Perfect Sets are Uncountable

Proof

We'll use the Nested Compact Set Property to show this, but let's prove it by contradiction.

Let $P$ be perfect and for the sake of contradiction let $P$ be countable (obviously it can't be infinite since then any point is isolated, so then $P$ wouldn't be perfect). Then:

We'll construct a sequence of compact subsets $K_n\subseteq P$ such that we'll have $x_1\notin K_2$, $x_2\notin K_3$, and so on. Some care should be taken to show that each $K_n$ is nonempty, but using the Nested Compact Set Property to produce an:

that cannot be on the list $\{x_1,x_2,x_3,\dots \}$.

Let $I_1$ be a closed interval that contains $x_1$. Then $x_1$ is not an endpoint of $I_1$ so then $x_1$ is not isolated so $\mathrm{\exists }{y}_1\ne x_1$ where $y_1\in P$ such that $y_1\in I_1$.

then let $I_2$ is the closed subinterval centered on $y_2$ of $I_1$. Namely if $I_1=[a,b]$ let:

then the interval $I_2=[y_1-\frac{\epsilon }{2},y_1+\frac{\epsilon }{2}]$ has the desired properties.

Repeat this process. Because $y_1$ is not isolated then $\mathrm{\exists }{y}_2$ in the interior of $I_2$ and we can say $y_2\ne x_2$. Now construct $I_3$ centered on $y_2$ and small enough so that $x_2\notin I_3$ and $I_3\subseteq I_2$. Observe that $I_3\cap P\ne \mathrm{\varnothing }$ because the set contains $y_2$.

Carrying out the construction inductively, the result is a sequence of $I_n$'s that are closed where:

1. $I_{n+1}\subseteq I_n$
2. $x_n\notin I_{n+1}$
3. $I_n\cap P\ne \mathrm{\varnothing }$

To finis the proof we let $K_n=I_n\cap P$. For each $n\in \mathbb{N}$ we have $K_n$ is closed because it is the intersection of closed sets, and bounded because it is contained in the bounded set $I_n$. Hence $K_n$ is compact, so by construction then $K_n$ is not empty and $K_{n+1}\subseteq K_n$. Thus using the Nested Compact Set Property then:

But since each $K_n\subseteq P$ and since $x_n\notin I_{n+1}$ then $\bigcap _{n=1}^{\mathrm{\infty }}{K}_n=\mathrm{\varnothing }$ which is our contradiction.

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