Does there exist a path connected metric space , in which at least one open ball is countable ?

Question

Does there exist a path connected metric space with more than one point , in which at least one open ball is countable ?

Answer 1

I don't think this is true. Take two points in an open ball. Since there is a path between the points and the image of the path is uncountable, the open ball is uncountable.

Answer 2

At least, if the open ball $B$ is connected, this is false because then if $B$ is countable, it has only has one point. To see this, suppose there are two distinct points $a,b\in B$. Then $d$ restricted to $B$ is a metric on $B$.

Consider S= $\left \{ d(a,x) :x\in B\right \}$. If $S$ contains every real number in $(0,d(a,b))$ then $B$ is uncountable, which is a contradiction.

If there is a real number $z\in (0,d(a,b))$ such that $d(a,x)\neq z$ for any $x\in B$, then the sets $\left \{ x\in B:d(a,x) >z\right \}$ and $\left \{ x\in B:d(a,x) <z\right \}$ provide a separation of $B$ and so $B$ cannot be connected.

Answer 3

Let $X$ be a path connected space with more than one point, and let $B \subset X$ be an open ball. If $B$ is just a single point then by an easy argument $B=X$, a contradiction.

So there are two points $x \ne y \in B$, and using them I'll prove that $B$ has contains a subset of the cardinality of the reals.

Applying path connectedness of $X$, let $\gamma : [0,1] \to X$ be a continuous path with $\gamma(0)=x$ and $\gamma(1)=y$. By continuity of $\gamma$ there exists $T \in (0,1]$ such that $\gamma[0,T] \subset B$. It cannot happen that $\gamma(T)=x$ for all such $T$, because if so then taking $T_0$ to be the supremum of such $T$ we would have $\gamma(T_0)=x \ne y$, and so $T_0 \ne 1$, but then by taking small enough $\delta>0$ we would have $\gamma[0,T_0+\delta] \subset B$ contradicting that $T_0$ is the supremum.

So we may therefore take $T \in (0,1]$ so that $\gamma[0,T] \subset B$ and $\gamma(T) \ne x$. Now consider the function $$f : [0,T] \to \mathbb{R}, \quad f(t) = d(x,\gamma(t)) $$ This is a continuous function. By the intermediate value theorem its image $f[0,T]$ contains the entire nontrivial closed interval $[0,d(x,f(T))]$. So for each $D \in [0,d(x,f(T))]$ there exists a point $z \in B$ such that $d(x,z)=D$. It follows that the image $\gamma[0,T] \subset B$ contains a subset of cardinality equal to the reals.

Answer 4

In fact, $X$ is not even connected:

Since $B$ contains more than one point, choose $b_{0}, b_{1}\in B$ and let $d(b_{0}, b_{1})=r>0$. Choose $0<s<r$, such that $d(b_{0},x)\neq s$ for any $x\in X$. This is possible since $B$ is countable and since $\left \{ x\in X:d(x,b_{0})=s \right \}\subseteq B$

But now, the sets $\left \{ x\in X:d(x,b_{0})<s \right \}$ and $\left \{ x\in X:d(x,b_{0})>s \right \}$ are open, disjoint and their union is $X$. That is, they form a separation of $X$.