1
$\begingroup$

A collection $\mathcal{O}$ of open sets is an open cover of $A$ (or, briefly, covers $A$) if every point $x\in A$ is in some open set in the collection $\mathcal{O}$. For example, if $\mathcal{O}$ is the collection of all open intervals $(a,a+1)$ for $a\in\Bbb R$, then $\mathcal{O}$ is a cover of $\Bbb R$. Clearly no finite number of the open sets in $\mathcal{O}$ will cover $\Bbb R$ or, for that matter, any unbounded subset of $\Bbb R$.

I don't get the last one (in italic).

Take $(0,1)$ (which is an unbounded subset of $\mathbb R$) then if we take $a=0$ then this set $\{(0,1)\}$ will cover the subset $(0,1)$.

$\endgroup$
2
$\begingroup$

You are misunderstanding the term "unbounded". This term means that the set contains elements of arbitrarily large magnitude, i.e., it is not contained in $[-N,N]$ for any positive number $N$.

Your comment suggests you are thinking of the term as meaning "not containing its boundary". This is incorrect. The set $(0,1)$ which you cite is indeed bounded (and not unbounded, contrary to your claim) since it is contained in (say) $[-17,17]$. You could, of course, choose the more efficient containing set $[1,1]$, but the point is there just has to be some bounding interval.

$\endgroup$
  • $\begingroup$ thank you mpw i cant add comments or register ,i didnt know the definition of unbounded (im not english so i thought it just means a normal open set ) $\endgroup$ – user171838 Aug 26 '14 at 19:02
  • $\begingroup$ If we are working in $\mathbb{R}$, a set $S$ is bounded if there are reals $a$ and $b$ such that $a\lt x\lt b$ for every $x$ in $S$. A set is unbounded if it is not bounded. In $\mathbb{R}^n$, a set $S$ is bounded if there is a ball that contains every element of $S$. $\endgroup$ – André Nicolas Aug 26 '14 at 19:28
  • $\begingroup$ @daclaro Take care of your id information then you can log in again. $\endgroup$ – AmirHosein SadeghiManesh Sep 20 '14 at 3:21

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.