In metric space, the continuity of a function depends on a metric. Can I define some metric $(\mathbb{R},d_1)$ and $(\mathbb{R},d_2)$ such that $f:\mathbb{R}\rightarrow \mathbb{R}$ given by $f(x)=x$ is descontinuous? I've tried zero-one metric for $d_1$ and maximum metric for $d_2$, but zero-one is all isolated points so is continuous at every point.
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All metrics are equivalent in finite dimensional spaces – user57 Nov 17 '22 at 17:35
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5@mnp That's not true at all. Consider the discrete metric versus the usual one. – Randall Nov 17 '22 at 17:35
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1Try $d_2$ the zero-one (a.k.a. discrete) metric, and $d_1 \neq d_2$. – azif00 Nov 17 '22 at 17:36
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5@mnp all norms are equivalent. Not all metrics – Luca.b Nov 17 '22 at 17:38
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Then, only the metrics that can be induced by norms are equivalent? – user57 Nov 17 '22 at 17:39
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1@mnp No (whatever that means), it is not an iff – Anne Bauval Nov 17 '22 at 17:44
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@azif00 Of course, thank you. – ends7 Nov 17 '22 at 17:45
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3@EmanuelDias This also answers your question Which two would be an example of non-equivalent metrics on $\mathbb{R}$. – Anne Bauval Nov 17 '22 at 17:47
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That's weird. All norms are equivalent in finite dimensional spaces. So, they all should induce equivalent metrics? @AnneBauval – user57 Nov 17 '22 at 17:47
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1@mnp Yes they do. – Anne Bauval Nov 17 '22 at 17:49
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Then we have a theorem that we can use to build a set S that contains only those metrics that can be induced by a norm. And so, only the metrics in S would be equivalent in any finite dimensional space? @AnneBauval – user57 Nov 17 '22 at 17:51
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2Given a metric you can always define a bounded euivalent metric. So given a norm induced metric you have an equivalent metric which is bounded hence not norm induced – Luca.b Nov 17 '22 at 17:54
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2@mnp No, again, it is not an if and only if: there are metrics on your finite dimensional space which are not induced by a norm and which are nevertheless equivalent. Moreover, I said "whatever it means" because there are numerous equivalence classes of metrics, and each of them is infinite. – Anne Bauval Nov 17 '22 at 17:54