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I was calculating a domain of a function $f(x,y)$ and I need to say if the domain is an open set or closed set, and if it is bounded.

At the end of my calculations, I got $xy \geq 1$, which is the correct domain.

The final answer in the book said it is closed and not bounded.

I wanted to ask you guys, how can a set of point be infinite and still be closed ?

The attached photo shows the domain of the function

From single variable calculus, I know that for example $[a, +\infty)$ is an open set, since we infinity can't be equal to anything, so why is it different with two variables. And if there is a mistake and it is not closed, than what's the difference between bounded and closed then ?

Thank you !

essay
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user149107
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  • There is a difference between "infinite" and "unbounded." Be careful not to confuse the two. – Thomas Andrews May 10 '14 at 14:39
  • When one say "closed/half-open/open interval" one mean something else, just that the endpoints belong or not to the interval. However, in topology the notion of open/closed set is more involved. For example, the "open interval" $(-\infty,+\infty)$ is actually a closed set. – A.Γ. Jul 13 '15 at 01:31

2 Answers2

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To figure out whether a domain is closed, ask yourself if it includes the boundary lines. If it does include boundary lines, the domain is closed. Because you found the domain to be xy≥1, you are including the boundary line xy=1. This domain is closed.

To figure out whether the domain is bounded or unbounded, ask yourself if you could draw a circle on the graph of the domain that would contain all of it. If you can't, the domain is unbounded. In this question, the domain is unbounded because it continues forever in the first and thirds quadrants. No circle you could draw would be able to "bound" the entire domain.

Riddhi Kumar
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From the axioms of topology, a set is closed iff its complement is open, and in your case the complement $\{(x,y)\in\mathbb{R}^2\mid xy < 1\}$ is open (eg, you can see it as the reciprocal image of the open set $(\infty, 1)$ by the continuous function $(x,y)\mapsto xy$).

Note that boundedness and being open/close are completely independent: $\mathbb{R}$ is by definition an open in $\mathbb{R}$ (endowed with its usual topology) (and also a closed set, btw), so is $\emptyset$. $(0,1)$ is bounded but open, $[0,1]$ is bounded but closed, $(0,1]$ is bounded but neither open nor closed.

Edit: incidentally, $[a,\infty)$ is a closed set: its complement is $(-\infty,a)$ which is open. If you need, I can give you pointer to several (equivalent) definitions of open/close, which may help you in your understanding.

Clement C.
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    And what about (inf,a], [a,inf), (-inf,inf) ? – user149107 May 10 '14 at 14:24
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    For any $a\in\mathbb{R}$: $(-\infty, a]$ is closed, so is $[a,\infty)$; and $(-\infty,\infty)=\mathbb{R}$ is both open and closed. – Clement C. May 10 '14 at 14:26
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    Thank you, and yes please, if you can give me a pointer it would be helpful. I have not studies topology yet, this comes as part of Calculus 2 course. – user149107 May 10 '14 at 14:30
  • As a (short) intro, you can read the first page of this: http://www.maths.bris.ac.uk/~maxmr/opt/closedandconvex.pdf ; or the first 4 pages of this: http://www.math.kit.edu/iana3/~schnaubelt/media/isem-appendix.pdf – Clement C. May 10 '14 at 14:36