Z is a non-bounded subset of the reals and hence, by Heine-Borel, not compact (in the usual topology), no?
Hello everybody,
I am reading a paper which currently gives me some confusion. The setup is like this (I do not need a prove for it. I am mostly curious if my intuition is right.)
I am looking at a set where is the set of all positive integers and the super index refers to an infinite dimension.
The vectors of this set are counting measures for another set ( is again the set of all integers.) A counting measure means: If some 's row element was with this would express that with this vecor 3 "entities" were counted (something of the model that is described in the paper) that have a value of . If it must be that for all vectors since no elements of this type can even exist.
The author now states that and must be chosen such that is compact. I am wondering: Isn't already compact? I darkly remember some theorem. In any case: is compact, or am I wrong? If so: Must be finite such that is automatically compact?
Help is very appreciated!
Best, Rafael
Outch, of course, thank you. At least I succeeded to prove that I should not work all night.
However, then the count for the element rows of any must be bounded too? Otherwise, this set would not be compact even if was compact? Just to be sure. Thank you!