Show that a complete metric space without isolated points is uncountable

Question

I've seen a few questions already posted on here, but they all deal with perfect subsets and being Hausdorff, both of which are topics we haven't covered yet.

I know that a point a is isolated if $\{a\}$ is open (aka there exists $\epsilon > 0$ s.t. $B_{\epsilon}(a) = \{a\}$. I am supposed to consider singletons and then use Baire Category Theorem.

Could someone help me with where to start?

Answer 1

Hint. Let $(X;d)$ be a (nonempty) complete metric space with no isolated points.

Suppose $X$ is countable and fix an enumeration $X = \{x_n \mid n \in \mathbb N\}$. For each $x \in X$ let $O_x := X \setminus \{x\}$.

Show that, for each $n \in \mathbb N$, $O_{x_n}$ is dense in $(X;d)$.

By the Baire Category Theorem this implies that $\bigcap_{n \in \mathbb N} O_{x_n}$ is dense in $(X;d)$. Now figure out why this is a contradiction.

Answer 2

Say $(X,d)$ is a countable complete metric space without isolated points.
Then, we can write $X=\{x_n :n \geq 1 \}$(as we assume it is countable).
Now, consider the Sets $U_n= X \backslash \{x_n\}$. Since $\{x_n \}$ is closed, $U_n $ is open. Also because $\{ x_n \}$ is not isolated(as given), for all $\varepsilon > 0$, $B(x_n,\varepsilon) \cap U_n \neq \phi$, Hence $U_n$ is dense in $X$(Any other point not equal to $X$ already in $U_n$ ).
Now, by Baire's Category Theorem $\bigcap_{i=1}^{\infty} U_n$ is dense in $X$. But since $\bigcap_{i=1}^{\infty} U_n \neq \phi$ is a contradiction so our assumption that $X$ is countable is not true hence, $X$ must by uncountable.

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Stefan Meskan also explains up to this point in his answer. Only point need to reconsider is why $\bigcap_{i=1}^{\infty} U_n \neq \phi$ is contradiction.
Let's consider a general case for $X$
Since $\mathbb{X}$ is countable we can write $\mathbb{X}=\{x_1,x_2,x_3 \dots,x_{n-1} x_n \}: n \in \mathbb{N} $ & as we defined $$U_1=\{\hphantom{x_1,}x_2,x_3,x_4 \dots,x_{n-1}, x_n\}$$ $$U_2=\{x_1,\hphantom{x_2,}x_3,x_4 \dots,x_{n-1}, x_n\}$$ $$U_3=\{x_1,x_2,\hphantom{x_3},x_4 \dots,x_{n-1}, x_n\} $$ $$\vdots \hphantom{U_4 = \{ x1x2} \vdots \hphantom{ \},\{ } \vdots \hphantom{ \},\{ } \vdots \hphantom{ \},\{ } \vdots $$ $$ U_n= \{ x_1,x_2,x_3,x_4 \dots x_{n-1},\hphantom{x_n} \}$$
So, cleary $\bigcap_{i=1}^{\infty} U_n = \phi$. (As $\forall x_n \in X \hspace{0.2cm} \exists \hspace{0.2cm} U_n \text{ and } \bigcup_{i=1}^{\infty} U_n=X$)