Subspaces that undo Products I have been working on Munkre's homework sets, and I have come across the following phenomenon: 
Let $\mathbb{R}_\ell$ be the lower limit topology on the real numbers. If you consider a line as a subspace of $\,\mathbb{R}_\ell\times \mathbb{R}$, it will be homeomorphic to one of the factors in the product(i.e. it will either be $\mathbb{R}_\ell$ or $\mathbb{R}$). In this way, the straight lines "undo" the product topology.
My question is this: Consider the product topology on a collection of spaces $\{X_\alpha\}_{\alpha\in A}$. For which subsets of the Cartesian product will the induced topology be homeomorphic to one of the $X_\alpha$ for some $\alpha\in A$?
 A: I'm not sure how familiar you are with the box/product topologies, so I'll keep it brief. I'll let Munkres fill in any gaps!
EDIT response: If the collection of spaces is finite, then the box and product topologies will agree, so the 'induced' topology of the subspace will still have  open $U \subset \prod_i X_i$ such that $X_n \cap U$ is open for some subspace $X_n \subset \prod_{i \in K} X_i$. Here, for some finite index $K$. Therefore the open sets in question would be the ones where we can identify $U = X_i$. If the collection is arbitrary, however, then these topologies won't agree; but in the box topology, such a $U_{\alpha} = X_{\alpha}$ would still be open.
Say we have an arbitrary index $J$ with some finite subset $H = \{\beta_1, \beta_2, \dots\}$ The product topology on such an arbitrary set is more interesting, because each subbasis element, by way of inverse projection mapping, generates an infinite number of basis elements $\prod_{\alpha \in J} U_{\alpha}$ where $U_{\alpha}$ is the ENTIRE space $X_{\alpha}$ as long as $\alpha \notin H$. So from your 'Cartesian' perspective, every basis element would contain a subset of the product space that is homeomorphic to an infinite number of such $X_{\alpha}$. [Note: for an uncountable product, this is pretty abstract, but by definition, there will always be an infinite number of such subsets open in this topology.] To tie it in with your example: consider a countable product $\mathbb{R}_{\ell} \times \mathbb{R} \times \mathbb{R} \times \dots$. Then, by the product topology, either $[a,b) \times \mathbb{R} \times \mathbb{R} \dots$ or $\mathbb{R}_{\ell} \times \mathbb{R} \times \dots \times (a,b) \times \mathbb{R} \times \dots$ are open, but each has subsets identical to the individual product spaces in question. The latter basis element is even homeomorphic to the whole space, just like what you were initially seeking! It's not fulfilling, given where I think you were headed with your initial question, but the coarseness of the product topology (rel the box topology) answers your question almost directly.
