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?

  • 2
    $\begingroup$ I note that no one has answered your first question: yes, your logic for the finite case is correct. The biggest quibble I find is with saying "either finite or countable" when you mean "either finite or countably infinite". Afterall, finite is countable. But that is forgivable. For your case 2, the fact that finite subsets are not connected is not helpful for proving the infinite case, as Henno Brandsma indicates. For the infinite case, you will need a different proof, such as the ones already provided. $\endgroup$ – Paul Sinclair Oct 3 '18 at 16:50

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.

  • $\begingroup$ +1 for actually answering the real question. $\endgroup$ – Arnaud D. Oct 5 '18 at 10:18
  • $\begingroup$ @user1551 I've made some edits to your post. Do you agree with these? $\endgroup$ – Saaqib Mahmood Nov 25 '18 at 7:41

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 $X$.

  • $\begingroup$ For the record, this is pretty much the same as the other answer, but providing an explicit function $f$. $\endgroup$ – Wojowu Oct 3 '18 at 13:25
  • $\begingroup$ This is the proof that appears at the link given in the question. $\endgroup$ – Arnaud D. Oct 3 '18 at 13:42
  • $\begingroup$ @KaviRamaMurthy how are you? Can we connect using phone, email, or WhatsApp? $\endgroup$ – Saaqib Mahmood Nov 25 '18 at 7:58
  • $\begingroup$ @KaviRamaMurthy the range is not necessarily the whole of the interval $[0, +\infty)$. $\endgroup$ – Saaqib Mahmood Nov 25 '18 at 8:00

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.


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