The topology induced by $p$-product norm

Question

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?

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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.

2. $\tau = \tau'$ 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.

Answer 1

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.