$ \newcommand{\N}{\mathbb{N}} $ Let $R$ be a transitive binary relation on $X$ and $(x_n)_{n\in\N}$ be a sequence in $X$.
Claim: If $\forall n \in \N: x_n R x_{n+1}$, $$ \forall m, n \in \N: [m < n \to x_m R x_n] $$
Attempted proof: Let $m \in \N$. Try proof by induction on $n$.
- Base step) $n = 1$. Since $m \in \N$, $m < 1 \to x_m R x_1$ is vacuously true.
- Inductive step) Suppose $m < n \to x_m R x_n$. Let $m < n+1$. Then, we have $m < n$ or $m = n$:
a) Given that $m < n$, $x_m R x_n$ by the hypothesis. From $x_n R x_{n+1}$, we have $x_m R x_{n+1}$.
b) Given that $m = n$, $x_n = x_m$. From $x_n R x_{n+1}$, we have $x_m R x_{n+1}$.
Therefore, for each $m \in \N$, $\forall n \in \N: [m < n \to x_m R x_n]$ holds.
Is this a valid proof? It seems I don't need to use the proof by induction on both $m$ and $n$ as in anon.
Additional) Did I implicitly use induction on $m$, e.g., to accept $\forall m \in \mathbb{N}: \neg m < 1 $? When do I need to use induction for both variables?
