Misc fixes, 11 part 1

docs/math/real-analysis/11-metric-spaces.md

@@ -0,0 +1,39 @@
+# Chapter 11 - Metric Spaces
+
+## Section 11.4 - Netric Spaces
+
+**Definition**. A *metric* on set $S$ is a function $d: S \otimes S \rightarrow \mathbb{R}$ that satifies the following properties for all $x, y, z \in S$,
+
+- $d(x, y) \geq 0$
+- $d(x, y) = 0 \; \text{ if and only if } x = y$
+- $d(x, y) = d(y, x)$
+- $d(x, y) \leq d(x, z) + d(z, y)$
+
+**Definition**. A *metric space* $(S, d)$ is a set $S$, with elements called *points*, together with a metric $d$.
+
+**Definition**. With metric space $(S, d)$, if $A \subset S$, then $(A, d)$ is a *subspace* of $(S, d)$.
+
+**Definition**. The *discrete metric* is provided by
+
+$$
+d(x, y) = \begin{cases}
+  0 \; \text{ if } x = y \\
+  1 \; \text{ if } x \neq y
+\end{cases}
+$$
+
+---
+
+**Definition**. Let $(S, d)$ be a metric space. Then, for each $\epsilon > 0$, the *$\epsilon$-neighborhood* or *$\epsilon$-ball* of a point $a \in S$ is the set
+
+$$
+V_\epsilon(a) = {x \in S | d(a, x) < \epsilon}
+$$
+
+**Definition**. Let $(S, d)$ be a metric space. Then, a subset $G \subseteq S$ is *open* if for each $x \in G$, there exists some $\epsilon > 0$ so that $V_\epsilon(x) \subseteq G$.
+
+**Definition**. Let $(S, d)$ be a metric space. Then, a subset $G \subseteq S$ is *closed* if its complement $C(G) = S - G = S \ F$ is closed.
+
+**Definition**. Let $(S, d)$ be a metric space. A point $c \in S$ is a *cluster point$ of a set $A \subseteq S$ if every $\epsilon$-neighborhood of $c$ contains some point $a \in A$ such that $a \neq c$.
+
+**Theorem**. Every $\epsilon$-neighborhood of a point is an open set.