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Theorem 20.3 on Page 123 of $\S$ 20 in J. Munkres' topology book is followed immediately by a proof. Within the proof, there is a line that I am having trouble understanding.

First the Theorem:

Theorem 20.3 The topologies on $\mathbb{R}^n$ induced by the euclidean metric $d$ and the square metric $\rho$ are the same as the product topology on $\mathbb{R}^{n}$.

And now some definitions are needed before I get to my actual question.

  • Euclidean metric d on $\mathbb{R}^n$ is defined by $d(\textbf{x},\textbf{y})=\parallel \textbf{x} - \textbf{y} \parallel =[ (x_1 - y_1)^2 + ... + (x_n - y_n)^2 ]^\frac{1}{2}$
  • Square metric $\rho$ on $\mathbb{R}^n$ is defined as $\rho (\textbf{x}, \textbf{y})=max \{ |x_1 - y_1| , ... , |x_n - y_n| \}$

Now for my actual question. The proof provided in the text starts off as follows:

Proof. Let $\textbf{x}=(x_1,...,x_n)$ and $\textbf{y}=(y_1,...,y_n)$ be two points of $\mathbb{R}^n$. It is simple to check that $\rho(\textbf{x}, \textbf{y})\leq d(\textbf{x}, \textbf{y})\leq \sqrt n \rho(\textbf{x},\textbf{y})$. The first inequality shows that $B_d(\textbf{x},\epsilon)\subset B_{\rho}(\textbf{x},\epsilon).$

That is where I am stuck. I don't see why containment holds in $\subset$ direction. If the book had not explicitly stated this, I would have assumed $B_d(\textbf{x},\epsilon)\supset B_{\rho}(\textbf{x},\epsilon)$.

I would appreciate some clarification on this portion of the proof.

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Just chase elements. Suppose that $\mathbf{y}\in B_d(\mathbf{x},\epsilon)$. Then

$$d_\rho(\mathbf{x},\mathbf{y})\le d(\mathbf{x},\mathbf{y})<\epsilon\;,$$

so $\mathbf{y}\in B_\rho(\mathbf{x},\epsilon)$ as well. Thus, $B_d(\mathbf{x},\epsilon)\subseteq B_\rho(\mathbf{x},\epsilon)$.

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Note that $d(x,y)<\epsilon \implies \rho(x,y)<\epsilon$.

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Note that if you assume that $d(x,y)< \epsilon$, that implies $\rho(x,y) \le d(x,y) < \epsilon$. Therefore, if $y \in B_d(x,\epsilon)$, then $y\in B_\rho(x,\epsilon)$.

Intuitively, the ball of radius 1 is contained within the box with corners $\{\pm 1\}^n$.

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  • $\begingroup$ That makes a lot of sense. I am starting with some $y$ in my $B_d$ ball, and by the inequality, it is also contained in the $B_\rho$ ball hence the $B_d$ ball is contained in the $B_\rho$ ball! Thank you! $\endgroup$ – Nidia Apr 9 '15 at 22:12

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