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Hi!

I've got a question concerning neighbourhoods of points in 2- and 3-dimensional space.

How can we explicitly show, using only the definition of a given metric, that the according neighbourhood is some figure?

For example the three metrics are given:

1) $d(x,y)=\sqrt{\sum_{i=1}^{n}{(x_i-y_i)^2}}$;

2) $d(x,y)=\displaystyle\max_i|x_i-y_i|$;

3) $d(x,y)=\sum_{i=1}^{n}|x_i-y_i|$

If $a\in{\mathbb{R}}^2$ and the metric is 1), then it's clear:

$d(a,x)=\sqrt{(a_1-x_1)^2+(a_2-x_2)^2}<\epsilon$ and it's clear that it's a circle...because one can rewrite:$(a_1-x_1)^2+(a_2-x_2)^2<{\epsilon}^2$

But I have problems to see a picture when looking at the other metrics:

$d(a,x)=\displaystyle\max_2\{|a_1-x_1|,|a_2-x_2|\}<\epsilon$;

$d(a,x)=|a_1-x_1|+|a_2-x_2|<\epsilon$ tell me nothing at the moment about what the according neighbourhood might look like.

Could you please enlighten me on this case? :smile:

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Do you still need an answer? For anyone else wondering you simply draw a picture and think about it for a bit. To be honest the *hardest* to understand is the first one, the one you think is the easiest, by which you simple mean the most familiar

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