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Let $(X,d)$ be a metric space. I wonder that is there a metric defined on $X \times X $ that does not make the metric function $d:X \times X \to \mathbb{R}$ continuous?

user1097620
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    A metric is always continuous. In fact, the metric topology is the coarsest topology that makes the metric continuous. – Ricky Oct 31 '23 at 12:54
  • I can't tell whether this question is about the possibility of a metric failing to be continuous with respect to some other metric, or about the relationship between the metric topology on $X$ versus $X\times X$. Can you clarify the question? – ziggurism Oct 31 '23 at 13:42
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    the usual way to think of it is that the metric is a function on $X\times X$ which defines a topology on $X$, and also on $X\times X$ via the product topology. It is with respect to this topology that the metric can be said to be continuous. Usually one does not introduce a metric on $X\times X$. But of course you could do so. – ziggurism Oct 31 '23 at 13:46
  • I think that the first sentence is potentially a misunderstanding of the relationship between a metric and the topology it induces, but the second sentence raises a valid question. – whpowell96 Oct 31 '23 at 14:18
  • @ziggurism I edited the question. – user1097620 Nov 01 '23 at 06:25
  • @user1097620 your edits did not clarify between the two alternate readings which I outlined. – ziggurism Nov 01 '23 at 15:04
  • @ziggurismI want to investigate the continuity of the $d$ function by defining a metric on $X\times X$. I know that there are metrics defined on $X\times X$ that make the function $d$ continuous. I wonder if a metric can be defined on $X\times X$ that does not make the $d$ function continuous? – user1097620 Nov 02 '23 at 11:18
  • @user1097620 while specific spaces may admit multiple incompatible metrics, and they also extend to the cartesian square, there are also spaces for which there is only a single metric, for example a singleton space. – ziggurism Nov 02 '23 at 11:28
  • @ziggurism Can you write this comment as an answer with an explicit example? – user1097620 Nov 02 '23 at 13:12
  • @user1097620 ok – ziggurism Nov 02 '23 at 14:11

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There do exist spaces with inequivalent metrics. For example the space $\mathbb{Q}$ of rational numbers has Euclidean metric $\lvert\cdot\rvert$ and for every prime $p$ a $p$-adic metric $\lvert\cdot\rvert_p$ given by $\lvert x\rvert=p^{-n}$ where $p^n$ is the largest power of $p$ in the reduced factorization of $x$.

The $p$-adic norm $\lvert \cdot\rvert_p$ is not equivalent to the Euclidean topology on $\mathbb{Q}$. Which is equivalent to saying that if you view the $p$-adic metric as a function on $\mathbb{Q}\times\mathbb{Q}$, $d(x,y)=\lvert y-x\rvert_p$ is not continuous with respect to the Euclidean topology.

Additionally, given any metric space $X$ you may consider the product metric on $X\times X$. This is not the same thing as viewing the metric on $X$ as a function of two variables, on $X\times X$, but the answer is the same. Since the $p$-adic metric is not continuous with respect to the Euclidean metric on $\mathbb{Q}$, its product metric is also not.

However, if the question is whether we can find such examples of inequivalent metrics for every metric space, the answer is no. One easy way to see why not is to observe that for example the empty set and the singleton space admit only one metric. If there aren't two metrics, then a foriori there are not two inequivalent metrics. More generally, while larger finite sets admit multiple metrics, every metric on a finite space is equivalent to the discrete metric. So no finite spaces admit two inequivelent metrics.

And if you do not have two inequivalent metrics on $X$, then you don't have a second metric to view as a function on $X\times X$ which is discontinuous. And you don't have a second metric to promote to a product metric on $X\times X$ which will be discontinuous.

ziggurism
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  • I realized that I used the words "metric" and "norm" as if they are interchangeable, however they are not quite the same thing. – ziggurism Nov 04 '23 at 05:25
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This is not possible since every metric satisfies the inequality, $$ |d(x,y)-d(u,v)|\leq d(x,u)+d(y,v),\quad x,y,u,v\in X. $$ Hence, it is Lipschitz continuous in bost components. To see the inequality just apply several time the triangle inequality.

El_mago
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