# Closure of $A = \{ (x, \frac{1}{x}) | x \in \mathbb{R} \backslash 0 \}$

What does this set look like? $A = \{ (x, \frac{1}{x}) | x \in \mathbb{R} \backslash 0 \}$, where $A$ is a subset of $\mathbb{R}^2$ with the Euclidean topology.

I thought $A = (-\infty, 0] \cup [0, \infty)$ but I know that its projection $(x,y) \rightarrow x$ is not closed, so this must be wrong.

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Interpreting your notation to mean that $A$ is a certain subset of $\mathbb{R}^2$...
Hint $A$ can be treated as the graph of some function.
Thanks, I can see that $A$ is the graph of $f(x) = \frac{1}{x}$, but what does this tell me? – Rudy the Reindeer Feb 1 '11 at 11:40
You want to know if $A$ is closed as a subset of ${\Bbb R}^2$ or not, and if not find its closure. But ${\Bbb R}^2$ is a metric space. What if one applies the definition of open sets in a metric space to the complement of $A$? – Andrea Mori Feb 1 '11 at 11:49
For a point not on the graph I can draw an epsilon ball around it! So it's closed! @Willie: I gave your answer a vote and accepted it because it answers my question. I forgot to include in my original question that I also wanted to know why $A$ is closed. My bad : ) – Rudy the Reindeer Feb 1 '11 at 19:06
@Matt: The set $A$ is closed so it is equal to its own closure. To show it is closed you can use the characterization of a closed set (in metric spaces) in terms of sequences.