# 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?

• First, do you think it could be open? Visualize what the curve looks like and what open balls look like in $\mathbb{R}^2$. Also, remember that to show something is closed, you can show that its complement is open. What would be the complement of a curve? Commented Apr 13, 2014 at 19:54
• Also, do you think this curve has any limit points in the plane that are outside of the curve itself? You can restate this as: can you find a sequence of points on this curve that converge to some point off of the the curve? If not, then the curve contains all of its limit points and so is closed.
– MPW
Commented Apr 13, 2014 at 20:09
• @RyanSullivant: Your first question is quite misleading. Closed is NOT not open. Commented Apr 13, 2014 at 20:18
• "Sets are not doors" :) Commented Apr 13, 2014 at 20:19
• @RyanSullivant, i'm not sure how to visualize or interpret the complement of a curve. Could you expand upon that. Also, I do realize sets are not doors. :P Commented Apr 13, 2014 at 21:00

The curve is the inverse image of a closed set by a continuous function: $$g(x,y)=xy,$$ $$\text{graph}=g^{-1}(\{1\}).$$
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 $G$.
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.)