Characterization of product topology Let $(E_i)_{i\in I}$ be a family of metric spaces. I know that a sequence $(x_n)_n$ in the product space $E:=\Pi_{i\in I} E_i$ converges in $E$ iff $(\pi_i(x_n))_n$ converges in $E_i$ for all $i$ in $I$ (see Convergence in product topology).
Question: is the product topology completely characterized by this property? Or there could be other topologies on $E$ with this property?
 A: Recall the following definition:

**Definition:**A sequence $(x_n)$ $\tau$-converges to $a$ if for every open $A\in \tau$ there is $N_A\in \mathbb{N}$ such that $x_n\in A$ for all $n\geq N_A$.

So, we want to analyze the status of the following property:

(P) A sequence of elements $(x_n)_n$ $\tau$-converges to $a$ in the product space $E$ if and only if $(\pi_i(x_n))_n$ converges to $\pi_i(a)$ for every $i\in\mathbb{N}$.

There are examples of intermediate topologies (i.e., topologies which are finer than the product topology but contained in the box topology) which also satisfies the property (P).
Suppose $I=\mathbb{N}$ and for every $i\in\mathbb{N}$, $E_i=\mathbb{R}$ with the usual metric topology.
Consider the topology $\tau$ on $E=\prod_{i\in \mathbb{N}}E_i$ whose basic open sets have the form $A=\prod_{i\in\mathbb{N}}A_i$, where each $A_i$ is an open interval and $A_i=A_j$ for every sufficiently large $i,j$. 
Note here that if $A,B$ are sets in the form above, then $A\cap B$ is either empty of has the form above.
Now, note that $\tau$ contains the product topology (as each of the basic open sets in the product topology have the required form), but the contenence is strict as for example the open $A=\prod_{i\in\mathbb{N}} (-2,2)$ does not contain any basic open of the product topology.
This topology is also strictly contained in the box topology because, for instance, $\prod_{i\in\mathbb{N}}\left(\frac{-1}{i},\frac{1}{i}\right)$ belongs to the box topology but does not contain any basic set of $\tau$.
Thus, it remains to show that $\tau$ satisfy the property (P):
$(\Leftarrow)$ Suppose that a sequence $(x_n)$ converges component-wise to an element $a$ in $E$, and let $A=\prod_{i\in\mathbb{N}}A_i$ be a basic open in $\tau$ containing $a$. Hence, there is $k_A$ such that $A_i=A_j$ for all $i,j\geq k_A$, and $\pi_i(a)\in A_i$ for each $i\in\mathbb{N}$.
Let $r>0$ be small enough such that $(\pi_i(a)-r,\pi_i(a)+r)\subseteq A_i$ for all $i\leq k_A$ (this can be chosen as the intervals $A_i$ are open and we only need to take care of finitely many components).
Since each sequence $(\pi_i(x_n))_n$ converges to $\pi_i(a)$, there are natural numbers $t_1,\ldots,t_{k_A}$ such that $n\geq t_j$ implies $\pi_i(x_n)\in (\pi_i(a)-r,\pi_i(a)+r)$. Thus, by taking $t=\max\{t_1,\ldots,t_{k_A}\}$, we conclude that for every $n\geq t$, $(\pi_i(x_n))\in (\pi_i(a)-r,\pi_i(a)+r)\subseteq A_i$ for every $i\in \mathbb{N}$, i.e., $x_n\in A$ for every $n\geq t$.
Therefore, $(x_n)$ $\tau$-converges to $a$.
$(\Rightarrow)$ Suppose now that $(x_n)$ $\tau$-converges to $a$, and fix a component $i\in \mathbb{N}$. 
Let $(r,s)$ be an open interval containing $\pi_i(a)$, and consider the $\tau$-open given by $U=\prod_{i\in\mathbb{N}}A_i$ where $A_i=(r,s)$ for every $i\in\mathbb{N}$. Notice that this is an open in the topology $\tau$, and since the sequence $(x_n)$ is $\tau$-convergent, there is $k\in\mathbb{N}$ such that $x_n\in U$ for every $n\geq k_U$. In particular, we have $\pi_i(x_n)\in \pi_i(U)=(r,s)$ for every $n\geq k_U$. 
Thus, we have shown that $(\pi_i(x_n))_n$ converges to $\pi_i(a)$ in $E_i$, for each $i\in\mathbb{N}$.
A: The (Tychonov) product topology is the initial or smallest
topology for the product that makes all the projections continuous.  
A: The answer: no, the product topology is not completely characterized by this property.
To prove it, we'll abstract from the setting for a moment.
Let $(X, \tau)$ be a topological space. We say that a subset $S \subseteq X$ is sequentially closed if it is closed under limits of sequences, i.e. whenever a sequence $(a_n) \subseteq S$ converges to $a \in X$, we must have that $a \in S$.
Let $\tau'$ be a topology on $X$ such that each $S \subseteq X$ is closed in $\tau'$ if and only if $S$ is sequentially closed in $\tau$. It's routine to check that $\tau'$ indeed satisfies the topology axioms and also $\tau \subseteq \tau'$.
Lemma. For each sequence $(a_n) \subseteq X$ and element $a \in X$:
$$a_n \to a \text{ in } \tau \iff a_n \to a \text{ in } \tau'.$$
Proof. $(\impliedby)$ is obvious since $\tau'$ is stronger than or equal to $\tau$.
$(\implies)$ Suppose that $a_n \to a$ in $\tau$ and let $V  \in \tau'$ be a neighborhood of $a$. If there were infinitely many terms of the sequence $a_n$ outside $V$, then they would form a subsequence $(a_{n_k}) \subseteq F = X \setminus V$ of $a_n$ which would converge to $a$. But $F$ is sequentially closed, so $a \in F$, which is not true. So almost all terms of the sequence $a_n$ lie in $V$, therefore $a_n \to a$ in $\tau'$. $\ \blacksquare$
So $\tau$ and $\tau'$ have exactly the same convergent sequences and their limits. Now if there is a subset $S \subseteq X$ sequentially closed in $\tau$ but not closed, then $\tau'$ is stricly stronger than $\tau$, so convergence of sequences doesn't characterize $\tau$ completely. Hence it suffices to show that in the product topology on $\displaystyle E = \prod_{i \in I} E_i$ there might be a subset sequentially closed  but not closed.
If among $E_i$ only countably many have at least two points, then $E$ is metrizable, hence closedness and sequential closedness are the same. So suppose each $E_i$ has at least two distinct points, or equivalently, there is an open subset $\varnothing \subsetneq U_i \subsetneq E_i$.
Now let
$$S = \{ f \in E : f(i) \in U_i \text{ for at most countably many } i \in I \}.$$

*

*$S$ is sequentially closed.
Let $(f_n) \subseteq S$ converge to $f \in E$ in the product topology. For each $n \in \mathbb{N}$ the set $I_n = \{ i \in I : f_n(i) \in U_i \}$ is countable. Let $J = \{ i \in I : f(i) \in U_i \}$. For $i \in J$, since $f_n(i) \to f(i) \in U_i$, there is $n \in \mathbb{N}$ such that $f_n(i) \in U_i$, thus $i \in I_n$. Therefore $\displaystyle J \subseteq \bigcup_{n=1}^{\infty} I_n$ is countable, i.e. $f \in S$.


*$S$ is not closed; moreover, $S$ is dense in $E$.
Choose any point $g \in E$ and its base neighborhood $\displaystyle g \in G = \bigcap_{k=1}^n \pi_{i_k}^{-1}[V_{i_k}]$, where $V_{i_k} \subseteq E_{i_k}$ are open and $\pi_i : E \to E_i$ is the natural projection. Let $s(i_k) = g(i_k)$ for $k = 1, \ldots, n$ and $s(i) \notin U_i$ for the rest of $i \in I$. Then clearly $s \in S \cap G$. Also $S \neq E$ because each $U_i$ is non-empty and $I$ is uncountable; therefore $S$ is not closed.
