# The topology induced by $p$-product norm

I am reading this answer in which the author said

There are many ways to equip $$X \times Y$$ with a norm, the most natural ones are $$\|(x,y)\| = \max{\{\|x\|,\|y\|\}}$$ and $$\|(x,y)\| = \|x\| + \|y\|$$ since they correspond to the categorical product and coproduct operations (in the category of Banach or normed linear spaces and linear maps of norm $$\leq 1$$). Be that as it may, it is a good exercise to check that all the $$p$$-norms $$\|(x,y)\|_{p} = \left(\|x\|^{p} + \|y\|^{p}\right)^{1/p}$$ on $$X \times Y$$ are equivalent,

I'm trying to do this exercise.

Let $$(X_i, \|\cdot\|_i)_{i=1}^n$$ be a finite sequence of n.v.s. Let $$\tau_i$$ be the topology induced by $$\|\cdot\|_i$$. Let $$\tau$$ be the product topology of $$\tau_1, \ldots, \tau_n$$. Let $$X := X_1 \times \cdots X_n$$. For $$p \in [1, \infty]$$, we define a norm $$[ \cdot]_p$$ on $$X$$ by $$[ x]_p := (\|x_1\|^p_1 + \cdots \|x_n\|_n^p)^{1/p} \quad \forall x=(x_1, \ldots, x_n) \in X.$$

1. $$[\cdot]_p$$ are equivalent for all $$p \in [1, \infty]$$.
2. $$\tau = \tau'$$ with $$\tau'$$ being the induced topology by $$[\cdot]_p$$.

Could you confirm if my proof is fine?

My atempt:

1. $$[\cdot]_p$$ are equivalent for all $$p \in [1, \infty]$$.

This is because $$[\cdot]_p$$ is induced by the canonical $$p$$-norm $$\| \cdot\|_{p, \mathbb R^n}$$ on $$\mathbb R^n$$. In fact, $$[x]_p = \big \| (\|x_1\|_1, \ldots, \|x_p\|_p) \big\|_{p, \mathbb R^n}$$

Moreover, all norms on finite-dimensional spaces are equivalent.

1. $$\tau = \tau'$$ with $$\tau'$$ being the induced topology by $$[\cdot]_p$$.

It suffices for us to consider $$p=\infty$$. In this case, $$[ (x_1, \ldots, x_n) - (y_1, \ldots, y_n) ]_\infty = \max \{\|x_1- y_1\|_1, \ldots , \|x_n- y_n\|_n\}.$$

It suffices to prove that $$\tau$$ and $$\tau'$$ have the same system of neighborhoods (nbh) at each point. Fix $$x \in X$$.

Let $$U$$ be a nbh of $$x$$ in $$\tau'$$. There is $$r \in \mathbb R_{>0}$$ such that $$B(x, r) := \{y \in X \mid \|y-x\|_\infty< r\} \subset U$$. Then \begin{align} B(x, r) &= \{y \in X \mid \forall i =1, \ldots, n: \|y_i - x_i \|_i < r\} \\ &= \bigcap_{i=1}^n \{y \in X \mid \|y_i - x_i \|_i < r\}. \end{align}

Notice that the map $$y \mapsto \|y_i - x_i \|_i$$ is continuous w.r.t. $$\tau$$, so $$\{y \in X \mid \|y_i - x_i \|_i < r\} \in \tau$$ for all $$i =1, \ldots, n$$. Hence $$B(x, r) \in \tau$$ and thus $$U$$ is a nbh of $$x$$ in $$\tau$$.

Let $$U$$ be a nbh of $$x$$ in $$\tau$$. Then there is nbh $$V_i$$ of $$x_i$$ in $$\tau_i$$ such that $$V_1 \times \cdots \times V_n \subset U.$$

Then there is $$r_i \in \mathbb R_{>0}$$ such that $$B(x_i, r_i) := \{y\in X_i \mid \|y - x_i \|_i < r_i\} \subset V_i.$$ Let $$r := \min \{r_1, \ldots, r_n\}$$. Then $$V := \{y \in X \mid \|y-x\|_\infty < r\} \subset U$$. Clearly, $$V \in \tau'$$. Then $$U$$ is a nbh of $$x$$ in $$\tau'$$. This completes the proof.

## 1 Answer

For part $$1$$, you assume that you are dealing with finite dimensional vector spaces which is not stated in the question. I do not know the full context, but I believe the theorem is true for any normed vector space, including infinite dimensional ones. To prove that $$[x]_p$$ is a norm, you'll basically only need to show the triangle inequality as the other parts are trivial.

For part $$2$$, this looks fine and does work in full generality.