Discrrete mathematics for computer science 11uncountable

Uncountable Sets

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Countably Infinite
There are as many natural numbers as
integers
0 1 2 3 4 5 6 7 8…
0, -1, 1, -2, 2, -3, 3, -4, 4 …
f(n) = n/2 if n is even, -(n+1)/2 if n is odd
is a bijection from Natural Numbers →
Integers

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Infinite Sizes
Are all infinite sets the same size?

NO!
Cantor’s Theorem
shows how to keep finding
bigger infinities.
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P(N)
• How many sets of natural numbers?
• The same as there are natural
numbers?
• Or more?

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Countably Infinite
Sets

::= {finite bit strings}
… is countably infinite
Proof: List strings shortest to
longest, and alphabetically
within strings of the same
length
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Countably infinite
Sets

= {e, 0, 1, 00, 01, 10,
11, …}
= { f(0),
= { e,
0, 1,

f(1), f(2),

00, 01, 10,

f(3), f(4), …}

11,
000,

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… }

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Uncountably Infinite Sets

What about infinitely long
bit strings? Like infinite
decimal fractions but with
bits

Claim:

::= {∞-bit
strings}
is uncountable.
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Diagonal Arguments
Suppose
0

1

2

3

.

.

.

n

n+1

.

.

.

s0

0

0

1

0

.

.

.

0

0

.

.

.

s1

0

1

1

0

.

.

.

0

1

.

.

.

s2

1

0

0

0

.

.

.

1

0

.

.

.

s3

1

0

1

1

.

.

.

1

1

.

.

.

.

.

.

.

.
.

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1
.

1
.

0
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Diagonal Arguments
• Suppose
0

1

2

3

.

.

.

n

n+1

.

.

.

s0

0
1

0

1

0

.

.

.

0

0

.

.

.

s1

0

1
0

1

0

.

.

.

0

1

.

.

.

s2

1

0

10

0

.

.

.

1

0

.

.

.

s3

1

0

1

1
0

.

.

.

1

1

.

.

.

.

.

.

.

.
.

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0
1
.

01
.

10
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Diagonal Arguments
Suppose
1

⋯

0
…differs
from every row!
1
So
cannot be listed:

0
this diagonal
0
sequence 0
1
will be missing

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Cantor’s Theorem

For every set, A
(finite or infinite),
there is no bijection
A↔P(A)
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There is no bijection A↔P(A)

Pf by
by contradiction:
contradiction: suppose
Pf
f:A↔P(A) is a bijection. Let

W::= {a 
A|a

f(a)}, so for any a,
a W iff a f(a).

f is a bijection, so W=f(a0), for some a0
A.

(∀a) a f(a0) iff a f(a ).

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There is no bijection A↔P(A)

Pf by
by contradiction:
contradiction: suppose
Pf
f:A↔P(A) is a bijection. Let
W::= {a 
A|a

f(a)}, so for any a,
a W iff a f(a).

f is a bijection, so W=f(a0), for some a0

A.
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a 0contradiction
f(a0) iff a 0f(a ). 0

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So P(N) is uncountable
P(N)
= set of subsets of N
↔ {0,1}ω
↔ infinite “binary decimals”
representing reals in the range
0..1

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