I wanted to prove the following result:
Let $\{X_\lambda\}$ be topological spaces and $X = \prod X_\lambda$ with the product topology. A net $(x_\alpha)_{\alpha\in A}$ on $X$ converges to $x\in X$ if and only if for each $\lambda$ the net $(\pi_\lambda(x_\alpha))_{\alpha \in A}$ converges to $\pi_\lambda(x)$.
I gave a proof for this, that I wanted to know if it is correct.
To give my proof, I first proved a quite simple lemma:
Let $X$ be a topological space with basis $\mathcal{B}$ for the topology and $(x_\alpha)_{\alpha \in A}$ a net in $X$, then $(x_\alpha)$ converges to $x$ if and only if for each $U\in \mathcal{B}$ with $x\in U$, there's $\gamma \in A$ such that $\gamma \leq \alpha$ implies $x_\alpha \in U$.
Proof: If $(x_\alpha)$ converges to $x$, the for any open set $U$ with $x\in U$ there's $\gamma \in A$ such that $\gamma \leq \alpha$ implies $x_\alpha \in U$. Since each element of $\mathcal{B}$ is open, it obviously holds for elements of $\mathcal{B}$.
On the other hand, suppose for each $U\in \mathcal{B}$ with $x\in U$ there's $\gamma \in A$ such that $\gamma \leq \alpha$ implies $x_\alpha \in U$. Let $V\subset X$ be open containing $x$. Since $V$ is open, because $\mathcal{B}$ is a basis for the topology, there's an element $U\in \mathcal{B}$ such that $x\in U\subset V$. But since $U\in \mathcal{B}$ we can find $\gamma \in A$ such that $\gamma \leq \alpha$ implies $x_\alpha \in U$, but since $U\subset V$ we have $x_\alpha\in V$ and hence $(x_\alpha)$ conveges to $x$.
With this lemma, I proved the proposition as follows:
Proof: First, if $(x_\alpha)$ converges to $x$, since $\pi_\lambda$ is continuous, we have $(\pi_\lambda(x_\alpha))$ converges to $\pi_\lambda(x)$ and there's nothing more to be shown.
To tackle the other direction, we know that a basis of the product topology is the set of sets of the form
$$U = \prod_{\lambda \in \Lambda} U_\lambda$$
such that there's $J\subset \Lambda$ finite with $U_\lambda = X_\lambda$ if $\lambda \in \Lambda \setminus J$ and $U_\lambda$ open in $X_\lambda$ if $\lambda \in J$.
In that case we use the above lemma. Suppose each $(\pi_\lambda(x_\alpha))$ converges to $x_\lambda$. We will show that $(x_\alpha)$ converges to $x$. For that, let $U$ be an element of the basis, containing $x$. In that case, $U$ is a product like we have already said.
Now, since $J$ is finite, let $J = \{\lambda_1,\dots, \lambda_n\}$. In that case, $x_{\lambda_i}\in U_{\lambda_i}$ which is open in $X_{\lambda_i}$. But the net $(\pi_{\lambda_i}(x_\alpha))$ converges to $x_{\lambda_{i}}$. Hence, there's $\gamma_i$ such that $\gamma_i \leq \alpha$ implies $\pi_{\lambda_i}(x_\alpha)\in U_{\lambda_i}$. Now, since $A$ is a directed set, there's $\gamma\in A$ such that $\gamma_i\leq \gamma$ for all $i=1,\dots,n$.
We claim now that if $\gamma\leq \alpha$ then $x_\alpha \in U$. Indeed, since $U$ is a product, to show that $x_\alpha\in U$ it suffices to show that $\pi_\lambda(x_\alpha)\in U_\lambda$. If $\lambda\in \Lambda\setminus J$ we have $U_\lambda = X_\lambda$ and there's nothing to be shown.
If $\lambda \in J$ then $\lambda = \lambda_i$ for some $i=1,\dots,n$. But then, $\pi_\lambda(x_\alpha)=\pi_{\lambda_i}(x_\alpha)$ and since $\gamma_i \leq \gamma$ and $\gamma \leq \alpha$ we have $\gamma_i\leq \alpha$ and by construction of $\gamma_i$ this means that $\pi_{\lambda_i}(x_\alpha)\in U_{\lambda_i}$.
Since no matter if $\lambda \in J$ or $\lambda \in \Lambda\setminus J$ we have $\pi_\lambda(x_\alpha)\in U_\lambda$, for $\gamma \leq \alpha$, we have shown that $x_\alpha \in U$ for $\gamma \leq \alpha$ and since $U$ is element of the basis, by the above lemma, $(x_\alpha)$ converges to $x$.
Now, is this proof correct? If it is, is there some simpler proof I could give, instead of this one? If it is not correct, where's my mistake?