Question

find disjoint neighborhood of√2 and√3​

Answer 1

Step-by-step explanation:

In any metric space, the set

Np(r)={x:d(x,p)<r}

is an open neighborhood of p. Given two points, p≠q, we have r=d(p,q)>0.

Suppose that s∈Np(r/3)∩Nq(r/3). The triangle inequality then says

r=d(p,q)≤d(p,s)+d(s,q)<r/3+r/3=23r

Contradiction. Thus, Np(r/3) and Nq(r/3) are disjoint.

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