Manifold Posted November 24, 2005 Share Posted November 24, 2005 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) [math]d(x,y)=\sqrt{\sum_{i=1}^{n}{(x_i-y_i)^2}}[/math]; 2) [math]d(x,y)=\displaystyle\max_i|x_i-y_i|[/math]; 3) [math]d(x,y)=\sum_{i=1}^{n}|x_i-y_i|[/math] If [math]a\in{\mathbb{R}}^2[/math] and the metric is 1), then it's clear: [math]d(a,x)=\sqrt{(a_1-x_1)^2+(a_2-x_2)^2}<\epsilon[/math] and it's clear that it's a circle...because one can rewrite:[math](a_1-x_1)^2+(a_2-x_2)^2<{\epsilon}^2[/math] But I have problems to see a picture when looking at the other metrics: [math]d(a,x)=\displaystyle\max_2\{|a_1-x_1|,|a_2-x_2|\}<\epsilon[/math]; [math]d(a,x)=|a_1-x_1|+|a_2-x_2|<\epsilon[/math] tell me nothing at the moment about what the according neighbourhood might look like. Could you please enlighten me on this case? :smile: Link to comment Share on other sites More sharing options...

matt grime Posted November 24, 2005 Share Posted November 24, 2005 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 Link to comment Share on other sites More sharing options...

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