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I have the following metric:

$d((a,b),(c,d))=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}$ or $0$ if $(a,b)=(c,d)$

the question is to show $R=\mathbb R^2\setminus\{(0,0)\}$ is disconnected with respect to $d$.

ive got 2 sets

$U=\{(x,y)\mid y<0\}$

$V=\{(x,y)\mid y\geq0\}\setminus\{(0,0)\}$

so i have $U\cup V=R$ so i just need to show $U$ and $V$ are open w.r.t the metric d

im having trouble finding open balls around points on the x-axis,say $(b,0)$. it seems to be the case if i need a ball of radius $\epsilon>0$ i only have the point $(b,0)$ when $\epsilon<b$. is this correct so i can say my sets are both open?

or maybe its easier to show they are both closed?

thanks

Lorenzo B.
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jiboom
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    You are on the right track. Try to prove that for any $P \neq (0,0)$ there exists an $\epsilon$ such that the open (or closed, it doesn't matter) ball around $P$ with radius $\epsilon$ is just $P$ (that's what you hinted at for $P=(b,0)$). This means any subset of $\mathbb R^2-{(0,0)}$ is open. Your division in $U$ and $V$ is therefore just as valied as any other. – Ingix May 28 '18 at 17:24

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