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However, since the \(p\)-adic norm assumes only discrete values \(\{p^{n}|\ n\in\mathbb{Z}\}\cup\{0\}\), we only need to consider balls of radii \(r=p^{n}\), where \(n\in\mathbb{Z}\). The _sphere_ in \(\mathbb{Q}_{p}\) is defined by:
\[S_{r}(a)_{p}:=\{x\in\mathbb{Q}_{p}:\ |x-a|_{p}=r\}.\]
There are several unexpected results that follow from the non-Archimedian property of \(|\cdot|_{p}\). In the following we give a sample of such results.
**Proposition 24**.: _Every triangle in \(\mathbb{Q}_{p}\) is an isosceles triangle. That is given \(x,y,z\in\mathbb{Q}_{p}\) at least two of the following distances are equal:_
\[|x-y|_{p},\quad|y-z|_{p},\quad|z-x|_{p}.\]
Proof
: We have already shown this result for ultrametric spaces and \(\mathbb{Q}_{p}\) is an ultrametric space.
**Proposition 25**.: _A sphere \(S_{r}(a)_{p}\) is an open set in \(\mathbb{Q}_{p}\)._
This is in sharp contrast with the situation in \(\mathbb{R}\) or \(\mathbb{R}^{n}\) where spheres are certainly not open sets.
Proof
: To show \(S_{r}(a)_{p}\) is an open set, for any \(x\in S_{r}(a)_{p}\) and \(\epsilon \[|x-y|_{p}<|x-a|_{p}=r.\] Then by the above proposition \[|y-a|_{p}=|x-a|_{p}=r,\] and thus \(y\in S_{r}(a)_{p}\). **Proposition 26**.: _Every point in a ball is a center of that ball. In other words if \(b\in B_{r}(a)_{p}\) then \(B_{r}(a)_{p}=B_{r}(b)_{p}\)._ : Let \(x\in B_{r}(b)_{p}\), then \(|x-b|_{p} \[|a-x|_{p}=|(a-b)+(b-x)|_{p}\leq\max\{|a-b|_{p},|b-x|_{p}\} hence \(B_{r}(b)_{p}\subset B_{r}(a)_{p}\). It is easy to see the other inclusion \(B_{r}(a)_{p}\subset B_{r}(b)_{p}\) **Proposition 27**.: _Two balls in \(\mathbb{Q}_{p}\) have a nonempty intersection if and only if one is contained in the other._ : We want to show \(B_{r}(a)_{p}\cap B_{s}(b)_{p}\neq\emptyset\) if and only if \(B_{s}(b)_{p}\subset B_{r}(a)_{p}\) or \(B_{r}(a)_{p}\subset B_{s}(b)_{p}\). Clearly if one ball is contained in the other then they have a nonempty intersection. Now suppose \(r\leq s\) and \(z\in B_{r}(a)_{p}\cap B_{s}(b)_{p}\). Since every point in a ball is its center, we have \[B_{r}(a)_{p}=B_{r}(z)_{p}\quad\text{and}\quad B_{s}(b)_{p}=B_{s}(z)_{p}.\] But \(B_{r}(z)_{p}\subset B_{s}(z)_{p}\), and the required inclusions are satisfied. **Remark 34**.: It is well known that the open balls in \(\mathbb{Q}_{p}\) are both open and closed. Furthermore, the field \(\mathbb{Q}_{p}\) has the Bolzano-Weierstrass property. That is if \(A\) is a bounded infinite subset of \(\mathbb{Q}_{p}\), then \(A\) has an accumulation point in \(\mathbb{Q}_{p}\). Additionally, \(\mathbb{Q}_{p}\) also has the Heine-Borel property: if \(A\) is a closed and bounded subset of \(\mathbb{Q}_{p}\), then \(A\) is compact. Note that \(\mathbb{Q}_{p}\) is in many ways analogous to the field \(\mathbb{R}\): both are complete metric spaces, both contain \(\mathbb{Q}\) as a dense subset, but (as illustrated in the above propositions) the ultrametric on \(\mathbb{Q}_{p}\) produces different properties for the topology of \(\mathbb{Q}_{p}\). ### Compact Sets in Metric Spaces **Definition 52**.: A subset \(S\) in a metric space \(M\) is called _sequentially compact_ if every sequence in \(S\) has a subsequence that converges to a point in \(S\). Let \(M\) be a metric space and \(S\subset M\) a subset. An _open cover_ for \(S\) is a collection of open sets \(\{U_{i}\}\) whose union also contains \(S\); it is a _finite cover_ if the collection contains only a finite number of sets. **Definition 53**.: A subset \(S\) in a metric space is called _compact_ if **every** open cover for \(S\) has a finite subcover. Note that in the above definition if one sets \(S=M\), then one can talk about _compact metric spaces_. In the following we list some important theorems, without proof, about compactness in a metric space. Any classical analysis book contains most of these proofs. For example see, e.g., [26], pp. 151-160. The link that connects compactness to sequential compactness is the following theorem. **Theorem 60** (Bolzano-Weierstrass Theorem).: _A subset of a metric space is compact if and only if it is sequentially compact._ It is relatively easy to see a sequentially compact set is both closed and bounded. To see that it is closed, let \(S\) be sequentially compact and take a sequence \(\{x_{n}\}\in S\) converging to \(x\in M\). Then since \(S\) is sequentially compact, there is a subsequence of \(\{x_{n}\}\) converging to some \(s\in S\). Uniqueness of the limit implies that \(s=x\), thus \(S\) is closed. A sequentially compact set must also be bounded. If \(S\) is sequentially compact and not bounded, then there is a sequence \(\{x_{n}\}\in S\) and a point \(s\in S\) with \(d(x_{n},s)\geq n\). Then \(x_{n}\) cannot have any convergent subsequence. **Example 64**.: In Euclidean space we can easily identify a compact set because the Heine-Borel theorem states that a set \(A\subset\mathbb{R}^{n}\) is compact if and only if it is closed and bounded. One half of this theorem is also true for a metric space. In fact a compact set is closed and bounded in a metric space too. But a set being closed and bounded does not imply compactness in an arbitrary metric space \(M\). Here is an example: let \(M\) be an infinite set with the discrete metric \(d\) defined by \[d(x,y)=\begin{cases}0&\text{ if }\ x=y\\ 1&\text{ if }\ x\neq y\end{cases}\quad.\] The set \(M\) is bounded because if \(x_{0}\in M\) be any point, then \(M\subset B_{2}(x_{0})\), where the open set \(B_{2}(x_{0})=\{y\in M|\ d(x_{0},y)<2\}\). Since M is the entire metric space it is also closed. However \(M\) is not compact. Indeed, for each \(x\in M\) we can form an open cover \(\{B_{\frac{1}{2}}(x)\}_{x\in M}\) for \(M\), but this open cover has no finite subcover, since we cannot delete any member \(\{B_{\frac{1}{2}}(x)\}_{x\in M}\) while maintaining the property of being a cover of \(M\). Thus it is natural to ask if there is another characterization of compactness which exists in metric spaces. The answer is yes. **Theorem 61**.: _A metric space is compact if and only if it is complete and totally bounded._ For proof and definition of totally bounded we again refer the reader to [26], p. 166. There is an important consequence of the Bolzano-Weierstrass theorem for metric spaces, called the Nested Sets Property. **Theorem 62**.: _Let \(K_{i}\) be a sequence of compact nonempty sets in a metric space \(M\) such that \(K_{i+1}\subset K_{i}\) for all \(i=1,2,\dots\). Then there is at least one point in \(\bigcap_{i=1}^{\infty}K_{i}\)._ This theorem is a consequence of the _finite intersection property_ of compact sets, which we have given a version for the compact subsets of \(\mathbb{R}\) in Section 1.5 of Chapter 1. **Remark 35**.: To describe the compact subsets of \(C[0,1]\) we need the concept of equicontinuous family of functions. A sequence of functions \(\{f_{n}\}\) defined on a set \(A\subset\mathbb{R}\) is called _equicontinuous_ if for every \(\epsilon>0\) there exists a \(\delta>0\) such that \(|f_{n}(x)-f_{n}(y)|<\epsilon\) for all \(n\in\mathbb{N}\) and \(|x-y|<\delta\) in \(A\). The Arzela-Acoli theorem states that any bounded equicontinuous collection of functions in \(C[0,1]\) must have a uniformly convergent subsequence. The proof of this theorem uses a clever diagonal selection process. See [26], p. 299-301. ### 2.2 Baire’s Category Theorem Baire’s Category Theorem is often regarded as one of the corner stones of functional analysis. The proofs of basic theorems in functional analysis, such as theuniform boundedness principle and open mapping theorem, require completeness and depend on Baire’s Category Theorem. There is also an unexpected consequence of this theorem: most continuous functions do not have derivatives at any point. Continuous functions that are differentiable at even one point in \([0,1]\) form a “sparse” subset. We first state the theorem for a complete metric space. **Theorem 63** (Baire’s Theorem for a complete metric space).: _Let \(M\) be a complete metric space. If \(M=\bigcup_{n=1}^{\infty}E_{n}\), then the closure of some \(E_{n}\) contains an open ball. Equivalently, if \((G_{n})\) is a sequence of dense open sets in \(M\), then \(\bigcap_{n=1}^{\infty}G_{n}\neq\emptyset\); in fact \(\bigcap_{n=1}^{\infty}G_{n}\) is dense in \(M\)._ For the proof of the above theorem we refer the reader to, e.g., [14], p. 133. In the following we state and give a proof of Baire’s theorem for \(\mathbb{R}\). There are many statements about the size of \(\mathbb{R}\): for example since \(\mathbb{R}\) is uncountable, we cannot write \(\mathbb{R}\) as a countable union of single points. Baire’s Theorem is an improved statement about the size of \(\mathbb{R}\) again, which states that it is impossible to write the real numbers \(\mathbb{R}\) as the countable union of nowhere-dense sets. The main ingredient to the proof of Baire’s Theorem is the completeness of \(\mathbb{R}\) which is given below. To understand what we mean by _nowhere-dense set_, and the motivation for defining such sets, we need to go back and see the formation of various subsets of \(\mathbb{R}\). Every open subset of \(\mathbb{R}\) is either a finite or countable union of open intervals, thus the structure of open sets in \(\mathbb{R}\) is well understood. On the other hand if one thinks about the Cantor set, which is a closed subset of \(\mathbb{R}\), it is uncountable and contains no open intervals of any type. Therefore we cannot expect a nice description of the closed subsets of \(\mathbb{R}\) as in the case of its open subsets. Recall also that an arbitrary intersection of closed sets is closed and an arbitrary union of open sets is open. Thus one way to enlarge the list of subsets of \(\mathbb{R}\) is to take the union of closed sets and intersection of open sets, which leads us to introduce the following concepts: **Definition 54**.: A subset \(S\subseteq\mathbb{R}\) is called a \(G_{\delta}\) set if it can be written as a countable intersection of open sets and a set \(K\subseteq\mathbb{R}\) is called \(F_{\sigma}\) if it can be written as a countable union of closed sets. Using De Morgan’s laws, it is not difficult to see that \(S\subseteq\mathbb{R}\) is a \(G_{\delta}\) set if and only if its complement \(S^{c}\) is an \(F_{\sigma}\) set. It turns out that a closed interval \([a,b]\) is a \(G_{\delta}\) set, and there are some sets which are both an \(F_{\sigma}\) set and a \(G_{\delta}\) set. The half-open interval \((a,b]\) is one such example. We leave it to the reader to prove that the set of rationals \(\mathbb{Q}\) is a \(F_{\sigma}\) set and thus the set of irrationals \(\mathbb{I}\) forms a \(G_{\delta}\) set. Baire’s Theorem makes it possible to claim that \(\mathbb{Q}\) is not a \(G_{\delta}\) set. Recall that the set \(S\subseteq\mathbb{R}\) is _dense_ in \(\mathbb{R}\) if given any two real numbers \(x 1. \(S\subseteq\mathbb{R}\) is dense in \(\mathbb{R}\) if and only if every point in \(\mathbb{R}\) is a limit point of \(S\). We now state a theorem (see, e.g., [14], p. 131). This theorem is about the dense open subsets of \(\mathbb{R}\), but its conclusion reminds us of the result we have seen about a nested sequence of compact sets having a nontrivial intersection. **Theorem 64**.: _If \(\{O_{n}\}\) is a sequence of dense open sets in \(\mathbb{R}\), then \(\bigcap_{n=1}^{\infty}O_{n}\) is not empty._ **Definition 55**.: A set \(S\) is _nowhere-dense_ in \(\mathbb{R}\) if \(\overline{S}\) contains no nonempty open intervals. Equivalently, a set \(S\) is _nowhere-dense_ in \(\mathbb{R}\) if \(\dot{\overline{S}}=\emptyset\) (its closure has empty interior).
Proof
Proof
2. \(S\subseteq\mathbb{R}\) is dense in \(\mathbb{R}\) if and only if \(\overline{S}=\mathbb{R}\).
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