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The box topology on the set $\mathbb R^\infty$ is defined to have sub-basis of sets $U_1 \times U_2 \times \ldots$ for each $U_i \subset \mathbb R$ open. Observe this is different from the product topology which also demands all but finitely many $U_i = \mathbb R$.

The box topology is known to be badly behaved. For example the reasonable-looking function $x \mapsto (x,x,\ldots)$ is not continuous when the codomain carries the box topology. On the other hand functions like $x \mapsto (x,1,1,\ldots)$ are in fact box-continuous.

There is an easy rule to check if any function $F \colon \mathbb R\to \mathbb R^\infty$ is product-continuous. Simply write $F(x) = (f_1(x),f_2(x),\ldots)$ then check all $f_i \colon \mathbb R \to \mathbb R$ are continuous. This condition is necessary and sufficient.

Is there a simple characterisation of exactly which functions are box-continuous? Failing such a characterisation, could someone provide me with a large family of box-continuous functions? That family has to at least include all functions like $G(x) =(g_1(x),g_2(x), \ldots, g_n(x), a_1,a_2,\ldots)$ for $g_i\colon \mathbb R \to \mathbb R$ continuous and $a_i$ constants.

Martin Sleziak
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Daron
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  • This is basically why the box topology is uninteresting. The product topology is characterized by how continuous maps behave with the projections. I don't know if there's an equivalent for the box topology though. – Matt Samuel Apr 19 '17 at 13:00
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    @MattSamuel Another perspective - this is basically why the box topology is interesting ;) – John Gowers Apr 19 '17 at 15:58
  • @John Sure, it's interesting if you're trying to study useless oddball topological spaces for the sake of it. Pure mathematician though I am, I don't think that's a productive thing to do. Perhaps there's pure and there's pure. – Matt Samuel Apr 20 '17 at 01:03

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Assume that $f:X\to\mathbb{R}^\infty$ is continous with $X$ connected.

Let $\pi_i:\mathbb{R}^{\infty}\to\mathbb{R}$ be the projection onto $i$-th coordinate and define $f_i=f\circ\pi_i$, i.e. $f_i$ is the $i$-th coordinate of the function $f$.

As you can read for example here connected components of $\mathbb{R}^{\infty}$ are characterized by the following property: $(a_n)$ and $(b_n)$ are in the same connected component if and only if $\{n\in\mathbb{N}\ |\ a_n\neq b_n\}$ is finite. In particular, since $X$ is connected then for any $x,y\in X$ we have that $f(x)$ and $f(y)$ differ at at most finitely many indexes.

So one property is: for any $x,y\in X$ there is only finitely many $i$ such that $f_i(x)\neq f_i(y)$. I think that the inverse also holds, i.e. if each $f_i$ is continous and for any $x,y\in X$ we have $f_i(x)\neq f_i(y)$ for at most finitely many $i$ then $f$ is continous.

Note that the property doesn't mean that all but finitely many $f_i$ are constant. The counterexample would be: take $\{p_i:\mathbb{R}\to\mathbb{R}\}$ to be an infinite partition of unity with compact supports. Then $x\to(p_1(x), p_2(x), p_3(x), \ldots)$ is continous. And it sill satisfies the property, indeed for any $x$ there is only finitely many $i$ with $p_i(x)\neq 0$.

This also explains why $x\mapsto(x,x,x,\ldots)$ is not continous.

That's all I can provide you with.

freakish
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