= Standard unit sequences in the real countable product
{title2=$(e_n)_j=\begin{cases}1&j=n,\\0&j\ne n\end{cases}$}
Let $\mathbb N=\{1,2,\ldots\}$ 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 $e_n\to0$ 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.
\Image[/product-topology-unit-sequences-cylinder.png]
{title=A moving unit entry escapes each fixed finite-coordinate cylinder}
{description=Only coordinates 1 and 3 are tested in the displayed cylinder, with values required to lie between minus one-half and one-half. All later coordinates remain unrestricted. The finite display illustrates the pattern; the threshold argument in the text applies to every finite coordinate set.}
In the <l-infinity sequence space> of bounded real <sequences>, however,
$$
\|e_n-0\|_\infty=\sup_{j\geq1}|(e_n)_j|=1.
$$
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> $\prod_{j\geq1}(-1/2,1/2)$ of zero excludes every $e_n$, so $e_n$ is not a <convergent sequence> with limit zero there either.
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