Essentially infinite continuous maps $\prod_{\mathbf{N}} \mathbf{N} \rightarrow \mathbf{N}$?

Question

Let $\mathbf{N}$ be the natural numbers with the discrete topology (or really, any countable set), and consider the space of natural number sequences $\prod_{\mathbf{N}} \mathbf{N}$ with the product topology (sometimes called "the Baire space"). Is there an "explicit" example of a continuous function $\prod_{\mathbf{N}} \mathbf{N} \rightarrow \mathbf{N}$ that does not factor through projection $\prod_{\mathbf N} \mathbf{N} \rightarrow \prod_{i=0}^{n} \mathbf{N}$ onto some finite product? I can't think of one off the top of my head. If no explicit example, does such a map at least exist abstractly?

Answer 1

For a continuous function $f:\mathbb{N}^\mathbb{N}\to \mathbb{N}$ and any $x$, there exists $n_x$ such that if $y(j)=x(j)$ for all $j\leqslant n_x$, then $f(y)=f(x)$, because the sets $$\{(y(1),y(2),\ldots):(\forall j\leqslant n)(x(j)=y(j))\}$$ form a neighborhood base at $x$. So locally $f$ only depends on finitely many coordinates. To come up with an example that doesn't depend only on finitely many coordinates, we need an example where $\sup n_x=\infty$. That's how we come up with this example.

Define $$M=\bigl\{(x(1),x(2),\ldots)\in \mathbb{N}^\mathbb{N}:x(1)=\max\{x(j):j\leqslant x(1)\}\bigr\}.$$

This is a clopen set. If $x\in M$, then the neighborhood $U$ of $x$ given by $$\{(y(1),y(2),\ldots):(\forall j\leqslant x(1))(y(j)=x(j))\}$$ is contained in $M$. Therefore $M$ is open. If $x\in \mathbb{N}^\mathbb{N}\setminus M$, then there exists $1<j\leqslant x(1)$ such that $x(j)>x(1)$. Then the open neighborhood $$V=\{(y(1),y(2),\ldots):y(1)=x(1),y(j)=x(j)\}$$ is contained in $\mathbb{N}^\mathbb{N}\setminus M$. So $M$ is closed.

Therefore the indicator $$f(x)=1_M(x)=\left\{\begin{array}{ll}1 & : x\in M \\ 0 & : x\in \mathbb{N}^\mathbb{N}\setminus M\end{array}\right.$$ is continuous. But it doesn't factor through any projection onto the first $n$-coordinates. Let $x$ be constantly $n+1$ and let $y$ be $n+1$ for the first $n$ coordinates and $n+2$ after. Then $x,y$ have the same first $n$ coordinates, but $f(x)=1$ and $f(y)=0$.

Answer 2

The answer by user469053 explains the right strategy for obtaining an example, but there's a slightly simpler implementation: $$ f(x(1),x(2),x(3),\dots)=x(x(1)). $$ This doesn't factor through a projection to any finite subproduct because $x(1)$ can be arbitrarily large. But it's continuous because $f(x)$ depends on only two components of $x$.