## Proposition 5.2.11: Perfect sets are 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 = { x*_{1}, x_{2}, x_{3}, ...}

*U*is a neighborhood of

_{1}= (x_{1}- 1, x_{1}+ 1)*x*. Since

_{1}*x*must be an accumulation point of

_{1}**S**, there are infinitely many elements of

**S**contained in

*U*.

_{1}
Take one of those elements, say *x _{2}* and take a
neighborhood

*U*of

_{2}*x*such that closure(

_{2}*U*) is contained in

_{2}*U*and

_{1}*x*is not contained in closure(

_{1}*U*). Again,

_{2}*x*is an accumulation point of

_{2}**S**, so that the neighborhood

*U*contains infinitely many elements of

_{2}**S**.

Select an element, say *x _{3}*, and take a neighborhood

*U*of

_{3}*x*such that closure(

_{3}*U*) is contained in

_{3}*U*but

_{2}*x*and

_{1}*x*are not contained in closure(

_{2}*U*) .

_{3}
Continue in that fashion: we can find sets
*U _{n}* and points

*x*such that:

_{n}

- closure(
*U*)_{n+1}*U*_{n} *x*is not contained in_{j}*U*for all_{n}*0 < j < n**x*is contained in_{n}*U*_{n}

**V**= ( closure(*U*)_{n}**S**)

*U*)

_{n}**S**) is closed and bounded, 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

*x*, because

_{1}*x*is not contained in closure(

_{1}*U*). It can not be

_{2}*x*because

_{2}*x*is not in closure(

_{2}*U*), and so forth.

_{3}
Hence, none of the elements
{ *x _{1}, x_{2}, x_{3}, ... * } 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.