Can anyone prove this?
There is no metric d,so that (Q,d) is a connected space
Q are rationals.
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1Fix $a$ and consider $f(x) = d(x,a)$. – Daniel Fischer Jan 23 '15 at 10:03
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I don't see a contradiction? :/ U suppose that (Q,d) is connected, then function f would be continuous,surjection,so R+ would be connected? – zariski Jan 23 '15 at 10:11
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Not necessarily a surjection. Think about cardinality. – user208259 Jan 23 '15 at 10:13
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Pick any point $x$. If for any $r\in\mathbb R_+$, the circle $\{y\mid d(x,y)=r\}$ is empty, then $$\{y\mid d(x,y)<r\}=\{y\mid d(x,y)\le r\},$$ so it must be clopen.
Assuming a connected space has at least $2$ points, all sufficiently small $r$ must find a point on the corresponding circle, proving there are uncountably (even $\frak c$) many.
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