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\[d(x,y)=\begin{cases}0&\text{if}\ \ x=y\\ 1&\text{if}\ \ x\neq y\end{cases}\quad.\]
This always available metric is called the _discrete metric_ and is a useful metric to remember if one is looking for a counterexample.
**Example 56**.: The usual absolute value defines a metric on \(\mathbb{R}\):
\[d(x,y)=|x-y|.\]
Anytime we refer to \(\mathbb{R}\) without explicitly naming a metric, the _absolute value metric_ is always understood to be the one we have in mind. It is useful to note that by restricting the metric to a subset, any subset of a metric space can be considered as a metric space.
**Example 57**.: Let \(A\) be a subset of \(M\); if we restrict \(d(x,y)\) to \(A\), we define \(d(x,y)\) on \(A\times A\) satisfying the properties \(a)\) to \(d)\) above using the same letter \(d\) and say \((A,d)\) is a metric space. In particular, when we consider \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\), and \(\mathbb{R}\setminus\mathbb{Q}\), each come already supplied with a natural metric, namely the restriction of the usual absolute value metric on \(\mathbb{R}\).
**Example 58**.: In \(\mathbb{R}^{n}\) the notion of \(n\)-dimensional Euclidean distance produces a metric. If \(x=(x_{1},x_{2},\dots,x_{n})\) and \(y=(y_{1},y_{2},\dots,y_{n})\) are elements in \(\mathbb{R}^{n}\), then
\[d(x,y)=\left((x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}+\dots+(x_{n}-y_{n})^{2}\right) ^{1/2}\]
defines a metric on \(\mathbb{R}^{n}\). This is the _usual metric_ on \(\mathbb{R}^{n}\). In order to justify this claim we need to check the above conditions from a) to d). The triangle inequality is the one which requires attention and we will give a proof of it when we see “Minkowski’s Inequality” later.
**Remark 31**.: There is an important collection of metrics on \(\mathbb{R}^{n}\). Let \(p\) be a real number such that \(p\geq 1\). For \(x=(x_{1},x_{2},\dots,x_{n})\) and \(y=(y_{1},y_{2},\dots,y_{n})\in\mathbb{R}^{n}\), we can define:
\[d_{p}(x,y)=||x-y||_{p}\]
where by \(||x||_{p}\), we mean
\[||x||_{p}=\left(\sum_{i=1}^{n}|x_{i}|^{p}\right)^{1/p}.\]
To show \(d_{p}\) is a metric on is a metric on \(\mathbb{R}^{n}\), we need the following _Holder’s Inequality_. See, e.g., [22], pp. 182-196 for details.
_Holder’s Inequality_: Let \(p\) and \(q\) be positive numbers such that \(\frac{1}{p}+\frac{1}{q}=1\). Then for arbitrary real numbers \(a_{1},a_{2},\cdots,a_{n}\) and \(b_{1},b_{2},\cdots,b_{n}\),
\[\left|\sum_{k=1}^{n}a_{k}b_{k}\right|\leq\left(\sum_{k=1}^{n}|a_{k}|^{p} \right)^{\frac{1}{p}}\left(\sum_{k=1}^{n}|b_{k}|^{q}\right)^{\frac{1}{q}}.\]
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### Topology of Metric Spaces
The fundamental and necessary notion in understanding metric spaces is an open ball.
**Definition 41**.: Suppose that \((M,d)\) is a metric space and \(a\in M\). If \(r\) is a positive real number, the _open ball_ centered at \(a\) of radius \(r\) is a subset of \(M\) defined by
\[B_{r}(a)=\{x\in M:\ d(x,a) The _closed ball_ centered at \(a\) of radius \(r\) is a subset of \(M\) defined by \[\overline{B_{r}(a)}=\{x\in M:\ d(x,a)\leq r\}.\] Notice that in \(\mathbb{R}\), open balls are open intervals \(B_{r}(a)=(a-r,a+r)\) and closed balls are closed intervals \(\overline{B_{r}(a)}=[a-r,a+r]\). If \((M,d)\) is a metric space with the discrete metric, then \(B_{r}(a)=\overline{B_{r}(a)}=\{a\}\) for all \(0 The following figure illustrates the unit balls (i.e., when \(r=1\)) centered at \((0,0)\) in \(\mathbb{R}^{2}\) with respect to several metrics (Figure 2.3): * \(d(x,y)=\left((x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}\right)^{1/2}\) (the usual metric). **Definition 42**.: Let \((M,d)\) be a metric space and \(A\subset M\). The set \(A\) is an _open_ set in \(M\) if, for each \(a\in A\) there is an \(r>0\) such that \(B_{r}(a)\subseteq A\). A set \(F\subseteq M\) is called closed if \(F^{c}:=M\setminus F\) is open (Figure 2.4). Figure 2.3: The unit balls in \(\mathbb{R}^{2}\) with respect to various metrics **Proposition 18**.: _An open ball in any metric space \((M,d)\) is an open set. Every closed ball is closed set (Figure 2.5)._ : Let \(B_{r}(a)=\{x\in M:d(x,a) \[d(a,z)\leq d(a,y)+d(y,z) which implies \(z\in B_{r}(a)\) and \(B_{\epsilon}(y)\subset B_{r}(a)\) as desired. Similarly, one can show that \(\{x\in M:\ d(x,a)>r\}\) is an open set. Hence the closed ball is closed. **Proposition 19**.: _In any metric space \(M\), the empty space \(\emptyset\) and the full space \(M\) are open sets._ : To show the empty set \(\emptyset\) is open, we must show that each point in \(\emptyset\) is the center of an open ball contained in \(\emptyset\). Since there are no points in \(\emptyset\), this requirement is automatically satisfied. On the other hand since every open ball centered on each point is clearly contained in \(M\), this implies that \(M\) is an open set. Figure 2.4: Radius \(r\) of \(B_{r}(a)\) depends on the point \(a\) Figure 2.5: An open ball is an open set **Remark 32**.: Note that being “open” and “closed” are related by complementation not with negation. If a set is not open, we cannot deduce it is closed or vice versa. Some sets are neither open nor closed, like the interval \((0,1]\). Moreover, in an arbitrary metric space \(M\), the empty set \(\emptyset\) and the whole space \(M\) are also both closed and for some metric spaces, such as \(\mathbb{R}^{n}\), these are only two sets which are simultaneously open and closed. For other metric spaces, the situation could be different as shown in the following proposition. **Proposition 20**.: _Consider \((\mathbb{R},d)\) where \(d(x,y)\) is the discrete metric. Then every set is both open and closed._ : It is sufficient to prove that every subset of \(\mathbb{R}\) is open with respect to the discrete metric. Let \(A\neq\emptyset\) and \(A\subseteq\mathbb{R}\). Let \(a\in A\) be an arbitrary point; since \(B_{1}(a)=\{a\}\subset A\), \(A\) is open. The fundamental properties of open sets in a metric space \(M\) are stated in the next theorem, which says that the class of all open sets in a metric space is closed under the formation of arbitrary unions and finite intersections. **Theorem 54**.: _Let \(M\) be a metric space. Then_ 1. _Any union of open sets in_ \(M\) _is open._ : Let \(\{G_{j}\}_{j}\) be a family of open sets in a metric space \((M,d)\) and suppose that \(x\in\bigcup_{j\in J}G_{j}\). Then \(x\in G_{k}\) for some \(k\in J\). Since \(G_{k}\) is open, there is a real number \(r>0\) such that \(B_{r}(x)\subset G_{k}\). Then \(B_{r}(x)\subseteq\bigcup_{j\in J}G_{j}\) and this completes the proof of part a). For part b), suppose \(x\in\bigcap_{j=1}^{n}G_{j}\), then \(x\in G_{j}\) for each \(j=1,2,\ldots,n\). Since \(G_{j}\) is open for each \(j\), there exists a \(r_{j}\) such that \(B_{r_{j}}(x)\subseteq G_{j}\). Let \(r=\min\{r_{1},r_{2},\ldots,r_{n}\}\). Then \(r>0\) and \(B_{r}(x)\subseteq\bigcap_{j=1}^{n}G_{j}\). Even though the structure of the open sets can be complicated in an arbitrary metric space, in the case of the real line one has the following intuitive description of open sets. **Theorem 55**.: _Every nonempty open set in \(\mathbb{R}\) is the union of a countable disjoint collection of open intervals._ For the proof of the Theorem 55 we refer the reader to the proof of the Theorem 27. ### Convergence in a Metric Space Next, we examine the convergence of sequences in arbitrary metric spaces. If you consider the similarity between a metric \(d(x,y)\) and the distance expressed as the absolute value, we see that we can transfer much of the theory of limits of sequences from \(\mathbb{R}\) to an arbitrary metric space. **Definition 43**.: Let \(\{x_{n}\}\) be a sequence in a metric space \(M\).
* \(d_{1}(x,y)=|x_{1}-y_{1}|+|x_{2}-y_{2}|\) (taxicab metric).
* \(d_{\infty}(x,y)=\max\{|x_{1}-y_{1}|,|x_{2}-y_{2}|\}\) (uniform metric).
Proof
Proof
Proof
2. _Any finite intersection of open sets in_ \(M\) _is open._Proof
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