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Bounded subset $S \subseteq R^n$ is bounded with respect to what metric ?

I have been reading several proofs where the theorem involves: "Suppose $S \subseteq R^n$ is a bounded subset in $R^n$". Recently, I saw this in a proof of the Bolzano Weierstrass theorem generalized to $R^n$. Here they used the fact that $S \subseteq R^n$ is bounded to state that any sequence $\{x_n\}$ in $S$ must have bounded coordinates $x_1 , x_2, \ldots, x_n$ so $|x_i| < K$ for $i = 1,2 \ldots, n$.

Indeed this make sense if the metric that the boundedness is with respect to is the Euclidean distance. However how can I be sure ? Is boundedness of $R^n$ always with respect to the Euclidean distance ? - Does all metrics satisfy that the coordinates are bounded in a sequence in a bounded subspace of $R^n$ ?

Shuzheng
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    If $R^n$ (or $\mathbb R^n$) is used, then metric properties should be assumed to refer to the usual metric. If another metric is intended, that should be stated. Similarly, mention of "measure" in $\mathbb R^n$ should be understood to refer to Lebesgue measure, unless otherwise stated. Addition in $\mathbb R^n$ should be understood as the usual addition. And so on. – GEdgar Dec 22 '13 at 20:17

3 Answers3

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You certainly know that all norms on $\mathbb{R}^n$ are equivalent so, if you are bounded for a metric coming from a norm, you are bounded for the euclidean metric.

Jeremy Daniel
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The implicit metric is always the Euclidean metric, unless stated otherwise.

For your last question, it is true that any reasonable metric on $\mathbb R^n$ will have the property that a bounded subset of $\mathbb R^n$ has elements with bounded coordinates. By 'reasonable', I mean that it comes from a norm. This follows from the fact that all norms on a finite dimensional vector space are equivalent. See here for a proof sketch, or consult Google.

Community
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Potato
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  • Thanks for your comment, I think I have to delve a little deeper into topics like Topology in order to understand the proof you are linking to. However it is nice to know that every metric is equivalent, if it comes from a norm ? – Shuzheng Dec 23 '13 at 07:14
  • @NicolasLykkeIversen Yes, that is true. – Potato Dec 23 '13 at 07:18
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To be a bit more explicit, there is a reasonable metric (that gets used in certain settings) on $\Bbb R^n$ in which every subset is bounded. In particular, given any metric $d$, we can define $\overline d(x,y) = \min\big(d(x,y),1\big)$, and so $\overline d(x,y)\le 1$ for all $x,y$. The topology induced by this metric is the same topology, but the sizes of big sets is significantly altered. This metric, of course, does not come from a norm.

Because of such matters, you will learn, as you proceed, that in a general metric space the criterion for compactness of a subset is that it be closed and totally bounded (because, as we've just seen, boundedness alone does not give enough control).

Ted Shifrin
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