Can metric space consists only of 0 as one element?
I think Yes, because we can take standard metric and all of axioms are done.
$d(0,0) =0$;
$d(0,0) = d(0,0) = 0$;
$d(0,0) \le d(0,0) + d(0,0)$
is it true?
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I need this because, I want to construct complete metric space X such that for all x,y from X: d(x,y) belong to Q – Dec 05 '21 at 21:26
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Yes, it's a metric space and is complete (any finite metric space is complete). – Arctic Char Dec 05 '21 at 21:27
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Yep, thanks, and another question: can we construct infinite complete space X such that for all x,y from X: d(x,y) belong to Q(rational numbers)? – Dec 05 '21 at 21:28
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Yes. (think of some subset in $\mathbb R$) – Arctic Char Dec 05 '21 at 21:29
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2Or think about the most trivial of all metrics. – Asaf Karagila Dec 05 '21 at 21:33
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Is it true that any space with discrete metric is complete? d(x,A) = 1 if d x != A – Dec 05 '21 at 21:37
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Yes, try to prove it.... – Arctic Char Dec 05 '21 at 21:41
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so the answers for for my question is: take any subset of R and discrete metric, so d(x,y) belong to Q and it is complete space? – Dec 05 '21 at 21:43
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@JonathanTaylor consider what sort of sequences the discrete metric can have. – CyclotomicField Dec 05 '21 at 21:46