In a proof of a theorem about nested intervals (see https://en.wikipedia.org/wiki/Nested_intervals#Theorem), I extracted following lemma, which was used without any proof. Therefore I tried to prove it.
Let $(I_n)$ with $I_n =[a_n, b_n]$ be a nested sequence of nonempty closed and bounded intervals in $\mathbb{R}$.
(Nested means: $\forall n \in \mathbb{N}: I_{n+1} \subset I_n\ \ \ (*)$)
Then $\forall k,n \in \mathbb{N}: k \le n \implies I_n \subset I_k$
How can I prove this statement?
Update: Here my proof attempt. Can anyone tell me if it is correct?
To prove: $\forall k \in \mathbb{N} (\ \forall n \in \mathbb{N} (k \le n \implies I_n \subset I_k)\ )$
Proof:
Fix an arbitrary $k \in \mathbb{N}$. Prove $\forall n \in \mathbb{N} (k \le n \implies I_n \subset I_k)$ by induction on n.
Base Case:
To show: $(k \le 1 \implies I_1 \subset I_k)$. True for k=1 since $I_1 \subset I_1$. Implication anyway true for k > 1.
Inductive Step:
Fix an arbitrary $n \in \mathbb{N}$.
Assume: $k \le n \implies I_n \subset I_k\ \ \ (**)$
To show: $k \le n+1 \implies I_{n+1} \subset I_k$
Proof:
Assume $k \le n+1$.
Case 1: $k \lt n+1$: Then $k \le n$. Then, by $(**)$, we have $I_n \subset I_k$. Since the intervals are nested, by $(*)$, we have $I_{n+1} \subset I_n$. Since $\subset$ is transitive, we have $I_{n+1} \subset I_k$.
Case 2: $k = n+1$: Then $I_{n+1} \subset I_{n+1}$. Thus $I_{n+1} \subset I_{k}$.