Let X = R2 and define a metric d by d((x1, y1), (x2, y2)) = max{|x1 - x2|, |y1 -

Question

Let X = R2 and define a metric d by d((x1, y1), (x2, y2)) = max{|x1 - x2|, |y1 - y2|) Verify that d is indeed a metric. What do balls B((x, y), r) look like? Draw a picture to illustrate. Consider the sequence (pn = (1/n, 1/n2)). Is this sequence convergent? Explain.

Explanation / Answer

a) its a metric: ie if d((x,y),(a,b)) = 0 iff x=a and y=b since d of 2 pts is max(the dist) also d((x,y),(a,b)) =d((a,b),(x,y)) this is also obvious also triangular inequality is easy d(p1,p2) + d(p2,p3) >= d(p1,p3) as |p1.x-p2.x| +|p2.x-p3.x| >= |p1.x-p3.x| and as |p1.y-p2.y| +|p2.y-p3.y| >= |p1.y-p3.y| hence also with max this holds hence the triangular inequality holds hence d is a metric. The balls B((x,y),r) look like squares of 2r * 2r which are centered at (x,y) and have their sides parallel to the axis c. Yes this sequence p_n converges to (0,0) as n tends to infinity since d(pn , (0,0)) = 1/n since 1/n > 1/n^2 hence d(pn , (0,0)) = 1/n and as n-> infinity 1/n -> 0 Hence the seq converges message me if you have any doubts

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