Finite distance between two points

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CCMarie
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If X is a metric space, what is the simplest sufficient condition for d(x,y)<∞, ∀ x, y ∈ X?
 
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Yes, it is the metric. And thank you!
 
CCMarie said:
If X is a metric space, what is the simplest sufficient condition for d(x,y)<∞, ∀ x, y ∈ X?
For X, it is maybe that X is compact. Being bounded may be too obvious/circular.
 
fresh_42 said:
For any two points the distance is always finite.
fresh_42 said:
For any two points the distance is always finite.
I think there are extended metrics that allow for infinite distance, but they are nonstandard.
 
fresh_42 said:
Interesting question: Do we have a metric on the projective line?
But these may be different questions: the Real line, nor Euclidean n-space is bounded.
 
fresh_42 said:
Sure. I was thinking about including infinite points and a non standard metric you mentioned.
But even with the standard metric, there is no Real M with d(x,y)<M for all x,y in ##\mathbb R^n ##Edit: Just choose the pair M,0 ( or generalized to n dimensions) for d(x,y).
 
fresh_42 said:
That's why I said projective space. Can we keep the metric via the transition from circle to line, sphere to plane?
What is the "standard" metric in Projective space? Is it the quotient metric?EDIT: Do we even know if the projective space is metrizable?
 
fresh_42 said:
I have no idea, that's why I asked.
BAH. It is a manifold and manifolds are metrizable. With weird exceptions for 1-2 people who allow non-Hausdorff.
 
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fresh_42 said:
When it comes to topology I need to see. If possible step by step. Those people are notoriously counterintuitive.
This is the standard case, don't know if there are many others "non-trivially" different: https://en.wikipedia.org/wiki/Non-Hausdorff_manifold#Line_with_two_origins :

EDIT: Proof of boundedness: d(x,y) is continuous on YxY for Y compact. Then d: YxY -->R is a continuous function on the compact set YxY, and so it is bounded.
 
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