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1. Is Met really Complete? Wikipedia argues that not, since indeed |R^|R isn’t even First-Countable so let alone Metrizable, when it would be the induced Topological Space underlying an Uncountable Product in Met…

GrothenDitQue:

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeMay 31st 2019

This may depend on some details of what is meant by “metric space”; in particular, whoever wrote that might well have meant “Lawvere metric space”.

A Lawvere metric space is a set $X$ equipped with a function $d: X \times X \to [0, \infty]$ such that $d(x, y) + d(y, z) \geq d(x, z)$ and $0 \geq d(x, x)$. Note that the value $\infty$ is allowed as a distance.

If $\{X_i\}_{i \in I}$ is a family of Lawvere metric spaces, then their cartesian product in the category of Lawvere metric spaces and short maps is given by the set $\prod_{i \in I} X_i$ equipped with the distance

$d((x_i), (y_i)) = \sup_i d(x_i, y_i)$

In any event, you have to be a little careful in interpreting these things. Whatever notion of metric space is used, letting the morphisms be continuous maps between metric spaces is very different from letting the morphisms be something more closely tied to the metrics (e.g., short maps, Lipschitz maps), and the fact that classical metric/metrizable spaces are not closed under topological products is completely different from the fact that products may be perfectly sensible when we change the notion of morphism.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeJun 1st 2019

Clarified the discussion of products.

2. Added section on injective objects in $Met_{ord}$.

3. Added reference to lecture notes of Anton Petrunin.

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