A subspace of $\mathbb R^n$ is compact if and only if it is closed and bounded in the Eulidean Metric or Square Metric or $l_1$ metric.

Is my statement correct?

Actually the statement - "A subspace of $\mathbb R^n$ is compact iff it is closed and bounded in the Euclidean Metric or Square Metric" was given in Mukresh. I think closed and bounded set in the $l_1$ metric would also be compact. As the topologies of three are same.

Am I correct? Please correct me if I am wrong.

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    $\begingroup$ Yes, the $\ell_1$ metric is equivalent to the euclidean metric. This follows from simple arithmetic inequalities. Moreover, there is only one norm-induced topology on R^n. $\endgroup$ – Lorenzo Quarisa Apr 8 '18 at 12:45
  • $\begingroup$ I did not get you @hardmath...All I want to know is whether my statement is correct or not. $\endgroup$ – cmi Apr 8 '18 at 12:50
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    $\begingroup$ The only compact subspace of $\mathbb R^n$ is the trivial subspace (consisting solely of the zero vector, one point). I'm concerned that you meant to ask about characterizing which subsets of $\mathbb R^n$ are compact rather than which subspaces are compact. But the statement is true (what applies to all subsets certainly applies to subspaces). $\endgroup$ – hardmath Apr 8 '18 at 12:56
  • $\begingroup$ yes..You are correct..But why Mukresh used Subspace ?@hardmath $\endgroup$ – cmi Apr 8 '18 at 12:59
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    $\begingroup$ Context matters. If one speaks of Euclidean space $\mathbb R^n$, most often one is thinking about it as a vector space that has a topology. In a general topology text one may instead think of the subsets of a topological spaces as inheriting a subspace topology, so I'm open to the author using "subspace" to mean topological subspace rather than vector subspace. $\endgroup$ – hardmath Apr 8 '18 at 13:04

As the topologies induced by the $\ell_1, \ell_2$ and $\ell_\infty$ metric are the same on $\mathbb{R}^n$, closedness under one of them is equivalent to closedness under any other of them.

Also, all these metrics are related by inequalities:

$d_1(x,y) \le d_2(x,y) \le \sqrt{n} d_\infty(x,y); d_\infty(x,y) \le d_1(x,y)$ etc. so that boundedness in one of them also implies boundedness in all others.

So the compactness characterisation holds in all 3 of the metrics, in fact in any metric derived from a norm. The topology being the same is not enough, as the truncated metric $d_{t}(x,y) = \min(d(x,y), 1)$ induces the same topology as $d$ does, but in $d_t$ all sets are bounded, while this need not hold for $d$. So one does require an argument to see that the combination of closedness and boundedness for a specific metric is preserved.


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