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In Theorem 20.3, page 123 Munkres Topology(second edition) the following is mentioned.

Inequality $\rho (x,y) \leq d(x,y) \leq n^{1/2}\rho (x,y)$ implies that $B_d(x,\epsilon)\subset B_\rho(x,\epsilon)$

Where $\rho$ and $d$ are square and euclidean metric on $\mathbb{R}^n$

I can prove the above implication by picking a point $y$ in ball $B_d(x,\epsilon)$ and showing that its in ball $ B_\rho(x,\epsilon)$. But I am having trouble to conceptually understand what it mean. My intuition says otherwise. Intuitively I feel that $B_\rho(x,\epsilon)\subset B_d(x,\epsilon)$. I would appreciate if someone can help me understand this.

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  • $\begingroup$ What is it you're finding conceptually difficult about this result? Try drawing a picture of both "balls" in $\mathbb{R}^2$. $\endgroup$ – anonymous Mar 1 '15 at 15:43
  • $\begingroup$ Intuitively it appears that if we have two balls $B_\rho(x,\epsilon)$ and $B_d(x,\epsilon)$ such that $\rho (x,y) \leq d(x,y)<\epsilon$ then the $B_\rho(x,\epsilon)$ ball should be inside $B_d(x,\epsilon)$ but the result shows otherwise. $\endgroup$ – Shrey Mar 1 '15 at 19:31
  • $\begingroup$ Right, so, think about it like this: $d$ and $\rho$ are two "measuring systems"; maybe $d$ gives you how far apart two points are in inches and $\rho$ tells you how far apart two points are in centimeters. An inch is larger than a centimeter, so if I wanted to look at all points within 1 unit of $x$, any point which is within one centimeter of $x$ is certainly within one inch of $x$ $\endgroup$ – anonymous Mar 1 '15 at 20:06
  • $\begingroup$ Thank You. That was helpful. $\endgroup$ – Shrey Mar 1 '15 at 21:57

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