embedding a T1 space into a product topology is the following statement true or false:
Let $X$ be a $T_1$ topological space, there exists a cofinite topological space $Y$, such that $X$ can be embedded into ${Y}^J$
 A: This is true by this paper (downloadable) where for a $T_1$ space a cardinal called the Fréchet character $\operatorname{frch}(X)$ is defined. A space $X$ has $\operatorname{frch}(X) \le \omega_\mu$ iff $X$ embeds into a power of spaces $F_\mu$, where this space is the set of ordinals $\le \omega_{\mu}$ in the cofinite topology. 
So we cannot find a single cofinite space that works, but one for every subclass of spaces with some bound on a cardinal. This in contrast to completely regular $T_1$ spaces, that all embed into powers of $[0,1]$ (a so-called generating space).
A: Here's an easy proof that you can always do this with $|Y|=|X|$.  For any closed set $C\subset X$ and $x\in X\setminus C$, define a map $f:X\to Y$ which maps all of $C$ to a single point and is otherwise injective (we can do this since $|Y|\geq |X|$).  If $X$ is $T_1$ and $Y$ has the cofinite topology, this $f$ will be continuous, since the inverse image of each point is closed.  Moreover, $f(x)\not\in\overline{f(C)}$.  That is, continuous functions $X\to Y$ separate points from closed sets.  It follows that if you take $J$ to be the set of all continuous maps $X\to Y$  and let $F:X\to Y^J$ be the canonical map, then $F$ is an embedding.
