Is $[0,1]^\omega$ homeomorphic to $D^\omega$? Let $n\in \mathbb N$ and let $D^n$ be the closed $1$-ball in $\left(\mathbb R^n, \|\,\cdot\,\|_1\right)$. It is not too hard to show that $[0,1]^n \cong D^n$ in this case. 
This observation leads to the question whether we also have $[0,1]^\omega \cong D^\omega$, where
$$D^\omega = \left\{(x_n)_{n\in \mathbb N} \in \ell^1 \; \Bigg| \; \sum_{n=1}^\infty |x_n|\le 1\right\}$$
is the closed unit ball in $\ell^1$. 
Now if we view $D^\omega$ as a subspace of $\ell^1$, then we can prove that it is not homeomorphic to $[0,1]^\omega$ by arguing that the latter is compact, while the former is not.
But what happens if we view $D^\omega$ as a subspace of $\mathbb R^\omega$?

Is $D^\omega$ homeomorphic to $[0,1]^\omega$, when both sets are endowed with the subspace topologies induced by $\mathbb R^\omega$ (in the product topology)?

In this case both spaces are compact ($D^\omega$ is the intersection over all $n\in \mathbb N$ of the compact sets $\{x\mid \sum_{i=1}^n |x_i| \le 1\} \cap [-1,1]^\omega$) and all topological properties I can think of are preserved when we go from $[-1,1]^\omega \cong [0,1]^\omega$ to the subspace $D^\omega$. I have also tried to explicitly construct a homeomorphism, but to no avail.
I hope some of the topologically savvy guys on this site could help me out here! Thanks =)
 A: The answer is yes, they are homeomorphic. 

Edit: Here's a better way to put the answer below:

Theorem (Klee, 1955). Let $K$ be a compact, convex and metrizable set in a locally convex space $E$. If $K$ is not contained in a finite-dimensional subspace of $E$ then $K$ is homeomorphic to the Hilbert cube $[0,1]^\omega$.

This applies directly to the compact convex set $K = D^\omega$ in the metrizable space $E = \mathbb{R}^\omega$. This result is not explicitly stated this way in Klee's work [1], but it follows immediately from Theorem (1.2) of that paper. The proof idea is outlined a bit further down.

Original Answer:
Klee proved in 1955, based on a theorem due to Keller, the following remarkable result (Theorem (1.2) of reference [1] below): 

Let $X$ be a separable normed space and let $E$ be either $X$ with the norm topology, $X$ with the weak topology or $X^\ast$ with the weak$^\ast$ topology. If $K$ is an infinite-dimensional compact convex subset of $E$ then $K$ is homeomorphic to the Hilbert cube $[0,1]^\omega$.

The idea is that in each case one can find a countable family of continuous linear functionals $(f_n)$ separating the points of $K$ and, normalizing them appropriately, the map $x \mapsto (f_1(x),f_2(x),\ldots)$ gives a linear homeomorphism from $K$ onto an infinite-dimensional norm-compact convex subset of $\ell^2$. Keller had previously shown that all infinite-dimensional compact convex sets in $\ell^2$ are homeomorphic to $[0,1]^\omega$.
Since $D^\omega$ is a compact convex subset of $\ell^1 = (c_0)^\ast$, Klee's theorem applies. To finish up, note that the topology on $D^\omega$ viewed as a subset of $(c_0)^\ast$ and the topology induced from $\mathbb{R}^\omega$ coincide by the standard proof of Alaoglu's theorem.
Here are the relevant papers:


*

*V. L. Klee, Some topological properties of convex sets, Trans. Amer. Math. Soc. 78 (1955), 30–45. MR0069388

*Ott-Heinrich Keller, Die Homoiomorphie der kompakten konvexen Mengen im Hilbertschen Raum, Math. Ann. vol. 105 (1931) pp. 748–758. MR1512740
