On the homotopy type of $\mathbb{QP}^\infty$ It can be shown that the infinite-dimensional rational projective space $\mathbb{QP}^\infty$ is a connected, Hausdorff topological space. What can be said about its homotopy type (is it simply connected, is there any hope to compute its cohomology algebra)?
 A: It is nice that you asked a question about the space $\mathbb Q P^\infty$. I have thought about this space for a long time and came to the conclusion that $\mathbb Q P^\infty$ is the most "regular" space among countable connected Hausdorff spaces.
It seems that $\mathbb Q P^\infty$ is a unique space among countable connected Hausdorff spaces that admits a simle topological characterization:
Theorem. A topological space $X$ is homeomorphic to $\mathbb QP^\infty$ if and only if $X$ is countable, Hausdorff, and has a countable base $\mathcal B$ of the topology such that for any $n\ge 2$ and basic open sets $U_1,\dots,U_n\in\mathcal B$ the intersection $\bar U_1\cap\dots\cap \bar U_n$ is connected, non-empty, and has zero-dimensional complement $X\setminus (\bar U_1\cap\dots\cap \bar U_n)$.
The proof can be done by a (more-or-less) standard back-and-forth argument. 
A: Any countable Hausdorff space $Q$ is totally path-disconnected.  Indeed, if $f:[0,1]\to Q$ is continuous, then its image $X$ is a countable connected compact Hausdorff space.  By Urysohn's lemma, then, continuous maps from $X$ to $[0,1]$ separate points.  But $X$ is connected, so the image of a continuous map from $X$ to $[0,1]$ is connected, and so must be just a single point since $X$ is countable.  Thus $X$ can only have one point, so $f$ is constant.
So, in particular, $\mathbb{Q}\mathbb{P}^\infty$ is totally path-disconnected, and has the weak homotopy type of a countable discrete space.
(In fact, more strongly, any countable $T_1$ space is totally path-disconnected.  See Why are the integers with the cofinite topology not path-connected?)
A: $QP^\infty$ has an open cover by copies of $Q^\infty$. If $\gamma: I \to QP^\infty$ is a path, this pulls back to an open cover of the interval $I$, so $I$ has a cover by open intervals, each of which is mapped into some $Q^\infty$, and then, as Fan Zheng points out, projects down to $Q$ and thus is constant. So the path is constant. Hence $QP^\infty$ is homotopically discrete.
A: It would be more appropriate to study it using algebraic geometry; say, etale homotopy theory. See:
Čech Theory: Its Past, Present and Future
https://doi.org/10.1216/RMJ-1980-10-3-429
Dave
http://alpha.math.uga.edu/~davide/
