Do we have $\sigma(X_1, X_2) = \sigma((X_1, X_2))$?

Question

Given two random variables $X_1, X_2: \Omega \rightarrow \mathbb{R}$, and $(X_1, X_2): \Omega \rightarrow \mathbb{R}^2$.

It is clear $\sigma(X_1, X_2) \subset \sigma((X_1, X_2))$ because we can take $(X_1, X_2)^{-1}(B\times \mathbb{R})$ or $(X_1, X_2)^{-1}(\mathbb{R} \times B)$.

For the other direction, we can use Dynkin's $\pi-\lambda$ theorem.

1. The collection $$P = \{(X_1, X_2)^{-1}((a,b)\times (c,d))\}$$ forms a $\pi$-system in $\sigma((X_1, X_2))$.

2. $P \subset \sigma(X_1, X_2)$.

3. Clearly $\sigma(X_1, X_2)$ is a $\lambda$-system.

By Dynkin's theorem, $\sigma(P) \subset \sigma(X_1, X_2)$.

Lastly, $\mathcal{B}({\mathbb{R}^2})$ can be generated by $(a,b)\times (c,d)$ so $$\sigma(P) = \{ \{(X_1, X_2)^{-1}(B), B\in \mathcal{B}({\mathbb{R}^2})\} = \sigma((X_1, X_2))$$

Is this argument okay?