Standard unit sequences in the real countable product ($(e_n{)}_j=\{\begin{array}{ll}1& j=n,\\ 0& j\ne n\end{array}$)

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.