Extreme points of Hilbert cube Define the set$$Q:=\prod \limits _{n=1}^\infty \left [0,\frac{1}{n}\right ]$$as a subspace of $X:=\mathbb{R}^{\mathbb{Z}_{>0}}$ with the product topology. Clearly $Q$ is a convex compact subset of $X$, hence by the Krein-Milman Theorem we have that $Q$ is the closed convex hull of the set of its extreme points $\mathcal{E}(Q)$.
In the book of Bühler and Salamon it is stated that$$\mathcal{E}(Q)=\prod \limits _{n=1}^\infty \left \{0,\frac{1}{n}\right \},$$that the convex hull of any finite subset of $\mathcal{E}(Q)$ is nowhere dense in $Q$ and that therefore the convex hull of $\mathcal{E}(Q)$ is not equal to $Q$. It also says that the last fact follows from Baire Category Theorem.
I proved that $\displaystyle \mathcal{E}(Q)=\prod \limits _{n=1}^\infty \left \{0,\frac{1}{n}\right \}$ and that $Q$ is a complete metric space, hence if I could express the convex hull of $\mathcal{E}(Q)$ as a countable union of convex hulls of finite subsets of $\mathcal{E}(Q)$ and prove that each of said convex hulls is nowhere dense in $Q$, then all the assertions would have been proved.
However, I couldn't prove that the convex hull of a finite subset is nowhere dense and I couldn't find a way of expressing the convex hull of $\mathcal{E}(Q)$ as a countable union (it is clearly equal to the union of the convex hulls of all of its subsets, but I don't know which subsets I should discard in order to obtain a countable union). How can I prove these two facts?
 A: Let $S$ be the span of the finite subset, this clearly has finite dimension.
The product of Hausdorff spaces is Hausdorff, we see that $S$ is closed. Hence $S \cap Q$ is closed.
The open sets in $Q$ are of the form $U \cap Q$, where $U$ is open in $X$, so suppose $U \cap Q \subset S \cap Q$.
Suppose $u \in U \cap Q$.
Note that $U$ is the union of sets of the form $\prod_n U_k$ where all but a finite number of the $U_k$ satisfy $U_k = [0, {1 \over k}]$. In particular, the set
$A=\{u_1,\} \times \cdots \times \{ u_N \} \times \prod_{k>N} [0, {1 \over k}]$ is contained in $U \cap Q$ for some $N$ and so $A \subset S \cap Q \subset S$.
Since $S$ is a subspace, it must contain the sets
$\{0\}^N \times \mathbb{R}^m \times \{0\} \times \cdots$, and hence it must have arbitrarily large dimension, which is a contradiction.
In particular, the closure of the convex hull of a finite number of points of $Q$ cannot contain any open set.
Aside:
If the goal is to show that $\operatorname{co} \mathcal{E}(Q) \neq Q$, consider the following.
Let $\alpha_k \in [0,1]$ be distinct and let $x^*_k = \alpha_k {1 \over k}$. We see that $x^* \in Q$. In particular, note that $k x^*_k$ takes on countably many different values.
Suppose $b_1,...,b_N \in \mathcal{E}(Q)$. Then
$(k [b_1]_k,...., k[b_N]_k) \in \{0,1\}^N$ and so can take on at most a finite number of values.
Now fix convex multipliers $\mu_1,...,\mu_N$ and consider
$\sum_i \mu_i b_i$, note that $k [\sum_i \mu_i b_i ]_k$ can only take on a finite number of values and hence $x^* \neq \sum_i \mu_i b_i$.
In particular, $x^* \notin \operatorname{co} \mathcal{E}(Q) $.
