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Let $J$ denote an infinite --- countable or uncountable --- index set. Let $\mathbb{R}^J$ denote the set of all $J$-tuples of real numbers (i.e. the set of all real-valued functions with domain $J$) under the box topology having as basis all Cartesian products of the form $$B \colon= \Pi_{\alpha \in J} \left(a_{\alpha}, b_{\alpha} \right) = \left\{ \ (x_{\alpha})_{\alpha \in J} \ \colon \ x_{\alpha} \in (a_\alpha, b_\alpha ) \ \right\},$$ where $$ \left(a_{\alpha}, b_{\alpha} \right) \colon= \left\{x_{\alpha} \in \mathbb{R} \ \colon \ a_{\alpha} < x_{\alpha} < b_{\alpha} \ \right\}.$$

Then is there an example of a "non-constant" continuous function $f$ from $\mathbb{R}$ to $\mathbb{R}^J$, especially when $J$ is uncountable?

Even the function $f \colon \mathbb{R} \to \mathbb{R}^{\mathbb{N}}$ defined by $$ t \mapsto (t, t, t, \ldots )$$ fails to be continuous!

What would be the form of such continuous functions, I wonder?

An afterthought:

Is it true that such functions necessarily have to consist of constant functions except in only finitely many coordinates? That is, is it true that if $f \colon \mathbb{R} \to \mathbb{R}^J$ is to be given by $$f(t) \colon = \left( f_\alpha (t) \right)_{\alpha \in J} \ \ \ \forall \ t \in \mathbb{R},$$ where $f_\alpha \colon \mathbb{R} \to \mathbb{R}$ for each $\alpha \in J$, then in order for $f$ to be continuous, all the functions $f_\alpha$ are constant functions except possibly for only finitely many indices $\alpha$?

Saaqib Mahmood
  • 26,762

1 Answers1

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There is a continuous $f:\Bbb R\to\Bbb R^{\Bbb N}$, where $\Bbb R^{\Bbb N}$ has the box topology, such that if $\pi_n:\Bbb R^{\Bbb N}\to\Bbb R$ is the usual projection map to the $n$-th factor, then $\pi_n\circ f$ is surjective for each $n\in\Bbb N$. However, even though the range of $f$ is big enough to project onto each factor of the box product, it’s still true that each $f(\alpha)$ is eventually constant – in fact, eventually $0$.

Let

$$f:\Bbb R\to\Bbb R^{\Bbb N}:\alpha\mapsto\big\langle\operatorname{sgn}(\alpha)\max\{|\alpha|-n,0\}:n\in\Bbb N\big\rangle\;.$$

For example, $f(\pi)=\langle\pi,\pi-1,\pi-2,\pi-3,0,0,0,\ldots\rangle$.

For $x\in\Bbb R^{\Bbb N}$ and any sequence $\epsilon=\langle\epsilon_n:n\in\Bbb N\rangle$ of positive real numbers let

$$B(x,\epsilon)=\left\{y\in\Bbb R^{\Bbb N}:|x_n-y_n|<\epsilon_n\text{ for each }n\in\Bbb N\right\}\;.$$

Let $\alpha\in\Bbb R$ with $\alpha\ge 0$, and let $\epsilon=\langle\epsilon_n:n\in\Bbb N\rangle$ be a sequence of positive real numbers. Without loss of generality assume that each $\epsilon_n<1$, and that if $\alpha\notin\Bbb Z$, then

$$\epsilon_n<\min\{\alpha-\lfloor\alpha\rfloor,\lceil\alpha\rceil-\alpha\}\;.$$

Let $\delta=\min\{\epsilon_n:n\le\lfloor\alpha\rfloor\}$, and suppose that $|\beta-\alpha|<\delta$.

If $n\ge\lfloor\alpha\rfloor+1$, then $\beta<\alpha+\epsilon_n<\lfloor\alpha\rfloor+1\le n$, so $$\max\{\beta-n,0\}=0=\max\{\alpha-n,0\}\;.$$

If $n\le\lfloor\alpha\rfloor$, then certainly $|(\beta-n)-(\alpha-n)|=|\beta-\alpha|<\delta\le\epsilon_n$, and we consider two cases.

  • If $\beta\ge\alpha$, then $\max\{\beta-n,0\}=\beta-n$, and $|(\beta-n)-(\alpha-n)|<\epsilon_n$.
  • If $\beta<n\le\alpha$, then $\max\{\beta-n,0\}=0$, and $|0-(\alpha-n)|=\alpha-n<\alpha-\beta<\delta<\epsilon_n$. If $n\le\beta<\alpha$, this is just like the first case.

It follows that $f(\beta)\in B\big(f(\alpha),\epsilon\big)$ and hence that $f$ is continuous at $\alpha$. The argument for negative $\alpha$ is entirely similar.

If I assume the continuum hypothesis, I can prove that this is the best we can do: if $f$ is a continuous map of $\Bbb R$ into the box product $\Bbb R^{\Bbb N}$, then each $f(\alpha)$ is eventually constant.

Saaqib Mahmood
  • 26,762
Brian M. Scott
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