distance from $x$ to the curve.)
