-
Notifications
You must be signed in to change notification settings - Fork 0
/
interval.html
119 lines (119 loc) · 6.35 KB
/
interval.html
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" lang="" xml:lang="">
<head>
<meta charset="utf-8" />
<meta name="generator" content="pandoc" />
<meta name="viewport" content="width=device-width, initial-scale=1.0, user-scalable=yes" />
<meta name="author" content="Ioannis Konstantoulas" />
<title>A problem on covering the interval</title>
<style>
code{white-space: pre-wrap;}
span.smallcaps{font-variant: small-caps;}
div.columns{display: flex; gap: min(4vw, 1.5em);}
div.column{flex: auto; overflow-x: auto;}
div.hanging-indent{margin-left: 1.5em; text-indent: -1.5em;}
/* The extra [class] is a hack that increases specificity enough to
override a similar rule in reveal.js */
ul.task-list[class]{list-style: none;}
ul.task-list li input[type="checkbox"] {
font-size: inherit;
width: 0.8em;
margin: 0 0.8em 0.2em -1.6em;
vertical-align: middle;
}
</style>
<link rel="stylesheet" href="styles/texstyle.css" />
<script src="https://polyfill.io/v3/polyfill.min.js?features=es6"></script>
<script
src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-chtml-full.js"
type="text/javascript"></script>
<!--[if lt IE 9]>
<script src="//cdnjs.cloudflare.com/ajax/libs/html5shiv/3.7.3/html5shiv-printshiv.min.js"></script>
<![endif]-->
</head>
<body>
<header id="title-block-header">
<h1 class="title">A problem on covering the interval</h1>
<p class="author">Ioannis Konstantoulas</p>
</header>
<p>This is a very nice problem I encountered in Analysis II as an
undergraduate. It has a two-line solution using a theorem from
Functional Analysis, but I didn’t know that at the time and managed to
solve it via an elementary contradiction.</p>
<p><strong>Problem:</strong> Show that the open unit interval <span
class="math inline">\((0,1)\)</span> is not the disjoint union of closed
intervals of positive length.</p>
<p><strong>Solution:</strong> Suppose that <span
class="math display">\[I = (0,1) = \bigsqcup_{J \in \mathcal{A}}
J\]</span> for some collection <span
class="math inline">\(\mathcal{A}\)</span> where all the <span
class="math inline">\(J\)</span> have positive length (in particular,
they have distinct endpoints). Then first of all, this collection of
intervals must be infinite, since the union of a finite number of closed
sets is closed whereas <span class="math inline">\(I\)</span> is not
closed.</p>
<p>Next, let <span class="math inline">\(D_I\)</span> be the set of all
distances between pairs of intervals <span class="math inline">\(J,
J'\)</span>. If this set had a lower bound <span
class="math inline">\(\epsilon\)</span> greater than zero, then <span
class="math inline">\(|\mathcal{A}| \leq \lceil 1/\epsilon
\rceil\)</span> since at most that number of gaps fit inside <span
class="math inline">\([0,1]\)</span>; this contradicts the fact that
<span class="math inline">\(\mathcal{A}\)</span> is infinite. Thus the
infimum of <span class="math inline">\(D_I\)</span> is <span
class="math inline">\(0\)</span>; so we can find two distinct intervals
<span class="math inline">\(L_1 = [a_1,l_1]\)</span> and <span
class="math inline">\(R_1 = [r_1,b_1]\)</span> with <span
class="math display">\[a_1 < l_1 < r_1 < b_1 \textrm{ and }
r_1 - l_1 < 1/2.\]</span></p>
<p>Let <span class="math inline">\(I_1 = (l_1, r_1)\)</span> be the open
interval between <span class="math inline">\(L_1\)</span> and <span
class="math inline">\(R_1\)</span>, and <span
class="math inline">\(\mathcal{A}_1\)</span> the subset of <span
class="math inline">\(\mathcal{A}\)</span> of intervals that intersect
<span class="math inline">\(I_1\)</span>. Then these cover <span
class="math inline">\(I_1\)</span>, are disjoint and do not intersect
<span class="math inline">\(L_1\)</span> or <span
class="math inline">\(R_1\)</span>, so <span class="math display">\[I_1
= \bigsqcup_{J \in \mathcal{A}_1}J.\]</span> As before, <span
class="math inline">\(\mathcal{A}_1\)</span> is infinite, so if <span
class="math inline">\(D_1\)</span> is the set of distances of members of
<span class="math inline">\(\mathcal{A}_1\)</span>, it too has zero
infimum. Therefore, we can repeat the previous construction to find
intervals <span class="math inline">\(L_2 = [a_2,l_2]\)</span> and <span
class="math inline">\(R_2 = [r_2,b_2]\)</span> bordering the open
interval <span class="math inline">\(I_2 = (l_2,r_2)\)</span> such that
<span class="math inline">\(|I_2| = r_2 - l_2 < 1/4\)</span>.</p>
<p>Continuing this process we find <span class="math inline">\(L_k, R_k
\in \mathcal{A}\)</span> and construct a nested sequence of intervals
<span class="math inline">\(I_k = (l_k, r_k)\)</span> of length <span
class="math inline">\(|I_k| < 2^{-k}\)</span>. The shrinking lengths
of the <span class="math inline">\(I_k\)</span> imply that the sequence
<span class="math display">\[s := l_1,\, r_1,\, l_2,\,
r_2,\,\ldots\]</span> is Cauchy; therefore it has a limit <span
class="math inline">\(x \in (0,1).\)</span> Furthermore, the
sub-sequence <span class="math inline">\(l_k\)</span> is strictly
increasing, so <span class="math inline">\(l_k \nearrow x\)</span>, and
the sub-sequence <span class="math inline">\(r_k\)</span> is strictly
decreasing, so <span class="math inline">\(r_k \searrow
x\)</span>.</p>
<p>Since <span class="math inline">\(\mathcal{A}\)</span> partitions
<span class="math inline">\((0,1)\)</span>, there exists a unique <span
class="math inline">\(J = [l,r] \in \mathcal{A}\)</span> such that <span
class="math inline">\(x \in J\)</span>. Let <span
class="math inline">\(\epsilon\)</span> be the length of <span
class="math inline">\(J\)</span>: <span class="math inline">\(\epsilon=r
-l\)</span>. Since the sequence <span class="math inline">\(s\)</span>
converges to <span class="math inline">\(x\)</span>, after a certain
<span class="math inline">\(k_0\)</span> we have <span
class="math inline">\(0 < x - l_k < \epsilon/2\)</span> and <span
class="math inline">\(0 < r_k - x < \epsilon/2\)</span>. Depending
on which half of <span class="math inline">\(J\)</span> contains <span
class="math inline">\(x\)</span>, either the first or the second (or
both) inequality gives a contradiction: infinitely many <span
class="math inline">\(L_k\)</span> or <span
class="math inline">\(R_k\)</span> intersect <span
class="math inline">\(J\)</span>, whereas <span
class="math inline">\(J\)</span> can be at most one of those.</p>
</body>
</html>