1. 设 `A, B, E` 是全集 `X` 的子集, 则 `B = (E nn A)^C nn (E^C uu A)` 当且仅当 `B^C = E`.

   `B = (E nn A)^C nn (E^C uu A)`
   `= (E^C uu A^C) nn (E^C uu A)`
   `= (E^C nn E^C) uu (E^C nn A) uu (A^C nn E^C) uu (A^C nn A)`
   `= E^C`. 这等价于 `B^C = E`.

2. 设 `A_1 sube A_2 sube cdots sube A_n sube cdots`, `B_1 sube B_2 sube cdots sube B_n sube cdots`, 则 `(uuu_(n=1)^oo A_n) nn (uuu_(n=1)^oo B_n)` `= uuu_(n=1)^oo (A_n nn B_n)`.

   记左边的集合为 `S`, 右边的集合为 `T`. 则 `S = {x | (EE n_1 in NN, x in A_(n_1)) and (EE n_2 in NN, x in B_(n_2))},`
   `T = {x | EE n in NN, x in A_n nn B_n}`. 显然 `T sube S`. 不妨设 `n_1 le n_2`, 则 `A_(n_1) sube A_(n_2)`, 此时 `S sube {x | EE n_2 in NN, x in A_(n_2) and x in B_(n_2)}=T`. 故 `S = T`.

3. 设 `A_1 supe A_2 supe cdots supe A_n supe cdots`, `B_1 supe B_2 supe cdots supe B_n supe cdots`, 则 `(nnn_(n=1)^oo A_n) uu (nnn_(n=1)^oo B_n)` `= nnn_(n=1)^oo (A_n uu B_n)`.
   记左边的集合为 `S`, 右边的集合为 `T`. 则 `S = {x | (AA n in NN, x in A_n) or (AA n in NN, x in B_n)},`
   `T = {x | AA n in NN, x in A_n uu B_n}`. 显然 `S sube T`. 现在取 `x in T`.
   1. 若 `EE k in NN, AA n ge k, x in A_n`, 则 `x in nnn_(n=k)^oo A_n = nnn_(n=1)^oo A_n sube S`.
   2. 若 `AA k in NN, EE n_k ge k, x !in A_(n_k)`, 则由于 `x in A_(n_k) uu B_(n_k)` 知 `x in B_(n_k) sube B_k`. 所以 `x in nnn_(n=1)^oo B_n sube S`. 故 `T sube S`.
   综上知 `S = T`.
4. 设有集合 `A, B, E` 和 `F`.
   1. 若 `A uu B = E uu F`, `A nn F = O/`, `B nn E = O/`, 则 `A = E` 且 `B = F`;
   2. 若 `A uu B = E uu F`, 令 `A_1 = A nn E`, `A_2 = A nn F`, 则 `A_1 uu A_2 = A`.
   1. 由 `A uu B = E uu F` 知, `(A uu B) nn A = (E uu F) nn A`, `(A uu B) nn E = (E uu F) nn E`. 化简得 `A = A nn E`, `A nn E = E`, 故 A = E. 同理 B = F.
   2. `(A nn E) uu (A nn F)`
      `= A nn (E uu F)`
      `= A nn (A uu B)`
      `= A`.

1. 设 `{f_n(x)}` 以及 `f(x)` 都是定义在 `RR` 上的实值函数, 且有 `lim_(n to oo) f_n(x) = f(x)`, `x in RR`, 则对 `t in RR`, 有 ` {x in RR | f(x) le t} = nnn_(k=1)^oo uuu_(m=1)^oo nnn_(n=m)^oo {x in RR | f_n(x) lt t + 1/k}`.

   只需证: 对任意 `t in RR`, `f(x) le t` 的充要条件是 `AA k in NN, EE m in NN, AA n ge m, f_n(x) lt t + 1/k`. 由 `lim_(n to oo) f_n(x) = f(x)` 知, `AA epsi gt 0`, `EE m in NN`, `AA n ge m`, `|f_n(x) - f(x)| lt epsi`.

   (1) 必要性. 对 `AA k in NN`, 取 `epsi = 1//k`, 由上式知 `EE m in NN`, `AA n ge m`, `f_n(x) lt f(x) + epsi le t + 1//k`.

   (2) 充分性. 若 `f(x) gt t`, 则可取 `epsi` 和 `1//k` 充分小, 使 `f(x) ge t + epsi + 1//k`, 但由条件知道, `EE m_1, m_2 in NN`, `AA n ge max(m_1, m_2)`, `f(x) lt f_n(x) + epsi lt t + 1//k + epsi`, 得到矛盾.

2. 设 `a_n to a` (`n to oo`), 则 ` nnn_(k=1)^oo uuu_(m=1)^oo nnn_(n=m)^oo (a_n - 1/k, a_n + 1/k) = {a}`.

   只需证: `x = a` 的充要条件是 `AA k in NN`, `EE m in NN`, `AA n ge m`, `|x - a_n| lt 1/k`. 由数列极限的唯一性, 这是显然的.

补充

1. `(AA i in I, A_i sube B) iff uuu_(i in I) A_i sube B`;
   `(AA i in I, B sube A_i) iff B sube nnn_(i in I) A_i`.