updates

content/Math4111/Math4111_L6.md

@@ -126,7 +126,8 @@ $A$ is countable, $n\in \mathbb{N}$,
 
 $\implies A^n=\{(a_{1},...,a_{n}):a_1\in A, a_n\in A\}$, is countable.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Induct on $n$,
 
@@ -138,7 +139,7 @@ Induction step: suppose $A^{n-1}$ is countable. Note $A^n=\{(b,a):b\in A^{n-1},a
 
 Since $b$ is fixed, so this is in 1-1 correspondence with $A$, so it's countable by Theorem 2.12.
 
-QED
+</details>
 
 #### Theorem 2.14