Extension of a uniformly continuous function, quibble

Question

I stumbled over this question which is slightly different from the usual version I remember, e.g. here: in one case one extends the uniformly continuous from a metric space (not necessarily complete) to its completion, and in the other case, from a dense set to the whole space.

Are the two versions equivalent? If I remember well

• a metric space is dense in its completion
• it seems that an arbitrary metric space is not the completion of a dense subset, so it seems to me that the version with the completion is less general, but I've been confusing myself...

Remark: the usual remark... completion is a notion from metric space, density (and closure) only require a topology.

Edit: indeed a metric space that is not complete is a dense subset of itself, and it is by assumption not the completion of itself, nevertheless can there be a dense subset such that it is its completion... probably not.

Answer 1

Finally I think I clarified my confusion: it seems to me that both are indeed equivalent but the formulation as an extension from a dense subset looks more general at first.

What could go wrong in the other case (but does not) is that a priori one could imagine that the completion of a dense subset does not yield the whole metric space. This is not the case because of the property that a complete subspace is closed, cf. this answer (and the closure of a dense subspace is the whole space). The completion of a dense subset of a non complete metric space would contain the whole space and actually correspond to the completion of the metric space itself.

So to recapitulate, "pour enfoncer le clou": in a not necessarily complete metric space, the completion of a subset is necessarily closed (but would not necessarily be a subset...take the intersection) but the closure is not necessarily complete.