A description/proof of that Euclidean topological space nested in Euclidean topological space is topological subspace
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of subspace topology.
Target Context
- The reader will have a description and a proof of the proposition that any Euclidean topological space nested in any Euclidean topological space is a topological subspace of the nesting Euclidean topological space.
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 any Euclidean topological spaces, \(\mathbb{R}^{d_1}\) and \(\mathbb{R}^{d_2}\), such that \(\mathbb{R}^{d_2} \subseteq \mathbb{R}^{d_1}\) where inevitably \(d_2 \leq d_1\), where it does not mean that \(\mathbb{R}^{d_1} = \mathbb{R}^{d_2} \times \mathbb{R}^{d_1 - d_2}\), because \(\mathbb{R}^{d_2}\) may have be turned and/or translated against \(\mathbb{R}^{d_1}\), the topology of \(\mathbb{R}^{d_2}\) is the subspace topology of \(\mathbb{R}^{d_1}\).
2: Proof
It suffices to show that any open ball, \(B_1\), on \(\mathbb{R}^{d_1}\) is an open ball, \(B_2 = B_1 \cap \mathbb{R}^{d_2}\), or the empty set on \(\mathbb{R}^{d_2}\), and for any open ball, \(B_2\), on \(\mathbb{R}^{d_2}\), there is an open ball, \(B_1\), on \(\mathbb{R}^{d_1}\) such that \(B_2 = B_1 \cap \mathbb{R}^{d_2}\), because then, any open set, \(U_2\), on \(\mathbb{R}^{d_2}\) is a union of some open balls, \(\cup_{\alpha} B_{2_\alpha}\), on \(\mathbb{R}^{d_2}\), and each of the open balls will be open in the subspace topology, so, \(U_2\) will be open in the subspace topology, while any subset, \(S_2\), on \(\mathbb{R}^{d_2}\) that is open in the subspace topology is \(S_2 = U_1 \cap \mathbb{R}^{d_2}\) where \(U_1\) is open on \(\mathbb{R}^{d_1}\), but \(U_1\) is a union of some open balls, \(\cup_{\alpha} B_{1_\alpha}\), on \(\mathbb{R}^{d_1}\), and \(S_2 = (\cup_{\alpha} B_{1_\alpha}) \cap \mathbb{R}^{d_2} = \cup_{\alpha} (B_{1_\alpha} \cap \mathbb{R}^{d_2})\), but as \(B_{1_\alpha} \cap \mathbb{R}^{d_2}\) will be an open ball or empty set on \(\mathbb{R}^{d_2}\), \(S_2\) will be open on \(\mathbb{R}^{d_2}\).
Now, \(\mathbb{R}^{d_2}\) is the subset of \(\mathbb{R}^{d_1}\) where the \(d_2 + 1, . . ., d_1\) components are 0, which (the subset) is afterward turned around the origin and then translated. So, we can take the global chart on \(\mathbb{R}^{d_1}\) that is the standard chart turned around the origin and then translated in the same way with for \(\mathbb{R}^{d_2}\). In the new chart, \(\mathbb{R}^{d_2}\) is the subset of \(\mathbb{R}^{d_1}\) where the \(d_2 + 1, . . ., d_1\) components are 0, and any point on \(\mathbb{R}^{d_2}\) has the same \(1, . . ., d_2\) components with the \(\mathbb{R}^{d_2}\) standard chart and with the \(\mathbb{R}^{d_1}\) new chart. Let us always use the new chart from now on.
\(B_1\) is \(\{x \in \mathbb{R}^{d_1}| \sum_{i = 1, . . ., d_1} (x_i - p_i)^{2} \lt \epsilon^2\}\) where \(p\) is the center of the ball. \(B_1 \cap \mathbb{R}^{d_2}\) is \(\{x \in \mathbb{R}^{d_2}| \sum_{i = 1, . . ., d_2} (x_i - p_i)^{2} + \sum_{i = d_2 + 1, . . ., d_1} (p_i)^{2} \lt \epsilon^2\} = \{x \in \mathbb{R}^{d_2}| \sum_{i = 1, . . ., d_2} (x_i - p_i)^{2} \lt \epsilon^2 - \sum_{i = d_2 + 1, . . ., d_1} (p_i)^{2}\}\), which is an open ball or the empty set on \(\mathbb{R}^{d_2}\).
\(B_2\) is \(\{x \in \mathbb{R}^{d_2}| \sum_{i = 1, . . ., d_2} (x_i - p_i)^{2} \lt \epsilon^2\}\) where \(p\) is the center of the ball. There is a subset on \(\mathbb{R}^{d_1}\), \(\{x \in \mathbb{R}^{d_1}| \sum_{i = 1, . . ., d_2} (x_i - p_i)^{2} + \sum_{i = d_2 + 1, . . ., d_1} (x_i - 0)^{2} \lt \epsilon^2\}\), which is an open ball, named \(B_1\), on \(\mathbb{R}^{d_1}\) with the center \((p_1, . . ., p_{d_2}, 0, . . ., 0)\), and \(B_1 \cap \mathbb{R}^{d_2}\) is \(B_2\).
