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Thread: Is this compact?

  1. #1
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    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 $\displaystyle S \subset Z^{\infty}_{+}$ where $\displaystyle Z_+$ 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 $\displaystyle O \subset Z$ ($\displaystyle Z$ is again the set of all integers.) A counting measure means: If some $\displaystyle s \in S$'s row element was $\displaystyle s_a = 3$ with $\displaystyle a \in O$ 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 $\displaystyle a$. If $\displaystyle b \not\in O$ it must be that $\displaystyle s_b = 0$ for all vectors since no elements of this type can even exist.

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

    Help is very appreciated!
    Best, Rafael
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  2. #2
    Guy
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    Z is a non-bounded subset of the reals and hence, by Heine-Borel, not compact (in the usual topology), no?
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  3. #3
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    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 $\displaystyle s \in S$ must be bounded too? Otherwise, this set would not be compact even if $\displaystyle O$ was compact? Just to be sure. Thank you!
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