I think it's ok.
I think I have the idea in my head, but I don't know if I am expressing it clearly in math terms.
The closure of , is defined to be (where are the limit points of )
totally bounded implies that finite number of points s.t
So we already have all the points in S, i.e the union of the balls is a cover for S. We just need to wory about S', the limit points.
s.t
so if any limit point is less than away from a point in S. We can say
Right? Is my thinking correct? The limit points are close to the set so we can make sure we get them by making the radius bigger.
My book defines totally bounded with an equal sign. Honestly "contained" makes much more sense though, it seems to be defined that way in every other source I've looked about. Is there a reason for this discrepancy in my book?
EDIT:
I think the whole space X is equal to the union, while a subset of X is contained.
Thanks for all your help!
Exactly. It is much more common in my experience that the definition involves equality instead of containment (despite what you indicate). This is just the difference of point of view where one usually thinks of topological (and in this case metric) spaces as living in ambient spaces (or not). So, if you're discussing the topology of a fixed space ( say) then since all the topology is done within this space a set contained in your ambient space can be said to be covered by if . But, if you think of as a space in it's own right (with the subspace topology say) then saying amounts to equality since (remembering now that we're thinking of as the ambient space itself) we have for every .