Review of metric spaces and point - set topology
نویسنده
چکیده
is not trivial to prove. In the one-dimensional case, the triangle inequality is an inequality on absolute values, and can be proven case-by-case. In R, it is best to use the following set-up. The usual inner product (or dot-product) on R is x · y = 〈x, y〉 = 〈(x1, . . . , xn), (y1, . . . , yn)〉 = x1y1 + . . .+ xnyn and |x| = 〈x, x〉. Context distinguishes the norm |x| of x ∈ R from the usual absolute value |c| on real or complex numbers c. The distance is expressible as
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