TheoremBase

The first clause writes the negative as the multiple by -1 and uses absolute homogeneity; the second unfolds the norm metric and cancels a.

Proof

(i). Let v∈Vv\in V. Since βˆ’v=(βˆ’1)v-v=(-1)v (Elementary Identities in a Vector Space (iv)), absolute homogeneity of the norm (Norms and Real Normed Spaces (iii)) gives

βˆ₯βˆ’vβˆ₯=βˆ₯(βˆ’1)vβˆ₯=βˆ£βˆ’1βˆ£β€‰βˆ₯vβˆ₯=βˆ₯vβˆ₯.\lVert -v\rVert=\lVert(-1)v\rVert=|-1|\,\lVert v\rVert=\lVert v\rVert .

(ii). Let a,w∈Va,w\in V. By the definition of the norm metric (The Norm Metric of a Real Normed Space (i)) and since (a+w)βˆ’a=w(a+w)-a=w (Rearranging Sums, Differences and Two-Term Combinations in a Real Vector Space (iv)),

dV(a+w,a)=βˆ₯(a+w)βˆ’aβˆ₯=βˆ₯wβˆ₯.d_V(a+w,a)=\lVert (a+w)-a\rVert=\lVert w\rVert .

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