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Define a extended metric space $X$ to be a metric space $X$ except the distance function $d$ maps from $X \times X\to [0,\infty]$ (Note that $\infty$ is allowed). Looking at this post Metric assuming the value infinity it doesn't seem that any difficulty can arise when switching to an extended metric, as far as theorem go. However, compact sets are not necessarily bounded in extended metric spaces. (consider two points with distance infinity) Is there a general theorem about when theorems in metric spaces can be converted into theorems in extended metric spaces?

It would also be nice to provide other examples of things failing when extended metric is used.

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davik
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  • The post you cite pretty well says it all. Extended metric spaces are a cute way of working with topological spaces whose connected components are ordinary metric spaces. – Rob Arthan Oct 11 '16 at 21:28
  • They are, yes, but also we can formulate analysis in terms of extended metric spaces. The linked post seems to suggest that nothing is lost, but in fact something is. – davik Oct 11 '16 at 21:31
  • Additionally, if someone has examples of theorems failing when a banach space is allowed to have an extended norm, defined similiarly, please share – davik Oct 11 '16 at 21:33
  • Of course, compact spaces are bounded - by the now allowed $\infty$. After all, the distance function $x\mapsto d(x_0,x)$ is continuous. – Hagen von Eitzen Oct 11 '16 at 21:36
  • I am not sure I understand your comment about formulating analysis. As for Banach spaces, I don't see how you can generalise to allow $|x| = \infty$: in a normed vector space, the vector space structure determines the norm on each line up to a constant factor dependent on the line. So $|x| = \infty$ would be incompatible with the requirement that the line ${\alpha x \mid \alpha \in \Bbb{R}}$ has its usual topology. – Rob Arthan Oct 11 '16 at 21:41
  • @HagenvonEitzen I don't understand, perhaps your definition of bounded is different? Your definition would have all sets be "bounded". Whereas I would have only ones bounded by a positive real number. Perhaps that only theorems on boundedness can be affected by the change in definition? – davik Oct 11 '16 at 21:45
  • @RobArthan Not sure if what I'm talking about makes sense, but consider $\mathbb{R}^{\mathbb{N}}$ with "euclidean norm" – davik Oct 11 '16 at 21:47
  • @davik That is just one problem that arises when changing one definition - one may need to reconsider other definitions as well – Hagen von Eitzen Oct 11 '16 at 21:49
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    @davik: I think I can guess the definition you have in mind: $|(x_1, x_2, \ldots)|$ would be the limit of $a_n = |(x_1, \ldots, x_n)|$ if that limit existed and would be $\infty$ otherwise. This "norm" would have some very odd properties (specifically relating to my point about about the induced topology on lines: your space would have a 2-dimensional subspace such that the topology on the $x$-axis is the usual topology on $\Bbb{R}$ and the topology on the $y$-axis is the discrete topology). – Rob Arthan Oct 11 '16 at 22:06

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The reason why extended metric spaces are so similar to usual metric spaces is that, in addition to the obvious formal analogy, they share even deeper similarities: given an extended metric $d$ in $X$, define $$ d_1(x,y) = \text{min}\{1,\ d(x,y)\}, \quad \text{for all }x,y\in X. $$ Then $d_1$ is a bona fide metric and the topology defined by $d_1$ is the same as that defined by $d$ (en passant, this is the usual trick for showing that every metric space is equivalent to a bounded one). However, anything related to boundedness will be left out of the analogy as such concepts are not intrinsic to topology.

Ruy
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