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 $(0,1)$.


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.

  • $\begingroup$ 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

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