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See if you could solve this, /sci/, as an exercise. It's not homework, don't worry. I already have my solution I will post afterwards.
Let (M, J) be any topological space. For generalized sequences (nets) (Fsubi) [for all i in I] of subsets of M one defines the limes inferior of Fsubi and limes superior of Fsubi and convergence analogously as for sequences. Show that every generalized sequence of subsets of M contains a converging generalized subsequence. (Hint: use Tychonoff's Theorem)
I didn't use Latex.
I said...
i is composed of nets, so the family I of M has the finite intersection property, and due to
A : Fsub1 ? Fsubi [for all values of 1 ? i ? I] = Ø
we have the topological space (M,J) being compact. Consider
Set of all subsequences where all A exists = B
If A is not true, then for a sufficiently high ?, it is true for all i ? ?, defining the Li(Fsubi) as the difference (Fsub?) - B = Li(Fsubi). In the sequence [Closure of]Fsub? U infinity [invoking Alexandroff compactification], clearly there are infinitely many values of i for which A is false, which defines Ls([Closure of]Fsub?).
Due to the compactness of (M,J), all subsequences are also compact, and due to Tychonoff's theorem, for ? sufficiently high,
? Fsub? = Fsub?
? Fsubi = Fsubi
Therefore, due to the properties of the Alexandroff compactification,
? [Closure of]Fsub? = ? Fsubi = Fsubi, defining a converging subsequence for ? sufficiently large, in relation to an Alexandroff compactification of Fsub?, converging on Fsubi. Therefore because ? has been chosen sufficiently large,
Ls(? [Closure of]Fsub?) = Li(? Fsubi) = Ls(Fsubi) = Li (Fsubi). Because Li (Fsubi) ? Ls (Fsubi), Fsubi has been defined as the Cauchy sequence, or closed convergence, of M.
