/sci/ - Science & Math

[Anonymous No.11488428]

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)

Penguins for motivation.

[Anonymous No.11488451]

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 ī, it is true for all i ≤ ī, defining the Li(Fsubi) as the difference (Fsubī) - B = Li(Fsubi). In the sequence [Closure of]Fsubī U infinity [invoking Alexandroff compactification], clearly there are infinitely many values of i for which A is false, which defines Ls([Closure of]Fsubī).

Due to the compactness of (M,J), all subsequences are also compact, and due to Tychonoff's theorem, for ī sufficiently high,

∏ Fsubī = Fsubī

∏ Fsubi = Fsubi

Therefore, due to the properties of the Alexandroff compactification,

∏ [Closure of]Fsubī = ∏ Fsubi = Fsubi, defining a converging subsequence for ī sufficiently large, in relation to an Alexandroff compactification of Fsubī, converging on Fsubi. Therefore because ī has been chosen sufficiently large,

Ls([Closure of]Fsubī) = Li(Fsubi) = Ls(Fsubi) = Li (Fsubi). Because Li (Fsubi) ⊆ Ls (Fsubi), Fsubi has been defined as the limit, or closed convergence, of M.

[Anonymous No.11488453]

>>11488451
>Ls(∏ [Closure of]Fsubī) = Li(∏ Fsubi) = Ls(Fsubi) = Li (Fsubi)

Sorry fixed.

[Anonymous No.11488469]

>>11488428
Very desperate, at least sage.