可數的含義
一個集合若是有限的，或能與自然數 ℕ 建立[[bijection|雙射]]，則稱為[[countable-set|可數]]。具體地說，可數意味著可列：你能把元素排成一個序列 a₀, a₁, a₂, …，使每個元素都出現在某個有限的位置上。自然數、整數、乃至有理數都是可數的——它們有相同的[[cardinality|基數]]，記作 ℵ₀，儘管有理數是[[density-of-the-rationals|稠密]]的而整數是分散的。
Why the rationals Q are countable: a snake through a grid.
List positive rationals p/q in a table, row p, column q:
q=1 q=2 q=3 q=4 ...
p=1 1/1 1/2 1/3 1/4
p=2 2/1 2/2 2/3 2/4
p=3 3/1 3/2 3/3 ...
p=4 4/1 ...
Traverse it along finite diagonals (p+q = 2, then 3, then 4, ...):
1/1 , 1/2 , 2/1 , 1/3 , 2/2 , 3/1 , 1/4 , ...
Skip any fraction not in lowest terms (2/2 already seen as 1/1).
Every positive rational sits on exactly one diagonal, so it
appears at some finite step. Interleave with 0 and negatives:
0, 1/1, -1/1, 1/2, -1/2, 2/1, -2/1, ...
That is an explicit listing of ALL of Q. Hence Q is countable.康托對角線論證
現在轉折來了。實數是[[uncountable-set|不可數]]的：不存在一張能把它們全部捕獲的列表。證明是[[cantor-diagonal-argument|康托對角線論證]]，它用反證法。為導出矛盾，假設 [0, 1) 中每個實數都能列成 x₀, x₁, x₂, …。我們將製造一個可證明不在你列表中的實數，從而與「列表完整」的假設矛盾。
Cantor's diagonal argument (proof by contradiction).
Assume ALL reals in [0,1) are listed by their decimal digits:
x0 = 0 . d00 d01 d02 d03 ...
x1 = 0 . d10 d11 d12 d13 ...
x2 = 0 . d20 d21 d22 d23 ...
x3 = 0 . d30 d31 d32 d33 ...
^
the diagonal digits are d00, d11, d22, d33, ...
Build a new number y = 0 . e0 e1 e2 e3 ... by the rule
e_k = 5 if d_kk is not 5
e_k = 4 if d_kk is 5
(any rule avoiding 0 and 9 dodges the 0.4999...=0.5000... issue).
Then y differs from x_k in its k-th digit, for EVERY k:
e_k != d_kk by construction.
So y is not x0, not x1, not x2, ... -- y is on NO row.
But y is a real in [0,1), so the list was not complete.
Contradiction. Therefore [0,1), and hence R, is uncountable.停下來欣賞其優雅。對角線數 y 被精心設計成在第 k 位與第 k 行不同，因此它不可能是任何一行。結論是永久的：ℝ——[[the-continuum|連續統]]——的基數嚴格大於 ℕ。實數確實比有理數多，儘管二者都是無窮。這一刻，無窮不再是單一的概念，而成為一個層級。