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.
Is every connected metric space with at least two points uncountable?
-
0@Mathematics: Yes, it avoids a singleton, and also the more pathological case of the empty space containing no points at all (which is connected by definition!). – 2012-11-18
3 Answers
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.
-
3@Saugata Honestly, defining the function $f$ as you did is unnecessarily obfuscating for the person just learning the subject. Taking $f(x)= d(x,x_0)$ and then showing $f(X)$ contains an closed interval of $\Bbb{R}$ suffices. – 2017-06-13
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.
-
0@BrianM.Scott The problem Kenny was most likely trying to point out is that $2^\omega=\omega$ by the standard ordinal exponentiation. That's why you should use $2^{\aleph_0}$. I believe I've seen this mistake in other posts of yours. – 2014-08-21
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 $|X| \geq \mathfrak{c}$.
-
0Yes, you a$r$e right. Since the problem was posed for a metric space, instead of invoking the Urysohn Lemma, I constructed a Urysohn function e$x$plicitly. – 2012-11-18