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Is it correct to say that this set $E=(0,1]$ where $E\subseteq R$ (Where $R$ is the set of real numbers) is not closed?

user11015000
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Yes. Closed sets are the complements of open sets. Therefore we check if the complement of $(0,1]$ is open or not. $$\mathbb{R}\backslash(0,1]=(-\infty,0] \cup (1,\infty)$$ An open set is a set for all points there exists a neighborhood contained in the set with positive radius. But, for point 0, no matter how little you go to $+$ side, you always go off the set. Therefore $\mathbb{R}\\(0,1]$ is a not open set, and therefore $(0,1]$ is a not closed set.

acarturk
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  • Could you explain further why (0,1] is not an open set? What would a similar example be of an open set? Thanks for your help by the way. – user11015000 May 18 '19 at 02:56
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    @user11015000 So, as in the example above, if a set $E$ includes a boundary point it is not open. By this logic, any purely open set and their unions are open sets. No problem, glad if I could help. (purely open: $(\cdot,\cdot)$) I advise you to check the linked thread as well, the OP there described open sets better than I did. – acarturk May 18 '19 at 03:02
  • and purely closed is [.,.] ? – user11015000 May 18 '19 at 06:23
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    @user11015000 Not exactly. For example, $[0,\infty)$ is a closed set. There are sets that are neither open or closed. The confusion is normal, but the main idea is: Open sets and closed sets are not opposite definitions. The reason $[0,\infty)$ is closed is that its complement $(-\infty,0)$ is open. See the definition. – acarturk May 18 '19 at 11:46