Def/Metric space

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Definition
Balls
Open sets

Definition of Metric space: A metric space is an ordered pair $(X, d)$, where $X$ is a set, and $d \colon X \times X \rightarrow \RR$ is a metric, that is:

$d(x,y) \geq 0$ for all $x,y \in X$.
$d(x,y) = 0 \Leftrightarrow x = y$.
$d(x,y) = d(y,x)$.
(The triangle inequality): If $x,y,z \in X$, then:

$$d(x,z) \leq d(x,y) + d(y,z).$$

In a metric space $(X,d)$, the nonnegative real number $d(x,y)$ is called the distance from $x$ to $y$ (or between $x$ and $y$).

Logical Connections

This definition logically relies on the following definitions and statements: Def/Real number, Def/Finite (set), State/Truth by false premise

The following statements and definitions logically rely on the material of this page: Def/Interior, and Def/Limit (sequence)

To visualize the logical connections between this definition and other items of mathematical knowledge, you can visit any of the following clusters, and click the "Visualize" tab: Clust/Metric geometry

If $(X,d)$ is a metric space, then $(X,d)$ has a natural topology $T_d$ with basis consisting of open balls. Specifically, we declare that a subset $U \subset X$ is open if: $$\forall u \in U, \exists r \in RR, \mbox{ such that } r > 0 \mbox{ and } B_r(u) \subset U.$$ In other words, a subset $U \subset X$ is open if, at every point of $U$, we can find an open ball centered at that point and contained within $U$.

Let $T_d$ be the set of open subsets of $X$, as defined above. The set $T_d$ is sometimes called the metric topology, or the topology induced by the metric $d$. This definition of openness satisfies the axioms for a topological space. Namely, we find the following about open sets:

The empty set is open.
The union of an arbitrary set of open sets is open.
The intersection of a finite number of open sets is open.

Indeed, the empty set is open, since the truth of the sentence "$\forall u \in \emptyset, \exists r \in \RR, \ldots$" is automatic (truth by false premise). Next, we check that the union of open sets is open:

	Let $F$ be a set of open subsets of $X$. Let $V = \bigcup F$ be their union.
	Suppose that $v \in V$. Then, there exists $U \in F$, such that $v \in U$. Hence, there exists a radius $r$, such that $B_r(v) \subset U$. Since $U \subset V$, we find that $B_r(v) \subset V$. Therefore, every point in $V$ is the center of an open ball contained in $V$
	Hence $V = \bigcup F$ is open.

Finally, we check that the finite intersection of open sets is open.

	Let $U_1, \ldots, U_n$ be a finite sequence of open subsets of $X$. Let $W = \bigcap_{i=1}^n U_i$.
	Suppose that $w \in W$. Then, $w \in U_i$, for all $i$ between $1$ and $n$. It follows that there is a finite sequence of radii $r_1, \ldots, r_n$, such that $B_{r_i}(w) \subset U_i$ for all $i$ between $1$ and $n$ (a finite sequence of choices does not require the axiom of choice). It follows that: $$\bigcap_{i=1}^n B_{r_1}(w) \subset \bigcap_{i=1}^n U_i = W.$$ On the other hand, if $r = min(r_1, \ldots, r_n)$, then we find that: $$B_r(w) = \bigcap_{i=1}^n B_{r_i}(w).$$ Hence, we find that: $$B_r(w) \subset W.$$
	Hence $W$ is open.

Def/Metric space

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