Topology induced by Eulidean metric is the same as product topology I have seen this theorem in Munkres where he proves that the topology induced by the Euclidean metric, the square metric and the  product topology on $\mathbb{R}^n$ are the same.
I know there is an answer to this question on here with a slightly different approach, But I was hopping someone could explain a few thing on this particular one to me.



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*I am having trouble verifying the Inequality for myself , any hints?

*Why is $B_d(x,\epsilon) \subset B_{\rho}(x,\epsilon)$?
And also the second inclusion. $ B_{\rho}(x,\epsilon /\sqrt{n}) \subset B_d(x,\epsilon)  $
Without looking, I would have written the first inclusion as $ B_{\rho}(x,\epsilon) \subset B_d(x,\epsilon)  $ since $\rho (x,y) \leq d(x,y)$ so I expect any ball defined by the metric $\rho$ to be smaller than a ball defined by $d$. and then we could conclude that the topology induced by $\rho$ is finer. But this seem to go in opposite direction to me.
 A: Let us recall that $$\rho(\mathbf{x}, \mathbf{y}) = \max(|x_1 - y_2|, \ldots,|x_n -y_n|)$$ and $$d(\mathbf{x}, \mathbf{y}) = \sqrt{\sum_{i=1}^n (x_i - y_i)^2}$$
As to 1. let  $\mathbf{x}, \mathbf{y}$ be arbitrary elements of $\mathbb{R}^n$.
We thus have that $\rho(\mathbf{x}, \mathbf{y}) = |x_m - y_m|$ for some $m \in \{1,\ldots,n\}$ (he maximum is assumed at some coordinate), and thus $\rho(\mathbf{x}, \mathbf{y})^2 = |x_m - y_m|^2 = (x_m - y_m)^2$ (as $|a|^2 = a^2$ for all $a \in \mathbb{R}$) and so 
$$\rho(\mathbf{x}, \mathbf{y})^2 \le \sum_{i=1}^n (x_i- y_i)^2 = d(\mathbf{x}, \mathbf{y})^2$$ because we just add extra some positive numbers $(x_j-y_j)^2$. As all numbers (metric values) are $\ge 0$ we can take square roots on both sides and get $\rho(\mathbf{x}, \mathbf{y}) \le d(\mathbf{x}, \mathbf{y})$ which is the first part of the inequality.
On the other hand, for each $j \in \{1,\ldots,n\}$ we have that (because $m$ is where the maximal absolute difference is and for positive numbers $a \le b \to a^2 \le b^2$) that $(x_j - y_j)^2 = |x_j - y_j|^2 \le |x_m - y_m|^2 = \rho(\mathbf{x}, \mathbf{y})^2$ and thus using this trivial upperbound for each of the $n$ terms in the sum:
$$d(\mathbf{x}, \mathbf{y})^2 = \sum_{j=1}^n (x_j - y_j)^2 \le n\rho(\mathbf{x}, \mathbf{y})^2$$
and again we take the square root on both sides to see that $\rho(\mathbf{x}, \mathbf{y}) \le \sqrt{n}d(\mathbf{x}, \mathbf{y})$, which is the final part of the inequality.
We want to see that $$B_d(\mathbf{x}, \varepsilon) \subseteq B_{\rho}(\mathbf{x},\varepsilon)$$ 
So let $\mathbf{y} \in B_d(\mathbf{x}, \varepsilon)$. This means exactly that $d(\mathbf{x},\mathbf{y}) < \varepsilon$, and thus we conclude by the first inequality that:
$$\rho(\mathbf{x},\mathbf{y}) \le d(\mathbf{x},\mathbf{y}) < \varepsilon$$
which again means $\mathbf{y} \in B_\rho(\mathbf{x}, \varepsilon)$. Hence the inclusion.
To see that $$B_\rho(\mathbf{x}, \frac{\varepsilon}{\sqrt{n}}) \subseteq B_d(\mathbf{x}, \varepsilon)$$
again pick an arbitrary $\mathbf{y} \in B_\rho(\mathbf{x}, \frac{\varepsilon}{\sqrt{n}})$ so that $\rho(\mathbf{x},\mathbf{y}) < \frac{\varepsilon}{\sqrt{n}}$. Using the second inequality we have 
$$d(\mathbf{x}, \mathbf{y}) \le \sqrt{n}\rho(\mathbf{x}, \mathbf{y}) < \sqrt{n} \frac{\varepsilon}{\sqrt{n}} = \varepsilon$$ so that $\mathbf{y} \in B_d(\mathbf{x}, \varepsilon)$ as required.
A: *

*You should verify for yourself that to show the inequality $$\rho(x,y) \leq d(x,y) \leq \sqrt{n}\rho(x,y)$$ holds, it's enough to show that it holds with $y = 0$. This should help simplify some of your work in proving the actual inequalities. To prove those, I'm not sure how far you've gone into analysis (I'm guessing not far), but you may find the Cauchy-Schwarz inequality useful at least for the inequality $d(x,y) \leq \sqrt{n}\rho(x,y)$. If not, they can both be done with clever algebraic tricks as Munkres suggests.

*I suggest proving these inclusions for yourself using element arguments.
