1. ## Is this compact?

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 $S \subset Z^{\infty}_{+}$ where $Z_+$ 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 $O \subset Z$ ( $Z$ is again the set of all integers.) A counting measure means: If some $s \in S$'s row element was $s_a = 3$ with $a \in O$ 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 $a$. If $b \not\in O$ it must be that $s_b = 0$ for all vectors since no elements of this type can even exist.

The author now states that $O$ and $S$ must be chosen such that $S$ is compact. I am wondering: Isn't $Z^{\infty}$ already compact? I darkly remember some theorem. In any case: $Z$ is compact, or am I wrong? If so: Must $O$ be finite such that $S$ is automatically compact?

Help is very appreciated!
Best, Rafael

2. Z is a non-bounded subset of the reals and hence, by Heine-Borel, not compact (in the usual topology), no?

3. 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 $s \in S$ must be bounded too? Otherwise, this set would not be compact even if $O$ was compact? Just to be sure. Thank you!