Assume $\mathbb{R}$ is equipped with standard Euclidean metric, I need to show that $[1, \infty)$ is a complete space, when $[1,\infty)$ is viewed as a subspace of $\mathbb{R}$. However, the complete subsets of a complete space are exactly those subsets that are closed, how can $[1,\infty)$ be complete if $[1,\infty)$ is not closed?
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3$\begingroup$ $[1,\infty )$ is closed $\endgroup$– ThePortakalJun 13, 2016 at 5:13
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$\begingroup$ @jordan A bounded interval is topologically closed if and only if it is a closed interval (it contains its extremal points). An unbounded interval is topologically closed if and only if it is $\Bbb R$, $[a,\infty)$ or $(-\infty,b]$. $\endgroup$– user228113Jun 13, 2016 at 5:18
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$\begingroup$ The complement $(-\infty,1)$ is open, hence $[1,\infty)$ is closed. $\endgroup$– copper.hatJun 13, 2016 at 5:31
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$\begingroup$ [1,inf) can't be complete if [1,inf) is not closed, but since [1,inf) * is* closed, [1,inf) must be complete. And [1,inf) is closed. $\endgroup$– fleabloodJun 13, 2016 at 15:19
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2 Answers
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Note: There is no contradiction since $[1,\infty)$ is closed in $\mathbb{R}$.
Your proof would then look something like this:
Let $[1,\infty)\subset \mathbb{R}$. We know that $\mathbb{R}$ is a complete metric space, and also that $[1,\infty)$ is closed, because its complement $(-\infty,1)$ is open. Since a closed subset of a complete metric space is complete (see here for a proof), it follows that $[1,\infty)$ is a complete subspace of $\mathbb{R}$.
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The complement is open, therefore the set is closed.
