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.
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.