Problem 992

最大的 $K$ 為 1011。

首先證明 $K \ge 1011$。令 $\epsilon \in (0, \frac{1}{1009})$, 並考慮集合 $\mathcal{A} = \{0, 2 + \epsilon, 4 + 2\epsilon, \dots, 2018 + 1009\epsilon\}$。顯然 $\mathcal{A}$ 的每個點都必須存在一個 $A_i$ 包含之, 且這些 $A_i$ 兩兩互斥 (因為 $A_i$ 的長度皆為 1), 故這邊共選到 1010 個 $A_i$。又基於 $\cup A_i = [0, 2021]$, 我們必須有一個 $A_j$ 是 $[2020, 2021]$, 而這與前述的 1010 個 $A_i$ 都不相同 (基於 $2018 + 1009\epsilon + 1 < 2020$), 故我們得到 1011 個兩兩互斥的 $A_i$。

接著我們證明 $K \le 1011$。考慮 $A_i = [i - 1, i]$, $i = 1, 2, \dots, 2021$。易知我們最多僅能找到 1011 個兩兩互斥的空集合 $[0, 1], [2, 3], \dots, [2020, 2021]$, 故 $K \le 1011$。

[The largest $K$ is 1011.

First we prove $K \ge 1011$. Let $\epsilon \in (0, \frac{1}{1009})$, and consider the set $\mathcal{A} = \{0, 2 + \epsilon, 4 + 2\epsilon, \dots, 2018 + 1009\epsilon\}$. Clearly, for each point of $\mathcal{A}$ there must exist some $A_i$ containing it, and these $A_i$ are pairwise disjoint (since the $A_i$ all have length 1), so here we obtain a total of 1010 such $A_i$. Also, since $\cup A_i = [0, 2021]$, there must be some $A_j$ equal to $[2020, 2021]$, and this is different from all the aforementioned 1010 $A_i$ (since $2018 + 1009\epsilon + 1 < 2020$), so we obtain 1011 pairwise disjoint $A_i$.

Next we prove $K \le 1011$. Consider $A_i = [i - 1, i]$, $i = 1, 2, \dots, 2021$. It is easy to see that we can find at most 1011 pairwise disjoint intervals, namely $[0, 1], [2, 3], \dots, [2020, 2021]$, so $K \le 1011$.]