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All topologies mentioned are on the cartesian product $\mathbb{R}^{[0,1]}$.

The sequence of functions $(f_k)$, for which, $f_k(x) = x^k, x\in [0,1]$, converges pointwise but not uniformly to $f(x)=\ \begin{cases} 0 & x\in [0,1), \\ 1&x=1\\ \end{cases}$

Therefore, since a sequence of functions $(f_k)$ converges pointwise to $f$ if and only if $(f_k)$ converges to $f$ in the product topology, we have that $f_k \rightarrow f$ in the product topology. Why can't $f$ converge in the box topology? Does that need to be shown every time using the fact that the sequence of functions doesn't converge uniformly or does that follow from the fact that the box topology is finer than the uniform topology?

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Let $$U=\left\{g\in\Bbb R^{[0,1]}:\forall n\ge 2\left(\left|g\left(\frac1n\right)\right|<\frac1{n^n}\right)\right\}\;$$

$U$ is open in the box topology on $\Bbb R^{[0,1]}$, and $f\in U$, but for $k\ge 2$ we have $f_k\left(\frac1k\right)=\frac1{k^k}$, so $f_k\notin U$. Thus, $\langle f_k:k\in\Bbb Z^+\rangle\not\to f$ in the box topology.

More generally, there are no non-trivial convergent sequences in the box topology on $\Bbb R^{[0,1]}$. (Non-trivial here means not eventually constant.) It suffices to show that there is no non-trivial sequence converging to the zero function, which I’ll call $z$. Let $\langle f_n:n\in\Bbb N\rangle$ be any sequence in $\Bbb R^{[0,1]}$ such that $A=\{n\in\Bbb N:f_n\ne z\}$ is infinite. For $n\in A$ let $\epsilon_n=|f_n(2^{-n})|$, and let

$$U=\left\{g\in\Bbb R^{[0,1]}:\forall n\in A\Big(\left|g\left(2^{-n}\right)\right|<\epsilon_n\Big)\right\}\;;$$

then $U$ is an open nbhd of $z$, and $f_n\notin U$ for each $n\in A$. For each $m\in\Bbb N$ there is an $n\ge m$ such that $n\in A$ and hence $f_n\notin U$, so $U$ witnesses the fact that $\langle f_n:n\in\Bbb N\rangle\not\to z$.

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Thanks for the answer. I know that $f_k$ doesn't converge in the uniform topology, but I am wondering if, since the box topology is finer than the uniform topology, that this implies that $f_k$ doesn't converge in the box. –  The Substitute Oct 12 '12 at 7:58
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@Broseph: Yes, it does. More generally, let $\tau,\tau'$ be topologies on $X$ such that $\tau\subseteq\tau'$. If $\sigma=\langle x_k:k\in\Bbb N\rangle$ does not converge with respect to $\tau$, then it cannot converge wrt $\tau'$, either. Since $\sigma$ doesn’t converge wrt $\tau$, each $x\in X$ has a nbhd $U(x)\in\tau$ that doesn’t contain a tail of $\sigma$; each $U(x)\in\tau'$ as well, so each $x\in X$ also has a $\tau'$-nbhd that doesn’t contain a tail of $\sigma$. Thus, $\sigma$ can’t converge wrt $\tau'$. –  Brian M. Scott Oct 12 '12 at 8:06
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