Metric Spaces Lecture Notes PDF

## Metric Spaces I by P.K. Jain & Khalil Ahmed Metric: distance function $(X, d)$ metric spaces * $i) d(x, y) \geq 0, \forall x, y$ * $ii) d(x, y) = 0 \iff x = y$ * $ iii) d(x, y) = d(y, x)$ * $ iv) d(x, y) \leq d(x, z) + d(z, y)$ $d: X \times X \to \mathbb{R}$ ### Chapter 1.1 Prelim...

## Metric Spaces I by P.K. Jain & Khalil Ahmed Metric: distance function $(X, d)$ metric spaces * $i) d(x, y) \geq 0, \forall x, y$ * $ii) d(x, y) = 0 \iff x = y$ * $ iii) d(x, y) = d(y, x)$ * $ iv) d(x, y) \leq d(x, z) + d(z, y)$ $d: X \times X \to \mathbb{R}$ ### Chapter 1.1 Preliminaries **Set:** Well-defined unordered collection of distinct objects **Synonyms:** Aggregate, class, family Tabulated form: $A = \{a, e, i, o, u\}$ Defining property method: $A =\{x | x \text{ is a vowel in English}\}$ $a \in A$ , $\cancel{a} \notin A$. **Subset:** $A \subset B$ if $x \in A \implies x \in B \qquad \phi \subset A, A \subset A$ **Proper Subset:** If $A \neq \phi, A \neq B$ and $A \subset B$ then A is called a proper subset of B and we denote it as $A \subseteq B$ $B = \{1, 2, 3, 4\} \qquad A = \{1, 2\}$ A is proper subset of B ### Operations on set $A \cup B = \{x | x \in A \text{ or } x \in B\}$ union $A \cap B = \{x | x \in A \text{ and } x \in B\}$ intersection $A - B = \{x | x \in A \text{ but } x \notin B\}$ $B - A = \{x | x \in B \text{ but } x \notin A\}$ $A - B \neq B - A$ $A^c = U - A$ complement $A - B = \{x \in U | x \in A \text{ but } x \notin B\}$ $= \{ x \in U | x \in A \} \cap \{ x \notin B \}$ $= A \cap B^c$ $A - B = A \cap B^c$ Let A, B, C be sets, then the following laws hold good: 1) Commutative property $A \cup B = B \cup A, \qquad A \cap B = B \cap A$ 2) Associative property $A \cup (B \cup C) = (A \cup B) \cup C, \qquad A \cap (B \cap C) = (A \cap B) \cap C$ 3) $A \cup U = U, \qquad A \cap U = A$ 4) $A \cup A^c = U, \qquad A \cap A^c = \Phi$ 5) Distributive Laws $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ 6) De Morgan's Laws $(A \cup B)^c = A^c \cap B^c$ $(A \cap B)^c = A^c \cup B^c$ Let $A_1, A_2, \dots A_n$ be finitely many sets then $A_1 \cup A_2 \cup \dots \cup A_n = \bigcup_{i=1}^{n} A_i = \{x | x \in A_i \text{ for at least one } i \}$ $A_1 \cap A_2 \cap \dots \cap A_n = \bigcap_{i=1}^{n} A_i = \{x | x \in A_i \text{ for each } i \}$ $\{1, 2, 3 \dots, n\}$ is indexing set Taking $N = \{1, 2, 3, 4, ...\}$ as indexing set $\bigcup_{i=1}^{\infty} A_i = \{x | x \in A_i \text{ for some } i \in N \}$ $\bigcap_{i=1}^{\infty} A_i = \{x | x \in A_i \text{ for each } i \in N \}$ Let $\Lambda$ be indexing set, $\{A_\alpha\}_{\alpha \in \Lambda}$ be indexed family of sets $\bigcup_{\alpha \in \Lambda} A_\alpha = \{x | x \in A_\alpha \text{ for some } \alpha \in \Lambda \}$ $\bigcap_{\alpha \in \Lambda} A_\alpha = \{x | x \in A_\alpha \text{ for each } \alpha \in \Lambda \}$ De Morgan's Laws $(\bigcup_{\alpha \in \Lambda} A_\alpha )^c = \bigcap_{\alpha \in \Lambda} A_\alpha^c$ $(\bigcap_{\alpha \in \Lambda} A_\alpha )^c = \bigcup_{\alpha \in \Lambda} A_\alpha^c$