# Product topology hinders me

Let $X= \{0,1\}^\Bbb N$ be endowed with the box product topology where each factor has the discrete topology. Find a space $Y$ and a function $f: Y \to X$ such that for each $n$ the composition $\pi_n\circ f: Y \to \{0,1\}$ is continuous while $f$ is not itself continuous.

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Is N a finite $N$ or $\mathbb N$, the set of natural numbers? –  Asaf Karagila Feb 7 '13 at 19:43
N is the the set of natural no –  math Feb 7 '13 at 19:51

Take Y to be $\{0,1\}^{\mathbb{N}}$ with the product topology and $f : Y \to X$ to be the identity. For each $n$, $\pi_n \circ f : Y \to \{0,1\}$ is just a projection all of which are continuous for the product topology. Also, $f$ is not continuous, because, for example, the set $Z$ consisting solely of the constant sequence $(0,0,\ldots)$ is open in $X$, but not in $Y$ (so that the inverse image of the open set $Z$ is not open).
The main point is that the box topology on $X$ is actually the same as the discrete topology, making finding non-continuous functions to $X$ very easy.