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COMPACT, a tutorial by Scott W. Williams 

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2. A SPECIAL EXAMPLE
There are some special compact sets  the convergent sequence and its limit , the unit interval [0,1], and the unit circle. These tend to dominate how WE think of compact objects. The "long" space with an endpoint is quite useful to However, the Cantor set is a fundamentally important compact object many people believe an aberration  it is not. Here's one reason why (Urysohn 1914):
2.1. THEOREM. Each compact metric space is the continuous image of the Cantor set.
THE CANTOR
(MIDDLETHIRDS) SET
is the set of all real numbers the limit of an infinite series
of the form ,
where each is 0 or
1. Note that if each =1
(=0), then the sum is 1 (0).
The construction of the Cantor MiddleThirds Set proceeds recursively via removing various intervals from [0,1]:
Step1: From
remove the middlethird open interval (1/3 , 2/3)  we get a closed set, picture
C_{1}:
From both parts of C_{1}
remove the middlethird open interval  what's left is a closed set, picture
C_{2}:
From each of the four parts of
C_{2}:
remove the middle thirds open interval  picture
C_{3}:
And continue ...
The Cantor set is the intersection C =C_{n}. As an intersection of closed sets in [0,1], C is compact. As a closed subset of the compact set [0,1], C is closed, and hence, compact.
Noticing that the adjacent pairs in the Cantor C set can be mapped in an order preserving manner onto the rationals in (0,1). We see C can be pictured by considering [0,1] and replacing each rational number in (0,1) by two adjacent points. Indeed that picture gives impetus to a special case of Theorem 2.1  [0,1] is the continuous image of C {just send adjacent points to one}.
Cantor Set
[0,1]
Concerning the Cantor set, one must be careful with intuition. First it is very thin, because the sum of the lengths of the deleted intervals is 1; i.e., its measure is zero. On the other hand it has the same size as the entire interval [0,1]. This is strengthened by the problem which appears in W. Rudin's textbook, and on some Ph.D. Qualifying Exams:
2.2. EXERCISE. Each real in the interval [0,2] is the sum of two
members of the Cantor set:
For b[0,2]. Consider
the graph L_{b}
of the intersection of line x+y=b with subsets of the square [0,1]^{2}.
Indeed, each C_{n}^{2}intersects L_{b} in a copy of some C_{k}. Thus,
L_{b}C^{2}=(L_{b}C_{n}^{2})
is not empty; i.e., there are elements x and y in the Cantor set
such that x+y=b.
In 1999 the great topologist Mary Ellen Rudin solved an outstanding problem generalizing 2.1, which asked for a kind of "triangular inequality" extension of metric. She proved the hard "sufficient" in:
2.3. THEOREM. In order for a compact space X to be the continuous image of a compact linear ordered space it is necessary and sufficient that for each pair consisting of a point xX and its neighborhood G, there exists an open set G_{x} satisfying two conditions:
1. xG_{x}G.
2. If G_{x}H_{y} , then either yG or xH.
A preprint of the paper Nikiel's problem is available at the web site of the journal TOPOLOGY ATLAS (http://at.yorku.ca/topology/)
{Note. The "trianguar inequality" comment is motivated by observing that in a metric space when G is the open ball about x of radius r, we may take G_{x} to be the open ball about x of radius r/3. Then the triangular inequality proves condition (2)}
page 1: the beginnings 
page 2: a special example 
page3: functional separation 

page4: the universe in a box 
page5: open covers 