A description/proof of that subset of \(R^{d-k}\) is open if the product of \(R^k\) and subset is open
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean topological space.
Target Context
- The reader will have a description and a proof of the proposition that any subset of \(R^{d-k}\) is open if the product of \(R^k\) and the subset is open.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For the Euclidean topological spaces, \(\mathbb{R}^k\) and \(\mathbb{R}^{d - k}\), any set, \(U \subseteq \mathbb{R}^{d - k}\), is open on \(\mathbb{R}^{d - k}\) if \(\mathbb{R}^k \times U\) is open on \(\mathbb{R}^d\).
2: Proof
For any point, \(p \in \mathbb{R}^k \times U\), there is an open ball, \(B_p := \{ \forall (x^1, . . ., x^d) | \sum_{i=1, . . ., d} (x^i - p^i)^2 \lt \varepsilon^2\}\), contained in \(\mathbb{R}^k \times U\). Then, \(\{ \forall (p^1, . . ., p^k, x^{k + 1}, . . ., x^d) | \sum_{i = k + 1, . . ., d} (x^i - p^i)^2 \lt \varepsilon^2\}\) is contained in \(\mathbb{R}^k \times U\) because it is contained in \(B_p\). For any \(p' \in U\), \((p^{k + 1}, . . ., p^d)\), there is a \(p \in \mathbb{R}^k \times U\), \((p^1, . . ., p^k, p^{k + 1}, . . ., p^{d})\), and \(\{ \forall (p^1, . . ., p^k, x^{k + 1}, . . ., x^d) | \sum_{i = k + 1, . . ., d} (x^i - p^i)^2 \lt \varepsilon^2\}\) is contained in \(\mathbb{R}^k \times U\), and that \({ (x^{k + 1}, . . ., x^d)}\) is an open ball contained in U.
