# Continuous functions in product topology

There is a nice theorem characterizing continuous functions $A \to Y^X$ where $Y^X$ is equipped with product topology (pointwise convergence topology).

Is there a similar theorem for functions of the form $Y^X \to A$? I am most interested in a case where $Y^X = 2^\mathbb{N}$ and $A = \mathbb{R}$ (with the standard topology). Could someone give me some intuitions how continuous functions in this space (that is $A^{(Y^X)}$) look like (or at least some references)?

Edit:

There are non-constant continuous functions in this space, e.g. let $$f(A) = \sum_{k \in A} \frac{1}{2^{k}},$$ then $\lim_n f(a_n) = f(\lim_n a_n)$ for every $a_n$ convergent in $2^\mathbb{N}$ (in product topology = pointwise convergence topology, that means every sub-coordinate is constant from some point). Take any $\varepsilon > 0$, then there is a finite number $k_0$ such that $\sum_{i > k_0}2^{-i} < \frac{1}{2}\varepsilon$, so there is $M_0$ such that $$\forall k \leq k_0.\ \left(\forall m > M_0.\ k \in a_m\right) \lor \left(\forall m > M_0.\ k \notin a_m\right),$$ and thus for all $m > M_0$ we have $|f(a_m) - f(a)| < \varepsilon$.

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There is no general characterization of maps whose domain is an infinite product. However, it is possible to characterize maps $2^{\mathbb{N}}\to\mathbb{R}$.

First, observe that $2^\mathbb{N}$ is homeomorphic to the Cantor set.

There are a very large number of different maps $2^\mathbb{N}\to\mathbb{R}$. For the following description, let $2^{<\omega}$ be the set of all finite binary sequences, including the empty sequence $\emptyset$. If $\alpha$ and $\beta$ are binary sequences, we write $\alpha\preceq\beta$ if $\alpha$ is an initial subsequence of $\beta$.

To define a map $2^\mathbb{N}\to \mathbb{R}$, let $\{I_\alpha\}_{\alpha\in 2^{<\omega}}$ be a family of closed intervals with the following properties:

1. For each $\alpha,\beta \in 2^{<\omega}$ with $\alpha\preceq\beta$, we have $I_\beta\subseteq I_\alpha$.
2. For each $\alpha$, the interval $I_\alpha$ is the smallest closed interval containing $I_{\alpha0}\cup I_{\alpha1}$.
3. For any $\omega\in 2^{\mathbb{N}}$, the intersection $\bigcap_{\alpha\prec\omega} I_\alpha$ is a single point.

Such families of intervals are in one-to-one correspondence with maps $f\colon2^{\mathbb{N}}\to\mathbb{R}$. In particular, given such a family, we can define a map $f$ by the rule $$\bigcap_{\alpha\prec\omega} I_\alpha \;=\; \{f(\omega)\}.$$ To get the standard homeomorphism from $2^\mathbb{N}$ to the Cantor set, we would use the family $\{I_\alpha\}_{\alpha\in 2^{<\omega}}$ defined as follows:

• $I_\emptyset=[0,1]$.
• For each $\alpha\in 2^{<\omega}$, the interval $I_{\alpha0}$ is the first third of $I_\alpha$, and $I_{\alpha1}$ is the last third of $I_\alpha$.

By the way, the image of a map $2^{\mathbb{N}}\to\mathbb{R}$ can be any compact subset of $\mathbb{R}$. For example, here is a family which defines a map whose image is $[0,1]$:

• $I_\emptyset=[0,1]$.
• For each $\alpha\in 2^{<\omega}$, the interval $I_{\alpha0}$ is the first half of $I_\alpha$, and $I_{\alpha1}$ is the second half of $I_\alpha$.

Essentially, this is the map that interprets each element of $2^{\mathbb{N}}$ as a binary number in $[0,1]$.

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Thank you for your answer. Yet, there is one thing I don't understand. Let $p : 2^{\{0,1,2,3\}} \to 2^{\{0,1,2,3\}}$ be some bijection. Then $A \mapsto f(p(A \cap \{0,1,2,3\}) \cup (A \cap \overline{\{0,1,2,3\}}))$ might not be representable in your way (e.g. because of point 2), but would still be continuous if $f$ was (please remember that I am considering product topology on $2^\mathbb{N}$). Where am I wrong? –  dtldarek Jun 21 '12 at 20:20
@dtldarek I don't understand the formula you have written (starting with "$A\mapsto$"). –  Jim Belk Jun 21 '12 at 22:47
Let partition $A \in 2^\mathbb{N}$ into $B = A \cap \{0,1,2,3\}$ and $C = A \cap (\mathbb{N}-\{0,1,2,3\})$. Then my function could be defined as $g(A) = f(p(B) \cup C)$, that is, before applying $f$, I may change (almost arbitrarily) the first four coordinates of vector representation of $A \in 2^\mathbb{N}$, but I don't touch the rest. –  dtldarek Jun 21 '12 at 23:25
@dtldarek Note that the intervals $I_\alpha$ with $\mathrm{length}(\alpha)<4$ can be reconstructed from the later intervals. For example, $I_\emptyset$ is just the smallest interval containing $I_0$ and $I_1$. The map you suggest just involves permuting the intervals $I_\alpha$ for which $\mathrm{length}(\alpha) = 4$. –  Jim Belk Jun 21 '12 at 23:53