Verify the remark following the definition of normed space: Every normed space \((X,\lVert\cdot\rVert)\) is a metric space by taking \(d(x,y)=\lVert x-y\rVert\), while the converse is not true in general—even if \(X\) is a vector space.
Solution
1. Every normed space is a metric space via \(d(x,y) = \lVert x-y\rVert\).
1.1. Non-negativity and separation: \(\lVert x-y\rVert \ge 0\) with equality iff \(x = y\) (norm axioms), so \(d(x,y) \ge 0\) and \(d(x,y) = 0\) iff \(x = y\).
1.2. Symmetry: \(d(x,y) = \lVert x-y\rVert = \lVert -(y-x)\rVert = \lVert y-x\rVert = d(y,x)\) (homogeneity with \(\lambda = -1\)).
1.3. Triangle inequality: \(d(x,z) = \lVert x-z\rVert = \lVert (x-y) + (y-z)\rVert \le \lVert x-y\rVert + \lVert y-z\rVert = d(x,y) + d(y,z)\).
2. The converse fails in general, even for a vector space.Proof: take \(X = \mathbb{R}\) with the discrete metric \(d(x,y) = 1\) for \(x \ne y\), \(d(x,x) = 0\).This is a metric on the vector space \(\mathbb{R}\).But no norm induces it: a norm-induced metric satisfies \(d(nx, 0) = \lVert nx\rVert = n\lVert x\rVert\), which grows with \(n\), whereas the discrete metric gives \(d(nx,0) = 1\) for every \(n \ge 1\) (with \(x \ne 0\)). Contradiction.(More conceptually, a norm-induced metric is translation-invariant and homogeneous; the discrete metric is neither.)
3. Q.E.D.