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The box topology on ${\prod }_{i\in I}{X}_i$ has a basis of a topology consisting of all Cartesian products ${\prod }_i{U}_i$ with every $U_i$ an open set in $X_i$; infinitely many coordinates may be restricted. It contains the product topology and agrees with it for finite $I$.^[ref]

For copies of the usual real line, the box topology is strictly finer when $I=\mathbb{N}$. The box $W={\prod }_{j\ge 1}(-1/2,1/2)$ is not an open set in the product topology: a basic cylinder set containing zero leaves some coordinate $j$ unrestricted, so it contains $e_j\notin W$. This is a construction for every finite coordinate set, not evidence from testing a finite prefix.

For a topological space $Y$ and topological spaces $X_i$, a function $f:Y\to {\prod }_i{X}_i$ is continuous into the product topology exactly when every coordinate function $f_i={\pi }_i\circ f$ is continuous.^[ref]

Proof outline: one direction uses function composition with the continuous projection maps. For the other, the preimage of a basic cylinder set isa finite set intersection of open sets in $Y$. This gives continuity on every basic test. It explains, for example, why $t\mapsto (t,t^2,t^3,\dots )$ from the real line to ${\mathbb{R}}^{\mathbb{N}}$ is continuous in the product topology: each coordinate is a polynomial function. The general coordinate criterion fails for the box topology, where infinitely many preimages can be required at once. For example, $g(t)=(t,t,\dots )$ has continuous coordinate functions, but the preimage of the box ${\prod }_{j\ge 1}(-1/j,1/j)$ is $\{0\}$, which is not an open set in the real line.

Let $\mathbb{N}=\{1,2,\dots \}$ and give each factor of ${\mathbb{R}}^{\mathbb{N}}$ the usual topology of the real line. The familiar standard basis vectors extend to unit sequences $e_n$ with $(e_n{)}_j=1$ when $j=n$ and zero otherwise. Their finite linear combinations are exactly the finitely supported sequences, rather than all of ${\mathbb{R}}^{\mathbb{N}}$.

For every fixed coordinate $j$, $(e_n{)}_j=0$ for all $n>j$. Therefore $e_n\to 0$ in the product topology. More directly, any basic cylinder set $B(F,U)$ containing zero contains every $e_n$ with $n>\max (F\cup \{0\})$: all of its tested coordinates are then zero. The moving unit entry escapes every fixed finite collection of observations.

In the l-infinity sequence space of bounded real sequences, however,Thus the same sequence does not converge to zero in the supremum norm. Here the comparison is between the subspace topology inherited from the product topology and the norm topology on ${\ell }^{\infty }$. The supremum norm is not a finite-valued norm on all of ${\mathbb{R}}^{\mathbb{N}}$, which also contains unbounded sequences. In the box topology on ${\mathbb{R}}^{\mathbb{N}}$, the neighborhood ${\prod }_{j\ge 1}(-1/2,1/2)$ of zero excludes every $e_n$, so $e_n$ is not a convergent sequence with limit zero there either.