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**Proposition 10**.: _The Cantor set \(C\) is uncountable._
Proof
: There are several proofs that \(C\) is uncountable, for example, in [20], pp. 309-313, and using the Schroder-Bernstein Theorem, it is proved that \(C\) and \([0,1]\) have the same cardinality. Here we start by claiming that every \(c\in C\) yields a sequence \((x_{1},x_{2},x_{3},\dots)\) of zeros and ones, likewise every such sequence corresponds to a point in \(C\). By the above lemma since the set of sequences of zeros and ones is uncountable, \(C\) must be uncountable. To see intuitively that there is a one-to-one correspondence between \(C\) and the sequence \(\{x_{n}\}\), where \(x_{n}=0\) or \(1\), first label the intervals \(C_{1},C_{2}\), and so on as left (L) and right (R) as shown in Figure 1.25.
For each \(c\in C\), set
\[x_{1}=0\text{ if }c\text{ is in the (L) component of }C_{1}.\]
\[x_{1}=1\text{ if }c\text{ is in the (R) component of }C_{1}.\]
Having established where in \(C_{1}\) the point \(c\) is located, now we have two possible components of \(C_{2}\) that might contain \(c\). Next we set
\[x_{2}=0\text{ if }c\text{ is in the (L) component of }C_{2}.\]
\[x_{2}=1\text{ if }c\text{ is in the (R) component of }C_{2}.\]
Continuing in this fashion, for every \(c\in C\) we can generate a sequence \(\{x_{1},x_{2},\)\(\dots\}\) of 0’s or 1’s, a set we already know to be uncountable.
Thus we can write
\[card(C)=card(2^{\mathbb{N}})=card([0,1]).\]
Here by “\(card(A)\)” we mean the cardinality of the set \(A\). This is hard to believe that the Cantor set is as big as \([0,1]\) but has length 0. No wonder the Cantor set sometimes referred to as “Cantor dust.”
To motivate Theorem 36, we consider \(x\in[0,1]\), where \(x\) can be written as \(x=0.a_{1}a_{2}a_{3}\dots\) (base 3) with each \(a_{n}=0,1,\) or 2. These three choices correspond to a three-way splitting of intervals. For example, the three intervals for \(C_{1}\) are \([0,\frac{1}{3}]\), \((\frac{1}{3},\frac{2}{3})\), and \([\frac{2}{3},1]\) (Figure 1.26).
We need to pay special attention to the end points, since \(1/3=0.1=0.0222\dots\) (base 3), \(2/3=0.2=0.1222\dots\) (base 3), and \(1=1.0=0.222\dots\)
Figure 1.26: Splitting of intervals in \(C_{1}\)
Figure 1.25: Labelling of the intervals as L or R
(base 3), but each has at least one representation with \(a_{1}\) in the proper range. Next we examine \(C_{2}\) but ignore the discarded intervals (Figure 1.27):
Again there is some ambiguity at the end points, for example,
\[7/9=0.21000\cdots=0.20222\ldots\]
These end points of removed intervals are called triadic rational numbers of the form \(\dfrac{m}{3^{n}}\). Even though these points have two distinct ternary expansions, exactly one of the expressions has all digits \(a_{k}=0\) or \(2\). These examples point us to the following which we state without the proof (see [14], p. 28).
**Theorem 36**.: \(c\in C\) _if and only if \(c\) can be written as_
\[c=0.a_{1}a_{2}a_{3}\cdots=\sum_{n=1}^{\infty}\dfrac{a_{n}}{3^{n}},\]
_where each \(a_{n}\) is either \(0\) or \(2\)._
**Remark 21** (Dimension of \(C\)).: The Cantor set is as “big” as the interval \([0,1]\) or the set of reals \(\mathbb{R}\) in the sense that they are all uncountable, but it is also as small as a point because a single point and the Cantor set have zero length. Does it make sense to ask about the dimension of \(C\)? Even though we did not give a proper definition of the dimension, we can have a discussion toward understanding its dimension. Actually it turns out that the dimension of \(C\) is \(\dfrac{\ln 2}{\ln 3}\), a non-integer, or fractional dimension. The notion of a non-integer or fractional dimension is the motivation behind the term “fractal.” The Cantor set \(C\) is a fractal. To calculate the dimension of \(C\), consider the following sets and their dimensions denoted \(d\) in Figure 1.28.
Figure 1.27: Splitting of intervals in \(C_{2}\)
Now think about what happens when you magnify a point by 3; nothing changes, you still have one single point or \(3^{0}=1\) copy of itself. However a segment magnified by 3 results in \(3^{1}=3\) copies of itself and a square, which is two-dimensional, magnified three times results in \(3^{2}=9\) copies. Now let us consider the Cantor set \(C\). We start with \([0,1]\); if we magnify by 3, we get \([0,3]\), and if you delete the middle third \((1,2)\), then you end up with \([0,1]\cup[2,3]\). Thus you get two of the original copies (Figure 1.29).
Let \(d\) denote the dimension of the Cantor set; then, solving \(2=3^{d}\) for \(d\) yields \(d=\frac{\ln 2}{\ln 3}\approx.631\).
Note also that there are other interesting objects whose construction is reminiscent of the Cantor set. For example, Sierpinski’s triangle is obtained by starting with a closed (filled) unit equilateral triangle and in the first step removing one open triangle of size \(1/2\), in the second step removing 3 open triangles of size \(1/4\), and thus at the \(n\)th step removing \(3^{n-1}\) open triangles of size \(1/2^{n}\). Sierpinski’s triangle also has a fractional dimension \(d=\frac{\ln 3^{n}}{\ln 2^{n}}=\frac{\ln 3}{\ln 2}=1.585\) (Figure 1.30).
Figure 1.30: Sierpinski’s triangle
Figure 1.28: A point, line segment, square, and cube
### 1.5 Topology of the real line
#### Exercises
1. Show that \(\emptyset\) and \(\mathbb{R}\) are the only subsets of \(\mathbb{R}\) that are both open and closed in \(\mathbb{R}\).
2. Discuss whether the following sets are open or closed. Determine the interiors, closures, and the boundaries of each set.
1. \((1,4)\) in \(\mathbb{R}\).
2. \([2,5]\) in \(\mathbb{R}\).
3. \(\{r\in(0,1):\ r\ \text{is rational}\}\) in \(\mathbb{R}\).
4. \(\bigcap_{n=1}^{\infty}\left[-1,\frac{1}{n}\right)\) in \(\mathbb{R}\).
3. Find the accumulation points of the following sets in \(\mathbb{R}\):
1. \(A=\mathbb{Q}\).
2. \(A=\mathbb{Z}\).
3. \(A=(0,1)\).
4. \(A=\big{\{}(-1)^{n}+\frac{1}{n}:\ n\in\mathbb{N}\big{\}}\).
4. Identify which of the following sets are compact. Which are connected?
1. A finite set in \(\mathbb{R}\)
2. \(\big{\{}\frac{1}{n}:\ n\in\mathbb{N}\big{\}}\cup\{0\}\)
3. \(\{x\in\mathbb{R}:\ 0\leq x\leq 1\ \text{and is irrational}\}\)
4. A closed set in \([0,1]\)
5. The boundary of a bounded set in \(\mathbb{R}\)
6. \(\mathbb{Z}\) in \(\mathbb{R}\)
5. Let \(A\) and \(B\) be compact subsets of \(\mathbb{R}\). Prove that \(A\cup B\) and \(A\cap B\) are compact.
6. Suppose \(K\subset\mathbb{R}\) is compact and nonempty. Prove that \(\sup K,\ \inf K\in K\).
7. Prove that the intersection of connected sets in \(\mathbb{R}\) is connected. Show that this is false if \(\mathbb{R}\) is replaced by \(\mathbb{R}^{2}\).
8. Let \(K\) be a nonempty compact set. Let \(\{A_{n}\}\) be a nonempty decreasing sequence of closed subsets of \(K\). Prove that \(\bigcap\limits_{n\geq 1}A_{n}\) is not empty (Cantor’s Intersection Theorem).
9. Let \(C\) be the Cantor set as defined above. Prove that \(C\) is totally disconnected, that is, if \(x,y\in C\) and \(x\neq y\), then \(x\in U\) and \(y\in V\), where \(U\) and \(V\) are open sets that disconnect \(C\).
10. Prove that if \(A\subset\mathbb{R}\) is connected and \(A\subset B\subset\overline{A}\), then \(B\) is connected.
11. Let \(A\) be a subset of \(\mathbb{R}\) and \(B\) be a set of points \(x\in\mathbb{R}\) with the property that \(A\cap(x-\delta,x+\delta)\) is uncountable for every \(\delta>0\). Show that \(A\setminus B\) is finite or countable.
12. Let \(A\subset\mathbb{R}\) be uncountable. Show that \(A\) has at least one accumulation point.
### 1.6 Continuous Functions
#### Functional Limits
Let \(f:S\subseteq\mathbb{R}\to\mathbb{R}\), and recall that a limit point \(a\) of \(S\) is a point where every \(\epsilon\)-neighborhood \(V_{\epsilon}(a)\) of \(a\) intersects \(S\) in some point other than \(a\). We showed in the section on the topology of \(\mathbb{R}\) that this is equivalent to the claim that \(a\) is a limit point of \(S\) if and only if \(a=\lim x_{n}\) for some sequence \(\{x_{n}\}\) in \(S\) with \(x_{n}\neq a\). Furthermore, limit points of \(S\) do not necessarily belong to \(S\) unless \(S\) is closed. Let \(a\) be a limit point of \(S\); then we write
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