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If I have $\{ (x, 1/x) \in \mathbb{R}^n : 0 < x \leq 1 \}$, is this set closed?

I know that almost every point is a limit point (I drew the graph in the first quadrant), but should I test whether 0 has a neighbourhood that contains other points in the set? Or is it okay I can forget about it since it isn't even in the set anyways?

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up vote 1 down vote accepted

Yes, since $\{ (x, 1/x) \in \mathbb{R}^n : 0 < x \leq 1 \} = ([0,1]\times\mathbb{R}) \cap \{(x,y) | x y = 1 \}$, and both of the sets on the right hand side are clearly closed (the latter set being $\phi^{-1}\{1\}$, where $\phi(x,y) = xy$).

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The set is closed. It suffices to show that its complement is open. This is easy to see because any point off the graph is contained in an open ball that does not intersect the graph.

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