Normed vector space, product topology Let $(V, ||\cdot||)$ be a normed vector space (not necessarily finite-dimensional) and consider the product topology on $V \times V$.
Why is the product topology on $V \times V$ the topology generated by the usual Euclidean metric?

The topology induced on $V$ by the norm $||\cdot||$, certainly doesn't have to be the Euclidean one, and my definition of product topology is the smallest topology for which the projection maps are continous.
Can anybody help me understand why the above is true? Thank you.

You have to be a bit more precise, there are several norms that induce the product topology on $V \times V$.
We have $$\|(v_1,v_2)\|_2 = \sqrt{\|v_1\|^2 + \|v_2\|^2}$$ as what you probably mean by the Euclidean one and $$\|(v_1, v_2)\|_1 = \|v_1\| + \|v_2\|$$
and $$\|(v_1,v_2)\|_\infty = \max\left(\|v_1\|,\|v_2\|\right)$$
as two other ones. All of these can checked to be valid norms on $V \times V$.
As we have $$\|(v_1,v_2)\|_\infty \le \|(v_1,v_2)\|_1 \le \|(v_1,v_2)\|_2 \le \sqrt{2}\|(v_1,v_2)\|_\infty$$ from the usual inequalities of reals, we can argue that all these norms on $V \times V$ are equivalent.
And as $B_{\infty}((v_1,v_2), r) = B(v_1,r) \times B(v_2,r)$ it's also not too hard to argue that the $\|\cdot\|_\infty$ surely induces the product topology. (open balls form a base, and products of open balls (from $V$) thus form a base for $V \times V$ etc.).
