Prob. 4, Sec. 27, in Munkres' TOPOLOGY, 2nd ed: Any connected metric space having more than one point is uncountable

Question

Here is Prob. 4, Sec. 27, in the book Topology by James R. Munkres, 2nd edition:

Show that a connected metric space having more than one point is uncountable.

Here is a solution. Although I do understand the proof at this URL [The gist of that proof is the fact that no finite or countably infinite subset of $[0, +\infty)$ can be connected in the usual space $\mathbb{R}$.], I'd like to attempt the following.

My Attempt:

Let $X$ be a set having more than one point. Suppose that $(X, d)$ is a metric space such that the set $X$ is either finite or countable.

Case 1.

If $X$ is finite, then we can suppose that $X = \left\{ \ x_1, \ldots, x_n \ \right\}$, where $n > 1$. Then, for each $j = 1, \ldots, n$, let us put $$ r_j \colon= \min \left\{ \ d \left( x_i, x_j \right) \ \colon \ i = 1, \ldots, n, i \neq j \ \right\}. \tag{1} $$ Then the open balls $$ B_d \left( x_j, r_j \right) \colon= \left\{ \ x \in X \ \colon \ d \left( x, x_j \right) < r_j \ \right\}, $$ for $j = 1, \ldots, n$, are open sets in $X$.

In fact, we also have $$ B_d \left( x_j, r_j \right) = \left\{ \ x_j \ \right\}, \tag{2} $$ because of our choice of $r_j$ in (1) above, for each $j = 1, \ldots, n$.

So a separation (also called disconnection) of $X$ is given by $$ X = C \cup D, $$ where $$ C \colon= B_d \left( x_1, r_1 \right) \ \qquad \ \mbox{ and } \ \qquad \ D \colon= \bigcup_{j=2}^n B_d \left( x_j, r_j \right). $$ Thus $X$ is not connected.

Is my logic correct?

Case 2.

If $X$ is countably infinite, then suppose that $X$ has points $x_1, x_2, x_3, \ldots$. That is, suppose $$ X = \left\{ \ x_1, x_2, x_3, \ldots \ \right\}. $$ Then, as in Case 1, we can show that every finite subset of $X$ having more than one points is not connected.

Am I right?

Can we show from here that $X$ is not connected?

Answer 1

The proof in the link can actually be adapted to follow your approach.

Enumerate the distinct members of $X$ as $x_1,x_2, x_3, \ldots$. Then the set $$C \colon= \big\{\ d(x_1,x_n) \ \colon \ n \in \mathbb{N}, n > 1 \ \big\} \tag{A} $$ is countable.

But the set of all the real numbers in the open interval $\big( \ 0, \ d \left(x_1, x_2 \right) \ \big)$ is uncountable. So there exists some real number $r$ such that $$ 0 < r < d\left(x_1, x_2 \right), \tag{1}$$ and also such that this $r \not\in C$. [Refer to (A) above.]

Therefore the sets $$ B_d \left(x_1, r\right) \colon= \big\{ \ x \in X \ \colon \ d \left(x, x_1 \right) < r \ \big\} $$ and $$ \big\{ \ y\in X \ \colon \ d\left( x_1,y \right) > r \ \big\} $$ form a separation of $X$, meaning that $X$ is disconnected. The first of these two disjoint open sets contains the point $x_1$, whereas the second set contains the point $x_2$, by virtue of (1) above.

Answer 2

Assume distinct $a,b$. By Urysohn's theorem there is a continuous $f:X \to [0,1]$, with $f(a) = 0$, $f(b) = 1$. Since $X$ is connected, so is $f(X)$. Thus $f(X) = [0,1]$. Whence $X$ cannot be countable.

Answer 3

Here is an argument that does not require Urysohn's Lemma: fix $x \in X$ and define $f(y)=d(x,y)$. Then $f:X \to [0,\infty)$ is continuous, $f(x)=0$ and $f(y) >0$ for some $y$. Since the range is an interval it is uncountable and so is $X$.

Answer 4

You are just using $T_2$-ness of metric spaces to show finite ones are disconnected.

But there are $T_2$ spaces (even $T_3$) that are countable and connected, so the last step cannot work based purely on case 1. You need to really use the metric (or normality) etc. to get the disconnectedness.