Connectedness in a Product Metric?

Question

Consider the proof below:

In the latter part of the proof: $X_{j,a}$ has been described as being an isometric copy of $X_j$ as per proof $1.6.4$. However $1.6.4$ is valid only when the metric involved is a conserving metric as per the proof below:

$\therefore$ I am unable to understand why would the author have considered the particular mapping as an isometric one when it's clearly mentioned that the metric is a product metric and not a conserving metric.

Note:Conserving metric: Suppose $(X,d) $ and $(X_i, \tau_i)$ are metric spaces for every $i$, where $X= \prod_{i=1} ^n X_i. d$ is termed as a Conserving metric iff $\max \{\tau_i(a_i,b_i)~|~i \in (1,\cdots ,n)\} \le d(a,b) \le \sum_{i=1}^n \tau_i (a_i,b_i)$

Any help would be appreciated. Thank you for reading through!

Answer 1

Connectedness is a topological property, so since all product metrics induce the same topology, it suffices to prove Theorem 11.4.4 for one particular choice of product metric. In particular, you can choose a product metric which is also conserving (e.g., $d(a,b)=\sum_i\tau_i(a_i,b_i)$).

(Really, the right way to prove this is to not talk about metrics at all and only talk about topologies: $X_{j,a}$ is always a homeomorphic copy of $X_j$ if you give $P$ the product topology, and that is all the argument needs.)