Prove that the projection $p_i$ from the product space $X_I S_i \tau$ into $S_i \tau_i$ is open for each $i \in I$ Full question: A function $f$ from a space $X, \tau$ to a space $Y, \tau'$ is said to be open if $f(V)$ is open in $Y$ whenever $V$ is an open subset of $X$. Prove that the projection $p_i$ from the product space $X_I S_i \tau$ into $S_i \tau_i$ is open for each $i \in I$.
Attempt: Let $S = \Pi_{i \in I} S_i$ be the product space. Then, let $p_i: S, \tau \to S_i, \tau_i$ be the projection $p_i(s) = s_i$ for each $s \in S$. Then, $p_i$ is continuous for all $i \in I$.
Then, if $V_i$ is open in $S_i$, then $p_i^{-1}(V_i) = S$ is one of the open sets in the basis, namely $\{\Pi_I V_i : V_i \text{ open in } S_i \text{ and } V_i = S_i \text{ for all but at most finitely many i}\}$, and we see clearly that $V_i$ fits the criteria for all $i \in I$ because we defined $V_i$ as being open in $S_i$ and $I$ is a finitely countable set, hence there will be at most finitely many $i$ for which $V_i = S_i$.
Thus, the projection $p_i$ is open for each $i \in I$.
I'm not sure if I tackled this correctly and wanted to make sure I was on the right track or got it right.
 A: A point $f \in X = \prod_i X_{i \in I}$ is by definition a function $f: I \mapsto \cup_{i \in I} X_i$ such that $\forall_{i\in I}: f(i) \in X_i$.
For every fixed $j \in I$, the projection $p_j: \prod_{i \in I} X_i \mapsto X_j$ on the $j$-th coordinate is defined by $p_j(f) = f(j)$.
By definition of the product topology on $X$, the standard base of that topology is given by all sets of the form $\cap_{i \in F} (p_i)^{-1}[U_i]$, where $F$ is any finite subset of $I$, and $U_i$ is an open subset of $X_i$ for every $ i \in F$.
Proposition: a function $f: X \mapsto Y$ between topological spaces is open iff for some fixed base $\mathcal{B}$ for the topology of $X$ we have that $f[O]$ is open in $Y$ for all members of $\mathcal{B}$. (This essentially means that we can check openness of a function by checking not on all open sets but it's enough to check for basic open sets, for some (convenient) base for $X$).
The proof of the proposition is simple: right to left is trivial, as basic open sets are open in particular, and if $O \subset X$ is open, we can write $O = \cup_{B \in \mathcal{B}'}$ for some subfamily $\mathcal{B}' \subset \mathcal{B}$. As direct function images commute with arbitrary unions we have $f[O] = f[\cup_{B \in \mathcal{B}'} B] = \cup_{B \in \mathcal{B}'} f[B]$, which is by assumption a union of open subsets of $Y$, hence open in $Y$ as well.
Now, the proof that projections are open just comes down to the observation that for $j \in I$, $F$ a finite subset of $I$, $U_i \subset X_i$ open for all $i \in F$ we have that 
$$p_j[\cap_{i \in F} (p_i)^{-1}[U_i]] = U_j \mbox{ if } j \in F, \mbox{ and } = X_j \mbox{ otherwise.}$$
And both alternatives are indeed open in $X_j$, as required. Here we do assume all $X_i$ are non-empty, to avoid trivialities.
