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Suppose $X$ is a Normed Vector Space with Norm $||\cdot||: X \to \mathbb{R}_{\geq 0}$. Then $X$ is also a Metric Space when endowed with the Distance Function: $$d(\mathbf{x}, \mathbf{y}) := ||\mathbf{x} - \mathbf{y}||.$$

We need only establish that $d$ satisfies the axioms of a Distance Function:

1. Observe that $0 = d(\mathbf{x}, \mathbf{y}) = ||\mathbf{x} - \mathbf{y}||$ If and Only If $\mathbf{x} = \mathbf{y}$ by Non-Degeneracy.
2. For any $\mathbf{x}, \mathbf{y} \in X$, $d(\mathbf{x}, \mathbf{y}) = ||\mathbf{x} - \mathbf{y}|| = |-1| ||\mathbf{y} - \mathbf{x}|| = ||\mathbf{y} - \mathbf{x}|| = d(\mathbf{y}, \mathbf{x})$
3. For any $\mathbf{x}, \mathbf{y}, \mathbf{z} \in X$, $$\begin{align*} d(\mathbf{x}, \mathbf{z}) &= ||\mathbf{x} - \mathbf{z}||\\ &= ||\mathbf{x} - \mathbf{y} + \mathbf{y} - \mathbf{z}||\\ &\leq ||\mathbf{x} - \mathbf{y}|| + ||\mathbf{y} - \mathbf{z}||\\ &=d(\mathbf{x}, \mathbf{y}) + d(\mathbf{y}, \mathbf{z}) \end{align*}$$ Thus $d$ is a Distance Function and $X$ is a Metric Space.