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**Example 34**.:
* If \(S=(a,b)\), then \(\overline{S}=[a,b]\).
* If \(S\) is a closed interval \([a,b]\), then \(\overline{S}=S\).
* If \(S=\left\{\frac{1}{n}:\ n\in\mathbb{N}\right\}\), then the closure of \(S\) is \(\overline{S}=S\cup\{0\}\).
* If the set \(S\) equals to \(\mathbb{Q}\), the set of rational numbers and \(y\in\mathbb{R}\) an arbitrary element, the density of rationals in \(\mathbb{R}\) allows us to find a rational number \(r\), where \(r\) is in any \(\epsilon\)-neighborhood of \(y\) with \(y\neq r\). Thus, \(y\) is an accumulation point of \(\mathbb{Q}\) and therefore \(\overline{\mathbb{Q}}=\mathbb{R}\).
**Definition 24**.: Let \(S\subseteq\mathbb{R}\). The _boundary_ of \(S\) is the set
\[\partial S:=\{x\in\mathbb{R}:\ \text{for all}\ \epsilon>0,\ V_{\epsilon}(x) \cap S\neq\emptyset\,\text{and}\,V_{\epsilon}(x)\cap S^{c}\neq\emptyset\}.\]
Let \(S\subseteq\mathbb{R}\). Recall that a point \(x\in\mathbb{R}\) is an _interior point_ of \(S\) if there exists an \(\epsilon\)-neighborhood \(V_{\epsilon}(x)\) of \(x\) such that \(V_{\epsilon}(x)\subset S\). The set of all interior points of \(S\) is denoted by \(\mathring{S}\). For the following sets, we find their interior and boundary points.
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However, \(y\in\overline{S}\), so every neighborhood of \(y\) intersects with \(S\). This means that there is a point \(z\in V_{\delta}(y)\cap S\) with \(z\neq y\). But then
\[z\in V_{\delta}(y)\subseteq V_{\epsilon}^{{}^{\prime}}(x),\]
so that \(x\in S^{*}\) and \(x\in\overline{S}\).
Proofs of parts c) and d) are left to the reader.
It turns out that the boundary of set can be described using the difference of two familiar sets, as stated in the next theorem.
**Theorem 30**.: _Let \(S\) be a subset of \(\mathbb{R}\). Then \(\partial S=\overline{S}\setminus\mathring{S}\)._
Proof
: Using the definition of the boundary of set, we must show that
\[x\in\overline{S}\quad\text{if and only if}\quad V_{r}(x)\cap S\neq\emptyset \quad\text{for all }r>0\]
and
\[x\notin\mathring{S}\quad\text{if and only if}\quad V_{r}(x)\cap S^{c}\neq \emptyset\quad\text{for all }r>0.\]
We will leave the details to the reader.
### 1.3 Compact Sets
Compactness is an important concept in the analysis. It is what reduces the infinite to the finite. There are a number of equivalent ways to describe compactness in \(\mathbb{R}\), and we can use whatever description of compactness is most appropriate for a given situation. Intuitively, a compact set in \(\mathbb{R}\) is closed and contained in a bounded region. This is called the Heine-Borel Theorem which has a number of profound and far reaching applications in the analysis. For example, if a function is defined on a compact set, then it is automatically bounded and uniformly continuous. See Theorem 42 in the next section.
**Definition 25**.: Let \(S\subseteq\mathbb{R}\). An _open cover_ for \(S\) is a collection of open sets \(\{O_{\alpha}:\ \alpha\in\Gamma\}\) whose union contains \(S\), that is, \(S\subset\bigcup_{\alpha\in\Gamma}O_{\alpha}\). A _finite subcover_ is a finite subcollection \(\{O_{\alpha_{i}}:\ \alpha_{i}\in\Gamma,\ i=1,2,\dots,m\}\) that covers \(S\).
For example, the collection of open sets \(\left\{\left(\frac{-1}{k},\frac{k+1}{k}\right)\right\}\) is a covering for the interval \((0,1)\). An open cover for a set may or may not have a finite subcover. If _every_ open cover for a set has a finite subcover, then the set is called a compact set.
**Definition 26**.: A subset \(S\) of \(\mathbb{R}\) is called _compact_ if every open cover for \(S\) contains a finite subcover.
To understand this definition, let us start with some examples.
**Example 36**.: ### Properties of Compact Sets **Theorem 31**.: _The following are basic properties of compact sets:_ 1. _Every compact set is closed._ : 1. Let \(S\) be a compact subset of \(\mathbb{R}\); we shall prove that the complement of \(S\) in \(\mathbb{R}\) is open. Let \(x\in\mathbb{R}\), \(x\notin S\). If \(y\in S\), let \(r<|x-y|\) and \(V_{x}\) and \(W_{y}\) be neighborhoods of \(x\) and \(y\), respectively, of radius less than \(r\). Since \(S\) is compact, there are finitely many points \(y_{1},y_{2},\ldots,y_{n}\) in \(S\) such that \[S\subset W_{y_{1}}\cup W_{y_{2}}\cup\cdots\cup W_{y_{n}}=W.\] If \(V=V_{x_{1}}\cap V_{x_{2}}\cap\cdots\cap V_{x_{n}}\), then \(V\) is a neighborhood of \(x\) which does not intersect with \(W\). Hence \(V\) is contained in the complement of \(S\), \(V\subset\mathbb{R}\setminus S=S^{c}\), \(S^{c}\) is open, and thus \(S\) is closed.
2. Let \(F\) be a closed subset of \(S\), where \(S\) is compact in \(\mathbb{R}\). Let \(\{O_{\alpha}:\ \alpha\in\Gamma\}\) be an open covering for \(F\). Now \(\mathbb{R}\setminus F=F^{c}\) is open; hence \(\{F^{c}\}\cup\{O_{\alpha}:\ \alpha\in\Gamma\}\) is an open covering of \(S\). Since \(S\) is compact, there is a finite subcover for \(S\) \[S\subseteq F^{c}\cup\left(\bigcup_{i=1}^{n}O_{\alpha_{i}}\right).\] But \(F^{c}\cap F=\emptyset\), so we may remove it from the above union and still retain an open cover for \(F\) with a finite subcover.
3. Since compact sets are closed and the intersection of two closed sets is closed, \(F\cap S\) is closed and \(F\cap S\subset S\); thus by item b) above \(F\cap S\) is compact.
4. Since \(S\) is nonempty and bounded above, we can appeal to the completeness axiom to conclude \(m=\sup S\) exists. Using the definition of the supremum, we know that given \(\epsilon>0\), \(m-\epsilon\) is not an upper bound for \(S\). Next, we claim \(m\in S\); thus \(m=\max S\). If \(m\notin S\), then there exists \(x\in S\) such that \(m-\epsilon It is often difficult to show directly that a given set satisfies the definition of compactness. The definition of a compact set requires us to consider _every_ open covering and show that it has finite subcover. We cannot replace _every_ by _some_ open cover either. Consider \(S=\mathbb{R}\) itself, and \(\mathbb{R}\) is not compact, but it is an open set and \(\mathbb{R}\subseteq\mathbb{R}\) (in which case an open cover is the same as finite subcover). Thus the definition of compact set is useful in showing a set is not compact. Then all we need is to find some open cover for the set which does not have a finite subcover as we have done in the above examples. Fortunately, the _Heine-Borel Theorem_ gives us a much easier characterization of compact subsets of \(\mathbb{R}^{n}\). Although this theorem is valid for \(\mathbb{R}^{n}\), we will state the theorem and give a proof for \(\mathbb{R}\). **Theorem 32** (Heine-Borel Theorem).: _A subset \(S\) of \(\mathbb{R}\) is compact if and only if it is closed and bounded._ : \(\Rightarrow\)): Suppose \(S\) is compact, and let \(\{O_{n}=(-n,n)\}\) be an open cover for \(S\). Since \(S\) is compact, it has a finite subcover, i.e., \(S\subset\bigcup_{j=1}^{k}(-n_{j},n_{j})\). Setting \(n_{0}=\max\{n_{1},n_{2},\ldots,n_{k}\}\), we get \(S\subset(-n_{0},n_{0})\); thus \(S\) is bounded. Suppose \(S\) is compact but not closed, i.e., \(S\neq\overline{S}:=S\cup\{\text{accumulation points of }S\}.\) Let \(x_{0}\in\overline{S}\setminus S\); then \(x_{0}\) is an accumulation point of \(S\), and thus for \(\epsilon>0\) there is a neighborhood \(V_{\epsilon}(x_{0})=(x_{0}-\epsilon,x_{0}+\epsilon)\), which contains \(y\in S\) and \(y\neq x\). For each \(n\in\mathbb{N}\), we let \(U_{n}=\mathbb{R}\setminus[x_{0}-\frac{1}{n},x_{0}+\frac{1}{n}]\). Now each \(U_{n}\) is an open set, and we have \[S\subseteq\bigcup_{n=1}^{\infty}\mathbb{R}\setminus\left[x_{0}-\frac{1}{n},x _{0}+\frac{1}{n}\right]=\mathbb{R}\setminus\{x_{0}\}.\] But \(S\) is compact; thus there exist \(n_{1} \[S\subseteq\bigcup_{j=1}^{k}U_{n_{j}}=\mathbb{R}\setminus\left[x_{0}-\frac{1}{ n_{0}},x_{0}+\frac{1}{n_{0}}\right],\] where \(n_{0}=\max\{n_{1},n_{2},\ldots,n_{k}\}\), which implies that \(S\cap[x_{0}-\frac{1}{n_{0}},x_{0}+\frac{1}{n_{0}}]=\emptyset\), contradicting our choice of \(x_{0}\in\overline{S}\setminus S\). \(\Leftarrow\)): Conversely, suppose \(S\) is a closed and bounded subset of \(\mathbb{R}\). If \(a=\sup S\) and \(b=\inf S\), then since \(S\) is closed, \(a\) and \(b\) are in both \(S\) and \(S\subseteq[a,b]\). Let \(\{U_{i}\}_{i\in I}\) be an open cover for \(S\). By adding the complement of \(S\) to this collection, we obtain a collection \(\mathcal{U}\) of open sets whose union covers \([a,b]\), i.e., \([a,b]\subseteq\left(\bigcup U_{i}\right)\cup S^{c}\). Let \[F=\{x\in[a,b]:\ [a,x]\,\text{is covered by the finite number of open sets in }\mathcal{U}\}.\] Since \(a\in F\), \(F\neq\emptyset\) and \(F\) is bounded above by \(b\). If we let \(c=\sup F\), then \(c\in F\) since \(F\) is closed. We want to show \(c=b\). If \(c\neq b\), there exists \(y\) such that \(c The following examples illustrate how useful this theorem is. **Example 37**.: In the section on Sequences and Series, (see Theorem 15), we have seen the proof of the _Bolzano-Weierstrass property_ which states: “Every bounded sequence of real numbers has a convergent subsequence.” Now consider a sequence \(\{x_{n}\}\subset S\), where \(S\) is a compact set. Since compact sets are bounded, then \(\{x_{n}\}\) is bounded, and by the Bolzano-Weierstrass property \(\{x_{n}\}\) has a subsequence \(\{x_{n_{k}}\}\) that converges to some point \(x\). But \(S\) is also closed; thus \(S\) contains the limit point. This gives us the following definition of compactness, sometimes referred to as _sequential compactness_. **Definition 27** (Sequential Compactness).: A set \(S\) is called _sequentially compact_ in \(\mathbb{R}\) if every sequence in \(S\) has a subsequence that converges to a limit that is also in \(S\). **Theorem 33**.: _Let \(S\) be a subset of \(\mathbb{R}\). Then the following are equivalent:_ 1. _Any open cover for_ \(S\) _has a finite subcover._ : Equivalence of a) and b) is the Heine-Borel Theorem. Equivalence of a) and c) is called the Bolzano-Weierstrass Theorem, and we already gave an idea why closed and bounded implies sequential compactness. We need to show sequential compactness implies closed and bounded. Assume c) is true. If \(S\) is unbounded, then \(S\) contains points \(x_{n}\) with \(|x_{n}|>n\) for \(n=1,2,\dots\). The set containing these points \(x_{n}\) is infinite and clearly has no limit point in \(\mathbb{R}\) and hence none in \(S\). This sequence could not have a convergent subsequence. Thus c) implies \(S\) is bounded. If \(S\) is not closed, then there is a point \(x_{0}\in\mathbb{R}\) which is a limit point of the subsequence but not a point of \(S\). For \(n=1,2,\dots\), there are points \(x_{n_{k}}\) such that \(|x_{n_{k}}-x_{0}|<1/n\). Let \(E\) be the set of these points \(x_{n_{k}}\). Then \(E\) is an infinite set and E has \(x_{0}\) as a limit point. If \(y\in\mathbb{R}\) with \(y\neq x_{0}\), then \[|x_{n_{k}}-y|\geq|x_{0}-y|-|x_{n_{k}}-x_{0}|>|x_{0}-y|-\frac{1}{n}\geq\frac{1 }{2}|x_{0}-y|\] for all but finitely many \(n\); thus \(y\) is not a limit point of \(E\) or \(\{x_{n_{k}}\}\) has no limit point in \(S\); hence \(S\) must be closed. **Remark 20**.: The equivalences given in the above theorem are still true if we replace \(\mathbb{R}\) by \(\mathbb{R}^{n}\). The equivalence of a) and c) still holds in metric spaces; however as we will see later that for an arbitrary metric space b) does not imply a) nor c). See Example 64 in Section 2.1.
1. The empty set and all finite subsets of \(\mathbb{R}\) are compact. The empty set needs no set to cover, and any finite set \(F\subset S\) can be covered by finitely many open sets, one for each element of \(F\).
2. The set \(S=(0,1)\) is not compact, because the open cover \(\{O_{n}=\left(\frac{1}{n},2\right)\}\) has no finite subcover. To see this let \(0
2. _A closed subset of a compact set is compact._
3. _If_ \(F\) _is closed and_ \(S\) _is compact, then_ \(F\cap S\) _is compact._
4. _If_ \(S\) _is a nonempty, closed, and bounded subset of_ \(\mathbb{R}\)_, then_ \(S\) _has a maximum and a minimum._Proof
Proof
1. The set \(S=[0,1]\cup[4,5]\) is compact because it is both closed and bounded.
2. The set \(S=\{x\in\mathbb{R}:\ x\geq 0\}\) is not compact because it is not bounded.
3. The set \(S=[1,2]\cap\mathbb{Q}^{c}\) is not compact because \(S\) is not closed. (Hint: Consider \(3/2\notin S\) and a neighborhood of \(3/2\) to show that \(S^{c}\) is not open).
2. \(S\) _is closed and bounded._
3. _Every sequence in_ \(S\) _has a convergent subsequence that converges to a limit in_ \(S\)_._Proof
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