#studyprep

[dreamstudiess]

02-20

studying in the library always makes me happier than studying at home. its nice to have a change of scenery but ap world can kindly go away, ick. boba is a good motivator too :p

[lifeofayoungdoctor]

Instead of Meal Prep...

Anyone else leave their study area prepared for the morning? Like a meal prep but with your study material 😂

[studysmartbegreat]

Perfect study-prep! Yes I post a lot of food, but you know what? I LOVE FOOD❤️😍

[fine-liner-junkie-blog]

My bujo layout from last week. I have gaps through the week but I'm working on it! 💪🏼💯

[maricine]

Preparing for finals. Wish me luck

[dreamstudiess]

02/10/19-

Today’s to-do list and a TBR list, all for a snowy day. (Gudetama’s such a mood)

🎶Lily- Alan Walker

[pico42]

The Suslin Theorem and Strong Separation Theorem

The former is a special case of the latter. also note that this is just DST prep for a potential meetup with a mathematician in Finland; they seem pretty strong in DST; obviously not 100% done.

Theorem 1. (Strong Separation Theorem) Suppose κ is an infinite cardinal and A, B are disjoint κ-Suslin subsets of some perfect product space χ. Then there exists a (κ+1)-Borel set C which separates A from B.

Proof (Sketch-ish). This proof simply relies on separating sets combinatorically; we use trees to accomplish this. Basically we use a tree to exhibit a (κ+1)-Borel set C that separates A from B. Suppose that A = ⋃_{i ∈ I} A_i, and B = ⋃_{j ∈ J} B_j. Note that the index sets I and J are arbitrary, and that for each i ∈ I and j ∈ J, there is a set C_{i,j} which separates A_i from B_j. We will then know that C = ⋃_{i ∈ I} ⋂_{j ∈ J} C_{i,j} separates A from B. If it is assumed that A and B are κ-Suslin sets of irrationals, then, at least under closure under Borel substitution and the perfect set theorem/property, we can study κ-Suslin sets in terms of trees. Borel substitution helps in that, given an arbitrary pointset P = 𝒜^κ_u P_u, with each P_u closed, and 𝒜 representing continuous Borel substitution, then f^-1[P] = 𝒜^κ_u. Since each f^-1[P_u] is Borel here, this is κ-Suslin. The perfect set theorem/property helps in that it (especially the theorem's results) holds for non-empty perfect subsets in particular, thus making it "nicer" to work with trees for this.

Not to get too detailed about the tree here, but here is the gist of it; the tree J on ω ⨉ κ ⨉ κ is defined as:

((t_0, ξ_0, η_0),...,(t_{n-1}, ξ_{n-1}, η_{n-1}) ∈ J iff (t_0, ξ_0,...,t_{n-1}, ξ_{n-1}) ∈ T and (t_0, η_0,..., t_{n-1}, η_{n+1})∈S.

And then one can define on the tree a function f: u -> C_u (in which u is a sequence from tree J) such that for each sequence u in ω ⨉ κ ⨉ κ:

(a) C_u is (κ+1)-Borel, and

(b) C_u separates A_τ(u) from B_σ(u),

in which A_τ(u) and B_σ(u) are all sets of sequences in ω ⨉ κ ⨉ κ. This then completes the proof. //

[scholarshipja]

Tomorrow!! AIM Boss @nicolemclarencampbell and @aimeduservices Chief Organization Officer Lesley will be dishing out the BEST tips to stay organized and keep on track during this COVID Quarantine for FREE! 📋 Sign up now at aimonlineacademy.com/schoolsoutseries to secure your spot in the Webinar! Swipe up in their story to register as well! 📱 #webinar #aimhugher #aimwebinar #studyprep #aimeducationalservices (at ScholarshipJamaica.com) https://www.instagram.com/p/B92XdjOh5nu/?igshid=bc8cpoql1ng7