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Proof
: For each \(\alpha\in\Gamma\), set \(O_{\alpha}=\mathbb{R}\setminus K_{\alpha}\) (or equivalently \(O_{\alpha}=K_{\alpha}^{c}\)), and fix a member \(K_{1}\) from \(\mathcal{U}\). Assume that no point of \(K_{1}\) belongs to every \(K_{\alpha}\). Then every point of \(K_{1}\) belongs to some \(O_{\alpha}\). That is, the sets \(O_{\alpha}\) form an open cover for \(K_{1}\); since \(K_{1}\) is compact, there are finitely many indices \(\alpha_{1},\alpha_{2},\dots,\alpha_{n}\) such that
\[K_{1}\subset O_{\alpha_{1}}\cup O_{\alpha_{2}}\cup\dots\cup O_{\alpha_{n}}.\]
Using the fact that
\[\mathbb{R}\setminus(K_{\alpha_{1}}\cap K_{\alpha_{2}}\cap\dots\cap K_{\alpha_ {n}})=(\mathbb{R}\setminus K_{\alpha_{1}})\cup(\mathbb{R}\setminus K_{ \alpha_{2}})\cup\dots\cup(\mathbb{R}\setminus K_{\alpha_{n}}),\]
we have
\[K_{1}\subset\mathbb{R}\setminus(K_{\alpha_{1}}\cap K_{\alpha_{2}}\cap\dots \cap K_{\alpha_{n}}),\]
and therefore
\[K_{1}\cap(K_{\alpha_{1}}\cap\dots\cap K_{\alpha_{n}})=\emptyset,\]
in contradiction to our hypothesis. Thus some point in \(K_{1}\) belongs to each \(K_{\alpha}\), and \(\bigcap\{K_{\alpha}:\ \alpha\in\Gamma\}\neq\emptyset\).
**Theorem 34** (Nested Set Property for Compact Sets).: _If \(\{K_{n}\}\) is a sequence of compact sets in \(\mathbb{R}\) such that \(K_{n+1}\subset K_{n}\) for all \(n=1,2,\dots\), then there is at least one point in \(\bigcap\limits_{n=1}^{\infty}K_{n}\)._
Proof
: The result follows from the finite intersection property. The set \(K_{1}\) is compact, and the sets \(K_{1},K_{2},\dots\) have the finite intersection property. Since the intersection of any finite collection equals \(K_{n}\) with the highest index,
\[K_{1}\bigcap\left(\bigcap\limits_{n=1}^{\infty}K_{n}\right)=\bigcap\limits_{n =1}^{\infty}K_{n}\neq\emptyset.\qed\]
### Connected Sets
We have introduced open and closed sets that are analogs of open and closed intervals in \(\mathbb{R}\), and compact subsets of \(\mathbb{R}\) are analogs of closed and bounded intervals. Next we define the concept of a _connected_ set. Loosely speaking a connected set is one piece and cannot be broken into nonempty open pieces which do not share any common points. For example, the set \(A=(0,1)\cup(1,2)\) has two pieces, although the two intervals \((0,1)\) and \((1,2)\) have a limit point \(x=1\) in common, there is still some “space” between them, meaning no limit point of one of these intervals is actually contained in the other. Said in another way, the closure of \((0,1)\) does not intersect \((1,2)\) and vice versa. Moreover, our intuition can fail in judging whether the set
\[A=\left\{(x,\sin(\frac{1}{x})):\ x>0\right\}\cup\{(0,y):\ y\in[-1,1]\right\} \subset\mathbb{R}^{2}\]
is broken or not. See Figure 1.23.
We must seek a sound mathematical definition that we can depend on. In later sections, we examine continuity and differentiability of functions defined on subsets of \(\mathbb{R}\), and there are some very important theorems such as the Intermediate Value Theorem which depend on the fact that an interval is connected. The definition of _connected_ is a bit tricky since it is stated as a negation of _disconnected_.
**Definition 28**.: A subset \(A\subset\mathbb{R}\) is said to be _disconnected_ (or not connected) if there exists a pair of open sets \(U\) and \(V\) in \(\mathbb{R}\) such that:
* \(U\cap A\neq\emptyset\), \(V\cap A\neq\emptyset\).
* \(U\cap V\cap A=\emptyset\).
* \(A\subset U\cup V\).
A set \(A\) is connected if it is not disconnected.
Note that basically two nonempty sets \(U\) and \(V\) separate \(A\subset\mathbb{R}\) if \(A=U\cup V\) and \(U\cap V=\emptyset\).
**Example 38**.: Figure 1.23: Is this set connected? * The set \(A=\{x\in\mathbb{R}:\ x\neq 0\}\) is also disconnected; in this case \(U=(-\infty,0)\) and \(V=(0,\infty)\) does the job. Note that although the above definition is valid for subsets of \(\mathbb{R}^{n}\) or even for an arbitrary metric space, one needs to pay attention to the metric. For example, \(\mathbb{R}\) with the usual metric (absolute value function) is connected, but if we put the _discrete_ metric on \(\mathbb{R}\), it is not connected. We will see later that in the discrete metric \((-\infty,0]\) and \((0,\infty)\) are both open sets. From example e) above we also learn that the open sets \(U\) and \(V\) need not be a positive distance apart. Even a small gap due to omitting the origin creates a disconnected set. The concept of connectedness is more subtle when working with subsets of the plane or higher dimensions. Luckily, connected sets have a simple description in \(\mathbb{R}\) which we give in the following theorem. **Theorem 35**.: _A subset \(A\) of \(\mathbb{R}\) is connected if and only if \(A\) is an interval._ : First a clarification: When we say an “interval” what we mean is a set of the form \([a,b],[a,b),(a,b]\), or \((a,b)\), where \(a\) and \(b\) can be \(\mp\infty\) on an open end of the interval. There is another concept of connectedness called _path connectedness_ which we will not cover here. But it can be shown that intervals are connected because they are path connected (for details see [18], p. 258). For the converse we assume \(A\) is not an interval. This means there are points \(x,y\), and \(z\) such that \(x * \(A\subset U\cup V\). Thus \(A\) is disconnected. ### The Cantor Set The _Cantor set_ or sometimes called the _Cantor ternary set_ is an intriguing subset of \(\mathbb{R}\) which will extend our understanding of the nature of subsets of \(\mathbb{R}\) and is a valuable source of counterexamples in the analysis. For example, the Cantor set is the basis of the construction of a function called “Devil’s staircase,” which is a continuous, non-decreasing function that is not constant, yet has zero derivative at almost every point. The following construction is due to Georg Cantor, whose name was already mentioned several times, especially in our discussion of uncountable sets. The Cantor set is a subset of \([0,1]\); it is “large” (uncountable), yet it is somewhat “small” (has length zero). Consider the process of successively removing “middle thirds” from the interval \([0,1]\). Let \(C_{0}=[0,1]\), and define \(C_{1}\) by removing the open interval \(I_{1}=\left(\frac{1}{3},\frac{2}{3}\right)\) from \(C_{0}\), that is, \[C_{1}=C_{0}\setminus\left(\frac{1}{3},\frac{2}{3}\right)=\left[0,\frac{1}{3} \right]\bigcup\left[\frac{2}{3},1\right].\]Now, construct \(C_{2}\) by removing middle thirds from \([0,\frac{1}{3}]\) and \([\frac{2}{3},1]\), let \(I_{2}=(\frac{1}{9},\frac{2}{9})\cup(\frac{7}{9},\frac{8}{9})\), and set \[C_{2}=C_{1}\setminus I_{2}=\left(\left[0,\frac{1}{9}\right]\cup\left[\frac{2}{9 },\frac{1}{3}\right]\right)\bigcup\left(\left[\frac{2}{3},\frac{7}{9}\right] \cup\left[\frac{8}{9},1\right]\right).\] Notice that \(C_{1}\) and \(C_{2}\) consist of two and four closed intervals each having length \(\frac{1}{3}\) and \(\frac{1}{9}\), respectively. If we continue this process inductively, then for each \(n=1,2,\dots\) we obtain a set \(C_{n}\) consisting of \(2^{n}\) closed intervals each having length \(\frac{1}{3^{n}}\). The Cantor set \(C\) is defined as (Figure 1.24) It follows from the nested interval property that \(C\neq\emptyset\). Notice that if we continue this process indefinitely, it seems like nothing is left, but if we keep track of the end points of each interval \(C_{n}\), we can see that \[0,1,\frac{1}{3},\frac{2}{3},\frac{1}{9},\frac{2}{9},\dots\in C.\] We refer to all of the points in \(C\) of the form \(\frac{a}{3^{n}}\) for some integers \(a\) and \(n\) as end points of \(C\). We will see that the Cantor set \(C\) contains end points and many other points as well. Surprisingly, \(C\) is uncountable. Before we prove that \(C\) is uncountable, we first prove \(C\) is “small”: **Proposition 8**.: _The Cantor set \(C\) has zero length._ : We start by observing the length of the intervals removed from \([0,1]\) to form \(C\). To form \(C_{1}\) we removed \(I_{1}=\left(\frac{1}{3},\frac{2}{3}\right)\) and the length of \(I_{1}=1/3\). In the second step we removed \(I_{2}=\left(\frac{1}{9},\frac{2}{9}\right)\cup\left(\frac{7}{9},\frac{8}{9}\right)\) and the length of \(I_{2}=2\cdot(1/9)\)To construct \(C_{n}\) we removed \(2^{n-1}\) middle thirds of length \(\dfrac{1}{3^{n}}\), and so the total length of \([0,1]\setminus C\) must be \[\frac{1}{3}+2\left(\frac{1}{9}\right)+4\left(\frac{1}{27}\right)+ \dots+2^{n-1}\left(\frac{1}{3^{n}}\right)+\dots = \sum_{n=1}^{\infty}\frac{2^{n-1}}{3^{n}}=\frac{1}{3}\sum_{n=1}^{ \infty}\left(\frac{2}{3}\right)^{n-1}\] \[= \frac{1}{3}\cdot\frac{1}{1-2/3}=1.\] Thus the Cantor set has zero length. Note that at this point we have not defined what we mean by a “measure zero” set. Thus, we cannot claim that the above argument implies that the Cantor set \(C\) has “measure zero.” However, if one assumes that “measure” is additive, then \(C\) has measure zero because its complement has measure \(1\). Later, in the section on the Riemann Integral, we define a “measure zero set” and show that the Cantor set is actually a measure zero set. **Proposition 9**.: _The Cantor set \(C\) is compact._ : The set \(C\) is closed because it is the intersection of closed sets; it is also bounded since it is contained in \([0,1]\). By the Heine-Borel Theorem, the cantor set \(C\) is compact. To prove the Cantor set is uncountable, we need the following lemma. **Lemma 7**.: _The collection of all sequences of \(0\)’s and \(1\)’s is uncountable._ : Take a sequence \(\{a_{n}\}\) of \(0\)’s and \(1\)’s, then \(\sum_{n=1}^{\infty}\dfrac{a_{n}}{2^{n}}\) represents an element in \([0,1]\), and conversely each element of \([0,1]\) can be so represented. That is, the map \[a_{n}\mapsto 0.a_{1}a_{2}a_{3}\dots\quad\text{(base $2$)}\,\,\,\text{is onto}.\] Hence the set of all \(0-1\) sequences, written as \(\{0,1\}^{\mathbb{N}}\), has cardinality at least that of the interval \([0,1]\).
* The empty set \(\emptyset\) is connected because it can never be written as the union of nonempty sets.
* The set \(A=[-1,0]\cup[5,6]\) is disconnected in \(\mathbb{R}\) because \(U=(-2,1)\) and \(V=(4,7)\) are disjoint open sets that each intersects with \(A\) and jointly contains \(A\).
* The set \(\mathbb{Z}\subset\mathbb{R}\) is disconnected. To see this let \(U=(1/2,\infty)\) and \(V=(-\infty,1/4)\), then \(\mathbb{Z}\subset U\cup V\), \(\mathbb{Z}\cap U=\{1,2,3,\dots\}\neq\emptyset\), \(\mathbb{Z}\cap V=\{\dots,-2,-1,0\}\neq\emptyset\), and \(\mathbb{Z}\cap U\cap V=\emptyset\).
* The set of rationals \(\mathbb{Q}\) is disconnected in \(\mathbb{R}\). Because if \(r_{1},r_{2}\in\mathbb{Q}\) with \(r_{1}
Proof
* \(U\cap A\neq\emptyset\) and \(V\cap A\neq\emptyset\).
* \(U\cap\ V\cap A=\emptyset\).
Proof
Proof
Proof
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