We define the \(n\)-dimensional Euclidean space as:
\[ \mathbb{R}^n = \{(x_1, \dots, x_n) \mid x_i \in \mathbb{R}\} \]On the real line \(\mathbb{R}\), topological space (t.s.) open sets are generated by open intervals. If \(A \subseteq \mathbb{R}^n\) and \(B \subseteq \mathbb{R}^m\), then:
\[ A \times B = \{(x,y) \mid x \in A, y \in B\} \subset \mathbb{R}^{n+m} \]More generally, if \(A_i \subseteq \mathbb{R}^{n_i}\), then \(A_1 \times A_2 \times \dots \times A_k \subset \mathbb{R}^{n_1 + \dots + n_k}\).
An open rectangle in \(\mathbb{R}^n\) is a cartesian product of open intervals \((a_1,b_1) \times \dots \times (a_n,b_n)\), where \(a_i, b_i \in \mathbb{R} \cup \{\pm \infty\}\).
The collection of all open rectangles in \(\mathbb{R}^n\) forms a basis for a topology on \(\mathbb{R}^n\). Verify that \((\mathbb{R}^n, \tau)\) is second countable and Hausdorff.
A set \(O \subset \mathbb{R}^n\) is open if for every \(x \in O\), there exists an open rectangle \(I \subset \mathbb{R}^n\) such that \(x \in I \subset O\).
For \(a_i < b_i \in \mathbb{R}\), we call \([a_1,b_1] \times \dots \times [a_n,b_n]\) a closed rectangle in \(\mathbb{R}^n\).
Show that a closed rectangle is closed in \((\mathbb{R}^n, \tau)\).
Recall that \([a,b] \subset \mathbb{R}\) (\(-\infty < a < b < \infty\)) is a compact subset of \(\mathbb{R}\). Therefore, \([a_1,b_1] \times \dots \times [a_n,b_n]\) is compact in \(\mathbb{R}^n\).
One equivalent way to think of boundedness in \(\mathbb{R}\) is that a set \(A \subset \mathbb{R}\) is bounded (bdd) if \(A \subset [a,b]\) for some \(-\infty < a < b < \infty\).
A set \(A \subset \mathbb{R}^n\) is said to be bounded if \(A \subset \prod_{i=1}^{n} [a_i, b_i]\) for finite \(a_i, b_i\).
Suppose \(K \subset \mathbb{R}^n\) is closed and bounded. Then \(K\) is compact in \(\mathbb{R}^n\).
Heine-Borel Theorem: A subset \(K \subset \mathbb{R}^n\) is compact if and only if it is closed and bounded. (Recall: A closed subset of a compact space is compact).
\((\mathbb{R}^n, +)\) is a real vector space of dimension \(n\). For \(x, y \in \mathbb{R}^n\), where \(x = (x_1, \dots, x_n)\) and \(y = (y_1, \dots, y_n)\), we define the inner product:
\[ \langle x,y \rangle := \sum_{i=1}^{n} x_i y_i \]This inner product defines the Euclidean norm (length) of \(x \in \mathbb{R}^n\):
\[ |x| = \left(\sum_{i=1}^{n} x_i^2\right)^{1/2} \]- \(|x| \ge 0\), and \(|x| = 0 \iff x = 0\).
- \(|\lambda x| = |\lambda| |x|\).
- Triangle Inequality: \(|x+y| \le |x| + |y|\).
This Euclidean norm gives rise to the Euclidean distance on \(\mathbb{R}^n\) via \(d(x,y) := |x-y|\) for \(x,y \in \mathbb{R}^n\). Thus, \((\mathbb{R}^n, |\cdot|)\) is a metric space.
Let \(X\) be any topological space. For a function \(f: X \to \mathbb{R}^n\), one can define continuity. If \(A \subset \mathbb{R}^n\) and \(f: A \to X\) is continuous, and \(A\) is a compact set, then \(f(A)\) is compact.
More generally, if \(f: A \to \mathbb{R}^n\) is continuous, then \(f(A)\) is compact if \(A\) is compact.
Let \(f: \Omega \to \mathbb{R}^m\), where \(\Omega \subseteq \mathbb{R}^n\) is open.
\[ \lim_{x \to a} f(x) = L \quad (a \in \mathbb{R}^n) \]This means: \(\forall \epsilon > 0, \exists \delta > 0\) such that \(|f(x) - L| < \epsilon\) whenever \(0 < |x-a| < \delta\).
Let \(T: \mathbb{R}^n \to \mathbb{R}^m\) be a linear transformation. Then \(T\) is continuous.
Proof: Let \(e_j := (0, \dots, 1, \dots, 0)\) be the standard basis in \(\mathbb{R}^n\). \(T(e_j) = \sum_{i=1}^{m} a_{ij} e_i\), where \(a_{ij} \in \mathbb{R}\). The matrix representation is \([T] = (a_{ij})_{m \times n}\).
Now, \(x = \sum_{j=1}^{n} \lambda_j e_j\).
\[ T(x) = \sum_{j=1}^{n} \lambda_j T(e_j) = \sum_{j=1}^{n} \lambda_j \left( \sum_{i=1}^{m} a_{ij} e_i \right) = \sum_{i=1}^{m} \left( \sum_{j=1}^{n} \lambda_j a_{ij} \right) e_i \] \[ |T(x)|^2 = \sum_{i=1}^{m} \left( \sum_{j=1}^{n} \lambda_j a_{ij} \right)^2 \]By Cauchy-Schwarz inequality:
\[ |T(x)|^2 \le \sum_{i=1}^{m} \left( |x|^2 \cdot \sum_{j=1}^{n} a_{ij}^2 \right) = \left( \sum_{i=1}^{m} \sum_{j=1}^{n} |a_{ij}|^2 \right) |x|^2 \]So, \(|T(x)| \le M|x|\) for all \(x \in \mathbb{R}^n\), where \(M = \left(\sum_{i=1}^{m} \sum_{j=1}^{n} |a_{ij}|^2\right)^{1/2}\). Here, \(M\) is the norm of the linear transformation \(T\).
For \(x, y \in \mathbb{R}^n\), by linearity \(T(x-y)\) is a vector:
\[ |T(x) - T(y)| = |T(x-y)| \le M|x-y| \]Therefore, \(T\) is Lipschitz continuous. Hence, \(T\) is continuous. ∎
If \(f, g\) are continuous, then by composition of functions:
- \((f \pm g)(x) = f(x) \pm g(x)\) is continuous.
- \((\lambda f)(x) = \lambda \cdot f(x)\) is continuous.
- \(S: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n, S(x,y) = x+y\) is continuous.
- \(m: \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n, m(\lambda, x) = \lambda x\) is continuous.
A vector-valued function \(f: \mathbb{R}^n \to \mathbb{R}^m\), written as \(f = (f_1, \dots, f_m)\) where \(f_i: \mathbb{R}^n \to \mathbb{R}\), is continuous if and only if each \(f_i\) is continuous for \(1 \le i \le m\).
Proof:
\((\implies)\) \(|f_i(x) - f_i(a)| \le \left( \sum_{j=1}^{m} |f_j(x) - f_j(a)|^2 \right)^{1/2} = |f(x) - f(a)|\).
As \(x \to a\), \(|f(x) - f(a)| \to 0\) (small), which implies \(|f_i(x) - f_i(a)| \to 0\).
Let \(f: \mathbb{R}^n \to \mathbb{R}\) be bounded near \(a\). For \(\delta > 0\), define:
\[ M_\delta := \sup \{f(x) \mid |x-a| < \delta\} \] \[ m_\delta := \inf \{f(x) \mid |x-a| < \delta\} \]The oscillation of \(f\) at \(a\) is:
\[ O(f,a) := \lim_{\delta \to 0^+} (M_\delta - m_\delta) \]Does this limit exist? Let \(0 < \delta_1 < \delta_2\). Then \(M_{\delta_2} \ge M_{\delta_1}\) and \(m_{\delta_2} \le m_{\delta_1}\), which implies \(-m_{\delta_2} \ge -m_{\delta_1}\).
So \(M_{\delta_2} - m_{\delta_2} \ge M_{\delta_1} - m_{\delta_1}\). The function \(M_\delta - m_\delta\) is monotonic as \(\delta \to 0^+\). Thus, the limit exists and \(O(f,a)\) is well-defined.
\(f: \mathbb{R}^n \to \mathbb{R}\) is continuous at \(x=a \iff O(f,a) = 0\).