# 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.

-
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.

-
@omer can u eleborate to some point so that i can understand better thanks – math Feb 7 '13 at 19:58
@omer that's fine I have one more question how could I show that f is non continuous in X could u please make me clear – math Feb 8 '13 at 13:55
I'm not sure what you are confused about, @motu, the answer I wrote includes a proof that the identity is not continuous (the sentence starting with "Also, f is not continuous, because"). Could you ask a specific question about it? – Omar Antolín-Camarena Feb 8 '13 at 15:32
I m facing problem in showing the composite map is continuous – math Feb 8 '13 at 17:26
The composition are the projection maps, which are always continuous for the product topology. In fact, the product topology can be defined to be the coarsest topology for which all projections are continuous. ("Coarsest topology" means the one with fewest open sets.) – Omar Antolín-Camarena Feb 11 '13 at 18:22