Thursday, May 1, 2025

Denumerability of Subsets

Given an enumeration a1,a2,,an, of a denumerable set A, and an enumeration b1,b2,,bn, of a denumerable set B, we can enumerate AB in the order a1,b1,a2,b2,,an,bn,. Write a more general version of this statement for the union of arbitrarily many denumerable sets.

More generally, one shows:

Theorem.
If {Am}mN is a countable family of countable (denumerable) sets, then
m=1Am is itself countable.

Proof sketch (by explicit enumeration).

  1. For each mN, choose an enumeration

Am={am,1,am,2,am,3,}.

  1. Recall that N×N is countable. Fix any bijection

π:NN×N,π(k)=(m(k),n(k)).

  1. Define a new sequence (ck)k=1 by

ck=am(k),n(k).

  1. Then {c1,c2,c3,} lists every element of mAm

(possibly with repetitions), showing that mAm is countable.


Equivalently, if you prefer a “diagonal” style ordering without naming a particular pairing function, you can say:

  1. List all pairs (m,n)N×N in a single sequence, for example by moving along the diagonals

(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),

  1. Whenever (m,n) appears in that list, write down the n-th element of Am, namely am,n.

Because every pair (m,n) will eventually appear, every element of every Am will be listed. Hence a countable union of countable sets is countable.

The set of all infinite subsets of N cannot be denumerable; show this by showing that the set of all finite subsets of N is denumerable, and then applying the above theorem.

First, partition the collection of all finite subsets of N by their cardinality. For each kN, let Fk={FN:|F|=k} be the family of all k-element subsets of N. Then:

  1. Each Fk is denumerable.
    An easy way to see this is to observe that any k-element subset can be written uniquely as an increasing k-tuple (n1,n2,,nk)withn1<n2<<nk,niN. The set of all such increasing k-tuples is a subset of Nk, and Nk is denumerable (e.g.\ by the diagonal‐pairing argument). Hence each Fk is denumerable.

  2. The set of all finite subsets is a countable union of the Fk.
    F={FN:|F|<}=k=0Fk. By the theorem you just proved, a countable union of denumerable sets is denumerable. Therefore F is denumerable.


Now let I={XN:|X|=} be the collection of infinite subsets of N. Notice that every subset of N is either finite or infinite, so P(N)=FI. Suppose, for the sake of contradiction, that I were denumerable. Then F is denumerable (as shown above), so by the union theorem again, P(N)=FI would be a union of two denumerable sets, hence itself denumerable. But this contradicts Cantor’s theorem that P(N) is uncountable. Therefore our assumption was false, and I cannot be denumerable. In other words, the family of all infinite subsets of N is uncountable.

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