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?
Best Answer
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 $X$.