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As the topic, how to prove that every connected metric space with at least two points uncountable? Of course i know the definition that a countable set mean there is a bijection between the set and the positive integer. Connected is opposite of disconnected where the set can partition into two disjoint open sets.

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marked as duplicate by 23rd, Lord_Farin, Stefan Hansen, Hanul Jeon, mau Oct 31 '13 at 10:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

So far I'm the only person who's up-voted this question. That often gets neglected. – Michael Hardy Nov 18 '12 at 3:58
For me, the main point is that a metric (or topological) space $X$ is connected if and only if the intermediate value theorem is valid on $X$; see user49521's answer. – Hendrik Vogt Nov 18 '12 at 7:11
@HendrikVogt But if the set X contains exactly 2 points, is the intermediate value theorem still valid? – Mathematics Nov 18 '12 at 7:38
@Mathematics: No, at least not if $X$ is a metric (or more generally a Hausdorff) space. And that's the point! The intermediate value theorem is not valid, and $X$ is not connected. (Take $f(x_0)=0$ and $f(x_1)=1$; then $f$ is continuous(!) and doesn't take the intermediate values in the open interval $(0,1)$.) – Hendrik Vogt Nov 18 '12 at 8:21
@HendrikVogt o i see. So is it always true that a connect sets always consist either infinitely many points or just singleton? – Mathematics Nov 18 '12 at 8:33
up vote 13 down vote accepted

Let us have another proof.

Since $X$ has at least two elements, let us choose $x_0,x_1\in X$, $x_0\neq x_1$. Define $f:X\rightarrow [0,1]$ by $$ f(x):=\frac{d(x,x_0)}{d(x,x_0)+d(x,x_1)},\text{ for all }x\in X. $$ Clearly, $f$ is continuous and $$f(x_0)=0\text{ and }f(x_1)=1.$$ Since $X$ is connected and the continuous image of connected space is connected (so called intermediate value theorem), it follows that $$ f(X)=[0,1], $$ which shows that $X$ is uncountable because $[0,1]$ is uncountable. This proves the result.

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i have a question, how do we know that $f(x)$ is continuous – Mathematics Nov 18 '12 at 4:32
f is continuous because x-->d(x,x_0) and x-->d(x,x_1) are continuous with d(x,x_0)+d(x,x_1)>0. That these functions are continuous follows from the triangle inequality satisfied by the metric. – Saugata Nov 18 '12 at 4:51
But if the set X only have 2 points, then is it still continuous? Also $f(X)$ seems $\ne[0,1 ]$ if $X$ has only 2 points – Mathematics Nov 18 '12 at 5:44
That $f(X)=[0,1]$ follows from the fact that X is connected. – Saugata Nov 18 '12 at 6:38

Not only must $X$ be uncountable, its cardinality must be at least $2^\omega=\mathfrak c$.

Let $\langle X,d\rangle$ be a metric space, and suppose that $|X|<2^\omega$. Fix $x,y\in X$ with $x\ne y$, and let $r=d(x,y)>0$. Let $D=\big\{d(x,z):z\in X\big\}$; $|D|\le|X|<2^\omega=|(0,r)|$, so there is a real number $s\in(0,r)\setminus D$. Show that $B_d(x,s)$ is a non-empty clopen subset of $X$ whose complement is also non-empty, and conclude that $X$ is not connected.

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What is $2^{\omega}=c$? What do $\omega$ and $c$ mean? – Mathematics Nov 18 '12 at 3:36
@Mathematics: $\omega$ is the cardinality of $\Bbb N$ (or any other countably infinite set); $2^\omega=\mathfrak c$ is the cardinality of $\Bbb R$. – Brian M. Scott Nov 18 '12 at 3:41
$\omega$ is the cardinality of the natural numbers. $\frak{c}$ is the cardinality of the real numbers. The statement $2^\omega = \frak{c}$ means that the power set of the natural numbers has the same cardinality as the real numbers. Whether or not there are infinities between the cardinality of the natural numbers and the cardinality of the real numbers depends on the axioms you choose, but Brian is saying that the size of $X$ is always at least $\frak{c}$, so even if there are smaller uncountable infinities, $X$ must in fact be at least the size of the real numbers. – Isaac Solomon Nov 18 '12 at 3:42
@Ricky: In normal mathematical contexts (like this one) I follow normal mathematical practice and take AC for granted. – Brian M. Scott Nov 18 '12 at 4:36
Shouldn't the correct symbol be $\aleph_0$? $\omega$ is an ordinal number. – kennytm Nov 18 '12 at 8:40

This is perhaps a generalisation of the method of user49521's answer above. Suppose now that $X$ is no longer a metric space but a normal space with at least two points that is connected. Call those two points $x_0$ and $x_1$. Then the Urysohn Lemma gives the existence of a continuous function $f : X \to [0,1]$ such that $f(x_0) = 0$, $f(x_1) = 1$. Now because $X$ is connected its image is also connected. Connected subsets of the reals are intervals and so we conclude that $X$ surjects via $f$ onto some interval that has cardinality $\mathfrak{c}$, so that $|X| \geq \mathfrak{c}$.

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Yes, you are right. Since the problem was posed for a metric space, instead of invoking the Urysohn Lemma, I constructed a Urysohn function explicitly. – Saugata Nov 18 '12 at 16:51

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