Determine if the set {0,1,1/2,1/3,1/4...} is closed, open or neither.
Justify your answer
What I was trying to prove that if the set A is closed if the A complement is open, but I could not figure it out
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1$\begingroup$ Open or closed in the usual topology of $\Bbb{R}$? $\endgroup$– YesNov 4, 2015 at 5:10
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$\begingroup$ If as Chou say is the ususal topology for real numbers and with Nitrogen's hint, now put in your mind what property/axiom in the definition of a topology you could use to prove that the complement in $\mathbb{R}$ of your genuine set, is open. $\endgroup$– user243301Nov 4, 2015 at 5:53
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1 Answer
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Hint: Look at neighborhoods of $0$. Can you find one that is contained in $K=\{0,1,1/2,...\}$?
Try to write $\Bbb{R}\backslash K$ as a union of open intervals and deduce that $K$ is closed.
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$\begingroup$ K complement =(0,1) union (1/2,1) union (1/3,1) union ......... Should I do something like this $\endgroup$ Nov 4, 2015 at 18:46
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$\begingroup$ More like $(-\infty,0)\cup (\frac{1}{2},1) \cup (\frac{1}{3},\frac{1}{2}) \cup ...$. $\endgroup$– NitrogenNov 4, 2015 at 18:50
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$\begingroup$ consider the set K={0,1,1/2,1/3,1/4,...}. The set K is said to be closed if Kc is open . Kc = (-∞,0)∪(1/2,1)∪(1/3,1/2)∪..... Thus, (-∞,0)∪{∪ (1/(j+1),1/j)} ∪ (1,∞) which is open. Hence, K is closed $\endgroup$ Nov 5, 2015 at 1:42