Prove that $(f_n:\Lambda\to\bigcup_{\lambda\in\Lambda}X_\lambda)$ converges iff $(f_n(\lambda))$ converges in $(X_\lambda,\tau_\lambda)$. Let $((X_\lambda,\tau_\lambda))_{\lambda\in\Lambda}$ be a family of compact topological spaces. It is needed to prove that a sequence $(f_n:\Lambda\to\bigcup_{\lambda\in\Lambda}X_\lambda)_{n\in\mathbb{N}}$ in $\prod_{\lambda\in\Lambda}X_\lambda$ converges in the product topology iff $(f_n(\lambda))$ converges in $(X_\lambda,\tau_\lambda)$ for all $\lambda\in\Lambda$. 
The following are the observations I've made so far.


*

*$\prod_{\lambda\in\Lambda}X_\lambda$ with the product topology is compact by the Tychonoff theorem. 

*The problem is talking about a sequence of $\Lambda$-tuples; so the sequence may be considered as ${(\mathbf{x_1,x_2,...})}$, where $\mathbf x_i$ is a $\Lambda$-tuple for each $i\in\mathbb{N}$.
Are these observations correct? 
I do not know how to proceed. A hint would be appreciated. Thanks.
 A: Compactness is unnecessary. All one needs to know is that the topology on $\prod_{\lambda \in \Lambda} X_\lambda$ is the minimal one that makes all projections $\pi_{\lambda_0}: \prod_{\lambda \in \Lambda} X_\lambda \to X_{\lambda_0}$ defined by $\pi_{\lambda_0}(f)  =f(\lambda_0)$ continuous, for all $\lambda_0 \in \Lambda$. The latter implies that all sets of the form $\cap_{i=1}^n \pi^{-1}_{\lambda_i}[O_{\lambda_i}]$, where $\lambda_1,\ldots,\lambda_n \in \Lambda$ and $O_{\lambda_i}$ is open in $X_{\lambda_i}$, for all $i \in \{1,\ldots n\}$ form a base for the topology on $\prod_{\lambda \in \Lambda} X_\lambda$.
If the $(f_n)$ converges to $f$ then the continuity of $\pi_{\lambda_0}$ implies that $f_n(\lambda_0) \to f(\lambda_0)$ for all $\lambda_0 \in \Lambda$. On the other hand, if we know the convergence in every coordinate and we take a basic neighbourhood $\cap_{i=1}^n \pi^{-1}_{\lambda_i}[O_{\lambda_i}]$, then we know that for all $i \in \{1,\ldots n\}$, $f_n(\lambda_i) \to f(\lambda_i)$, and as $f(\lambda_i) \in O_{\lambda_i}$ we can find $N_i \in \mathbb{N}$ such that for all $n \ge N_i$ we know that $f_n(\lambda_i) \in O_{\lambda_i}$. Now, if we set $N = \max(N_1,\ldots,N_n) \in \mathbb{N}$ (a maximum of finitely many numbers so well-defined) we have that for all $n \ge N$, $f_n \in \cap_{i=1}^n \pi^{-1}_{\lambda_i}[O_{\lambda_i}]$ as required. So $f_n \to f$ as these sets form a base for the topology on $\prod_{\lambda \in \Lambda} X_\lambda$.
