Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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 $\phi(x,y) = xy$).

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.