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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.

The summability domain is the inverse image of the convergent sequence space under the bounded matrix operator on the l-infinity sequence space. It is therefore a closed linear subspace. The bounded divergence for every regular summability matrix makes it proper, so it has empty interior and is a nowhere dense set. The Baire category theorem then ensures that countably many regular methods miss some common bounded sequence.

Coordinate evaluations on the l-infinity sequence space form a countable separating family, so a weakly compact set $K$ is metrizable by countable separating family metrizes a weakly compact set. Choose a countable weakly dense subset $D$. Then $K$ lies in the weak closure of $\mathrm{conv}D$, which equals its norm closure by Mazur theorem. Rational convex combinations make that norm closure separable, and every subset of a separable metric space is separable. The weakly dense subset itself need not be norm dense in a nonconvex $K$.