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? 

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.)
