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Consider the set $[a,\infty)$ where a is a real number. Then this is not open since $a$ is included.The complement of the set is $(-\infty,a)$, which is open, so it is a closed set. Is this correct? Also, for $\frac{1}{n}$ where $n$ is a positive integer. The set is in the interval $(0,1]$. This is not open, but I don't think it's closed either since the complement is $(-\infty,0]\cup(1,\infty)$.

Any feedback is appreciated, thanks.

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All of this is correct. –  Brian M. Scott Nov 24 '12 at 5:58
    
@BrianM.Scott Thanks! –  Alti Nov 24 '12 at 6:04
    
You're welcome! –  Brian M. Scott Nov 24 '12 at 6:04
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You should probably point out why having $a$ in the set is a problem; why does $a$'s presence mean that $[a, \infty)$ is not open? –  Benjamin Dickman Nov 24 '12 at 8:05
    
Thanks, yes I we had a theorem in class that a set is closed if and only if the complement is open. –  Alti Nov 25 '12 at 2:49
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3 Answers 3

up vote 2 down vote accepted

yes. what you say is right. First $\left (-\infty, a \right]$ is closed, because the complement of the set is open. The set $(0,1]$ is not open or close. The set $\left(-\infty,0\right]\cup \left[1,+\infty \right)$ is not open or close.

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A set is closed if and only if its complement is open. Sets like [a,b) or (a,b] are not open nor closed, because they not compliant with the definitions for that kind of sets: not every point are interior points and not every cluster point are contained in it. I hope that help you.

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While checking for open and closed sets, a check if the boundary is included suffices.

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