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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.

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  • $\begingroup$ By "usual" Euclidean metric, you mean the metric $d((x_1, x_2),(y_1,y_2)) = \sqrt{\|x_1 - y_1\|^2 + \|x_2 - y_2\|^2}$? $\endgroup$ – Nate Eldredge Jun 29 '19 at 21:10
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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.).

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Both the product topology and the topology induced by the Euclidean norm are valid topologies on $V \times V$ and both are first countable. To show that they are equal we can use sequences.

Let $((x_n, y_n))_n$ be a sequence in $V \times V$.

Assume $(x_n, y_n) \to (x,y)$ in the product topology. Then since projections are continuous, we get $x_n \to x$ and $y_n \to y$ in $V$ with respect to $\|\cdot\|$ so $$\|(x_n, y_n) - (x,y)\| = \sqrt{\|x_n-x\|^2 + \|y_n-y\|^2} \to 0$$ so $(x_n, y_n) \to (x,y)$ w.r.t the Euclidean norm.

Conversely, assume $(x_n, y_n)\to (x,y)$ w.r.t. the Euclidean norm. Then from $$\|x_n - x\| \le \sqrt{\|x_n-x\|^2 + \|y_n-y\|^2} = \|(x_n, y_n) - (x,y)\| \to 0 $$follows $x_n \to x$ in $V$ w.r.t. $\|\cdot\|$ and similarly $y_n \to y$. By the property of the product topology it follows that $(x_n, y_n) \to (x,y)$ in the product topology.

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