A closed set is one where it contains all it's limit points, even though the end is 'open' as in the traditional sense any sequence tending to infinity will never leave the subset; therefore it's closed. Is my logic sound? Thanks!
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9You should specify if you mean "closed as a subset of $\mathbb R$ with the usual topology". Is the complement of $[0,\infty)$ open in $\mathbb R$? – Matt Mar 07 '12 at 11:23
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1Yes, I mean as a subset of $\mathbb{R}$ with the standard topology. – Mar 07 '12 at 11:24
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@Matt The complement of $[0,\infty)$ is $(-\infty,0)$, right? – The Doctor May 31 '17 at 05:47
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This set is indeed closed. Note that $+\infty$ is not a real number, sequences which tend to it are therefore non-convergent and have no limit in $\mathbb R$.
From this we can easily infer that $[0,\infty)$ is closed, since every sequence of positive numbers converging to a limit would have a non-negative limit which is in $[0,\infty)$.
Asaf Karagila
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Note that the complement of $[0, \infty)$ is $(-\infty, 0)$, which is open in the usual topology on $\mathbb{R}$. Therefore $[0, \infty)$ is closed. I often find looking at the complement easier than thinking of limit points.
Michael McGowan
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It depends on the topology: so $[0,\infty)$ is closed for the standard topology on $\mathbb{R}$.
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1Also: $[0,\infty)$ is a closed subset of $[0,\infty)$, regardless of the topology used. – GEdgar Mar 07 '12 at 14:52
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I suppose you are talking about $[0,\infty)\subseteq\mathbb{R}$ in the natural topology? Then the answer is yes, this subset is closed.
Willie Wong
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InvisiblePanda
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