PUMaC 2024 · 加试 · 第 5 题

Problem 5.0.2. Find a and a . (Take your time with this and draw some pictures! 1 2 Doing this problem slowly will help a lot with the other problems in this section) s Theorem 5.0.2. H ( ) ≤ a . n Proof. The original proof for this is easy, but a tiny-bit more than elementary, so we provide a more elementary proof sketch instead. s For each δ , we will provde a covering based on a such that H ( ) ≤ a . We will start n n δ s with showing it for H ( ). 1 Since a is the finite minimum over non-empty subsets of T , there is some collection n n { ∆ } , i ≥ 1, which minimizes a . Call this collection U , with # U = k . Lastly, let U = i i n S ∆ . i ∆ ∈ U i If U = , we are done; it is already a cover for , and a calculation shows the theorem is satisfied. n Otherwise, let V = T \ U ; # V = 3 − k . Each element of V will be covered in a n ′ scaled down way like we covered itself: if S ( ) = ∆ ∈ V , then scale down each I n m element of U and place it in V at the 2 n -level of the construction of . That is, if we ′ ′ ′ − n denote our scaled-down U as U and notating U similarly, then | U | = 2 | U | . There will n ′ be # V = 3 − k of these U s, one for each missed element of T that U does not hit. Our n n ′ covering so far consists of U , and the 3 − k sets congruent to U . Now, since U ̸ = , our ′′ new covering also doesn’t cover , so we create a U by scaling down U again, this time ′′ − n ′′ − 2 n 2 so that | U | = 2 | U | = 2 | U | , and there will be (# V ) of these, and so on. n Letting this process run to infinity, we get a cover of consisting of 1 copy of U , 3 − k ′ n 2 ′′ copies of U , (3 − k ) copies of U , and so on; thus, ∞ X s n i ( i ) s H ( ) ≤ (3 − k ) | U | 1 i =0 ∞ X n i − in s = (3 − k ) (2 | U | ) i =0 ∞ i n s s X 3 − k | U | | U | | U | s = | U | = = = = a n n n 3 − k k 3 μ ( U ) 1 − n n 3 3 i =0 s Where we use the definition of H and s = log 3 / log 2, the definition of a , and the geometric n δ series formula. − m Now, for δ < 1, there exists some m ∈ N such that 2 < δ . This part is fairly easy. m − m Instead of starting with the cover U , simply start with 3 copies of U scaled down by 2 , one at each level m tile of . Then perform the process as stated above for each of the n ( n ) scaled down copies of U . At each stage we will simply get 3 extra copies of U , and each − n of them will have a diameter scaled by 2 . So we get " # ∞ X s m n i − n ( n ) s H ( ) ≤ 3 (3 − k ) (2 | U | ) δ i =0 " # ∞ ∞ X X m − n log 3 / log 2 n i ( n ) s n i ( n ) s = 3 2 (3 − k ) ( | U | ) = (3 − k ) ( | U | ) = a n i =0 i =0 Using the definition of s and the earlier calculation. Since the largest cover we used this − n s time was size 2 < δ , this is a valid covering for H , and taking the limit gives δ s H ( ) ≤ a n