The Hilbert Cube in the Product Topology and Box Topology

I am working toward showing that the Hilbert cube is closed in both the box and topologies, and then I have to show their respective interiors. I am pretty confident about how I showed it was closed, but I am a little uncertain about my interiors. Either, I might be misunderstanding how an open subset of $H$ would look like, and if my unioning is appropriate with respect to the final interior. This is my attempt:

The subset of $\mathbb{R}^{\omega}$ defined by

\begin{equation*} H = \prod_{n\geq 1} \left[0,\frac{1}{n}\right] \end{equation*} is called the Hilbert cube.

Show that $H$ is closed in $\mathbb{R}^{\omega}$ with respect to both the box and the product topologies.

Because the box topology is finer than the product topology, it will be enough to show that the complement of $H$ in the product topology is open since it will then also be open in the box topology. To begin, recall that projections are continuous \begin{equation*} \pi_{j}:\mathbb{R}^{\omega}\rightarrow R_{j}\quad\mathrm{where\ } \forall j\in\mathbb{N}\backslash\{0\}\ R_{j} = \mathbb{R}, \end{equation*} in the product topology. From this it follows that preimages of closed sets in $R_{j}$ are closed in $\mathbb{R}^{\omega}.$ Thus, $\pi_{j}^{-1}\left(\left[0,\frac{1}{j}\right]\right)$ is closed. In particular, \begin{equation*} \pi_{j}^{-1}\left(\left[0,\frac{1}{j}\right]\right) = \prod_{n\in\mathbb{N}\backslash\{0\}}R_{n},\quad R_{n} = \mathbb{R}\ \ \ n\neq j\quad\mathrm{and}\ \ R_{j} = \left[0,\frac{1}{j}\right] \end{equation*} is a closed set in $\mathbb{R}^{\omega}$ with the product topology. Moreover, arbitrary intersections of closed sets are closed. Thus, \begin{equation*} \bigcap_{j\in\mathbb{N}\backslash\{0\}} \pi_{j}^{-1}\left(\left[0,\frac{1}{j}\right]\right) = \prod_{j\in\mathbb{N}\backslash\{0\}} \left[0,\frac{1}{j}\right] = H \end{equation*} and indeed $H$ is a closed set in $\mathbb{R}^{\omega}$ with the product topology. From this it follows that $\mathbb{R}^{\omega}\backslash H$ is open in $\mathscr{T}_{\mathrm{pro}}$, and therefore in $\mathscr{T}_{\mathrm{box}}$ as well. But if $\mathbb{R}^{\omega}\backslash H$ is open in $\mathscr{T}_{\mathrm{box}}$, then it means $H$ is closed in the box topology.

Find $\mathrm{int}(H)$ if $\mathbb{R}^{\omega}$ has the box topology and find $\mathrm{int}(H)$ if $\mathbb{R}^{\omega}$ has the product topology

The open sets of the product topology are infinite products with every component being the whole corresponding space except for finitely many components. Thus, for $U\in\mathscr{T}_{\mathrm{pro}}$ to be an open subset of $H$ it must be that $\forall j\in\mathbb{N}\backslash\{0\}$ we have $\pi_{j}(U)\subseteq \left[0,\frac{1}{j}\right].$ However, only finitely many of these projections will not be the whole space. Those which are the whole space will certainly not be a subset of a closed interval. That is, without loss of generality if the first $N$ components are not all of $\mathbb{R},$ then for every $k>N$ $\pi_{k}(U) = \mathbb{R}\not\subseteq \left[0,\frac{1}{k}\right].$ Thus, the only open subset of $H$ is the empty set and as a result, $\mathrm{int}(H) = \varnothing$ in the product topology. In the box topology, however, open sets are infinite products where each component is simply an open set in the corresponding component's topology. Thus, for $U\in\mathscr{T}_{\mathrm{box}}$ to be an open subset of $H,$ it follows that $\forall j\in\mathbb{N}\backslash\{0\}$ we have $\pi_{j}(U) = U_{j}\subseteq\left[0,\frac{1}{j}\right].$ Thus, $U_{j}$ must be of the form $(a,b)$ with $0<a<b<\frac{1}{j}.$ And if we union all such $U$ with these components, then the interior of $H$ is \begin{equation*} \mathrm{int}(H) = \prod_{n\in\mathbb{N}\backslash\{0\}} \left(0,\frac{1}{n}\right) \end{equation*}

• Looks fine to me.The product topology is also known as the Tychonoff product topology. In practice it has been more fruitful than the box topology.I am more used to defining the Hilbert cube as $I^{\omega}$ where $I=[0,1]$ although it's homeomorphic to your $H$ so it's a moot point. – DanielWainfleet Oct 17 '15 at 5:46