Ytukyg

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?

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
$$C colon=big{ d(x_1,x_n) colon n in mathbb{N}, n > 1 big} $$
is countable. But the set of all the real numbers in the open interval $left( 0, d left(x_1, x_2 right) right)$ is uncountable. So there exists some real number $r$ such that
$$ 0 < dleft(x_1, x_2 right) < r,$$ and also $rnotin C$.

Therefore the sets
$$ B_d left(x_1, rright) colon= left{ x in X colon d left(x, x_1 right) < r right} $$ and
$$ left{ yin X colon dleft( x_1,y right) > r right}$$ form a separation of $X$, meaning that $X$ is disconnected.

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 $X$.

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.

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.