Definition: Let $X$ be a set, and let $d \colon X \times X \to \R$ be a function such that for all $x,y,z \in X$

1. $d(x,y) \geq 0$. (nonnegativity)
2. $d(x,y) = 0$ if and only if $x = y$. (identity of indiscernibles)
3. $d(x,y) = d(y,x)$. (symmetry)
4. $d(x,z) \leq d(x,y)+ d(y,z)$. (triangle inequality)

The pair $(X,d)$ is called a metric space. The function $d$ is called the metric or the distance function. Sometimes we write just $X$ as the metric space instead of $(X,d)$ if the metric is clear from context.

Lemma (Cauchy-Schwarz inequality): If $x =(x_1,x_2,\ldots,x_n) \in \R^n$ and $y =(y_1,y_2,\ldots,y_n) \in \R^n$, then \begin{equation*} {\biggl( \sum_{j=1}^n x_j y_j \biggr)}^2 \leq \biggl(\sum_{j=1}^n x_j^2 \biggr) \biggl(\sum_{j=1}^n y_j^2 \biggr) . \end{equation*}

Theorem: Let $(X,d)$ be a metric space and $Y \subset X$. Then the restriction $d|_{Y \times Y}$ is a metric on $Y$.

Definition: If $(X,d)$ is a metric space, $Y \subset X$, and $d' := d|_{Y \times Y}$, then $(Y,d')$ is said to be a subspace of $(X,d)$.

Definition: Let $(X,d)$ be a metric space. A subset $S \subset X$ is said to be bounded if there exists a $p \in X$ and a $B \in \R$ such that \begin{equation*} d(p,x) \leq B \quad \text{for all } x \in S. \end{equation*} We say $(X,d)$ is bounded if $X$ itself is a bounded subset.

Theorem: Suppose $(X,d)$ is a metric space, and $Y \subset X$. Then $U \subset Y$ is open in $Y$ (in the subspace topology), if and only if there exists an open set $V \subset X$ (so open in $X$), such that $V \cap Y = U$.

Theorem: Suppose $(X,d)$ is a metric space, $V \subset X$ is open, and $E \subset X$ is closed.

1. $U \subset V$ is open in the subspace topology if and only if $U$ is open in $X$.
2. $F \subset E$ is closed in the subspace topology if and only if $F$ is closed in $X$.

Definition: A sequence in a metric space $(X,d)$ is a function $x \colon \N \to X$. As before we write $x_n$ for the $n$th element in the sequence and use the notation $\{ x_n \}$, or more precisely \begin{equation*} \{ x_n \}_{n=1}^\infty . \end{equation*}

A sequence $\{ x_n \}$ is bounded if there exists a point $p \in X$ and $B \in \R$ such that \begin{equation*} d(p,x_n) \leq B \qquad \text{for all } n \in \N. \end{equation*} In other words, the sequence $\{x_n\}$ is bounded whenever the set $\{ x_n : n \in \N \}$ is bounded.

Definition: A sequence $\{ x_n \}$ in a metric space $(X,d)$ is said to converge to a point $p \in X$ if for every $\epsilon \gt 0$, there exists an $M \in \N$ such that $d(x_n,p) \lt \epsilon$ for all $n \geq M$. The point $p$ is said to be the limit of $\{ x_n \}$. We write \begin{equation*} \lim_{n\to \infty} x_n := p . \end{equation*}

A sequence that converges is convergent. Otherwise, the sequence is divergent.

Note: Compare this with the definition of convergence in $\R$

Theorem: A convergent sequence in a metric space has a unique limit.

Theorem: A convergent sequence in a metric space is bounded.

Theorem: Let $\{ x_j \}_{j=1}^\infty$ be a sequence in $\R^n$, where we write $x_j = \bigl(x_{j,1},x_{j,2},\ldots,x_{j,n}\bigr) \in \R^n$. Then $\{ x_j \}_{j=1}^\infty$ converges if and only if $\{ x_{j,k} \}_{j=1}^\infty$ converges for every $k=1,2,\ldots,n$, in which case \begin{equation*} \lim_{j\to\infty} x_j = \Bigl( \lim_{j\to\infty} x_{j,1}, \lim_{j\to\infty} x_{j,2}, \ldots, \lim_{j\to\infty} x_{j,n} \Bigr) . \end{equation*}

Definition: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and $c \in X$. Then $f \colon X \to Y$ is continuous at $c$ if for every $\epsilon \gt 0$ there is a $\delta \gt 0$ such that whenever $x \in X$ and $d_X(x,c) \lt \delta$, then $d_Y\bigl(f(x),f(c)\bigr) \lt \epsilon$.

When $f \colon X \to Y$ is continuous at all $c \in X$, then we simply say that $f$ is a continuous function.

Note: Compare this with the definition of continuity in $\R$

Theorem: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. Then $f \colon X \to Y$ is continuous at $c \in X$ if and only if for every sequence $\{ x_n \}$ in $X$ converging to $c$, the sequence $\bigl\{ f(x_n) \bigr\}$ converges to $f(c)$.

Note: Compare this Theorem with its version in $\R$

Lemma: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A function $f \colon X \to Y$ is continuous at $c \in X$ if and only if for every open neighborhood $U$ of $f(c)$ in $Y$, the set $f^{-1}(U)$ contains an open neighborhood of $c$ in $X$.

Theorem: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A function $f \colon X \to Y$ is continuous if and only if for every open $U \subset Y$, $f^{-1}(U)$ is open in $X$.