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Let $x, y \in \mathbb{R}^{d}$. I want to know whether it is true that $$|x-y|<r \text{ iff } |x_i -y_i|<r.$$

My attempt: Assume that $|x_i-y_i|<r$ for all $1\leq i\leq d.$ Then,$$ \sqrt{(x_1-y_1)^{2}+\ldots+(x_d -y_d)^2} \leq |x_1-y_1| +\ldots+|x_d-y_d|< \frac{d}{r}.$$

It seems it is not, but on the other hand, all distances are equivalent, which implies that the statement is true.

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  • $\begingroup$ Possibly you could replace $(x-y)$ with $a\in\mathbb R^d$ and each $(x_i-y_i)$ with $a_i$...? This would simplify the notation WLOG. $\endgroup$
    – CiaPan
    Commented Aug 27, 2022 at 20:27
  • $\begingroup$ What do you think that "equivalent distances" mean? $\endgroup$
    – jjagmath
    Commented Aug 27, 2022 at 20:44

1 Answer 1

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You can have a counter-example in two dimensions, choosing $x$ to be the origin. Here is the unit circle (i.e. $r=1$) and $y$ in the first quadrant:

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  • $\begingroup$ A picture worth a thousand words. :) Superb! $\endgroup$
    – CiaPan
    Commented Feb 17, 2023 at 9:36

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