Perfect Sets are Uncountable

Proposition: Perfect Sets are Uncountable

Every non-empty perfect set must be uncountable.

Proof:

If S is perfect, it consists of accumulation points, and therefore can not be finite. Therefore it is either countable or uncountable. Suppose S was countable and could be written as

S = { x₁, x₂, x₃, ...}

The interval U₁ = (x₁ - 1, x₁ + 1) is a neighborhood of x₁. Since x₁ must be an accumulation point of S, there are infinitely many elements of S contained in U₁.

Take one of those elements, say x₂ and take a neighborhood U₂ of x₂ such that closure( U₂ ) is contained in U₁ and x₁ is not contained in closure( U₂ ). Again, x₂ is an accumulation point of S, so that the neighborhood U₂ contains infinitely many elements of S.

Select an element, say x₃, and take a neighborhood U₃ of x₃ such that closure( U₃ ) is contained in U₂ but x₁ and x₂ are not contained in closure( U₃ ) .

Continue in that fashion: we can find sets U_n and points x_n such that:

closure( U_n+1 ) U_n
x_j is not contained in U_n for all 0 < j < n
x_n is contained in U_n

Now consider the set

V = ( closure( U_n ) S )

Then each set closure( U_n )

S ) is closed and bouned, hence compact. Also, by construction, ( closure( U_n+1 )

S )

(closure( U_n)

S ). Therefore, by the above result, V is not empty. But which element of S should be contained in V ? It can not be x₁, because x₁ is not contained in closure( U₂ ). It can not be x₂ because x₂ is not in closure( U₃ ), and so forth.

Hence, none of the elements { x₁, x₂, x₃, ... } can be contained in V. But V is non-empty, so that it must contain an element not in this list. That means, however, that S is not countable.

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