2
$\begingroup$

I am supposed to give an example of a closed set that is not bounded in $\mathbb{R}^2$. My idea was the graph of $y=1/x, \forall x$. If I take the complement of it, I get an open set. So the graph of $1/x$ is closed, but not bounded. But I am not sure of it. Could you please elaborate on it and give me a clue how to approach?

Thanks in advance!

$\endgroup$
  • 4
    $\begingroup$ There's $\mathbb{R}^2$ itself; or, say, the $x$-axis. Your idea works as well. $\endgroup$ – Jordan Oct 3 '14 at 5:37
2
$\begingroup$

Hint: Let $(x_n,y_n)$ be convergent sequence of elements such that $(x_n,y_n) \in Graph$. Prove that $\lim (x_n,y_n) \in Graph$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Should i take a specific graph? @jonasgomes $\endgroup$ – Marion Crane Oct 3 '14 at 5:37
  • 1
    $\begingroup$ Graph in this case is the set $(x,\frac{1}{x})$, the graph of the function you suggested. $\endgroup$ – Jonas Gomes Oct 3 '14 at 5:37
2
$\begingroup$

An easier way to think (at least for me) is to consider a bunch of isolated points. They must form a closed set as a set of isolated points contains all (there are none) its limit points. So you could just pick an unbounded set of isolated points.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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