Paths in a connected subset of $\mathbb R^n$

Question

Prove that every connected subset $X$ of $\mathbb R^n$ with more than one point has the continuum cardinality.

In order to use Baire Lemma, I would like to demonstrate that there is a compact (or just complete) and connected subset of $X$ with at least two points. That would have cardinality $>\aleph_0$, because its points are not open (by connectedness) and completeness allows us to use Baire Lemma.

If I knew that $X$ contains the image of a nonconstant path, then I would be ready to conclude. How can I prove this, if it is true?

(And more: is the statement true without the Continuum Hypotesis?)

Thank you in advance.

Answer 1

This is a consequence of Urysohn's Lemma.

Consider different points $a,b \in X$. Urysohn's Lemma produces a continuous function $f : X \to \mathbb{R}$ such that $f(a) \ne f(b)$. Let $J \subset X$ be the interval with endpoints $f(a),f(b)$.

To prove that $X$ has the cardinality of the reals it suffices to prove that $J \subset \text{image}(f)$, for then by choosing a point in the inverse image of each element of $J$ one obtains an uncountable subset of $X$.

So if $J \not\subset \text{image}(f)$, pick $t \in J - \text{image}(f)$. Thus $\text{image}(f) \subset \mathbb{R}-\{t\}$, and $\text{image}(f)$ contains points in both components $(-\infty,t)$ and $(t,+\infty)$ of $\mathbb{R}-\{t\}$. Therefore $X = f^{-1}(-\infty,t) \cup f^{-1}(t,+\infty)$ produces a disconnection of $X$, contradicting that $X$ is connected.

Certainly I did not use the continuum hypothesis in this proof (although the "choosing" operation seems to have used the axiom of choice; perhaps that can be avoided too?).

Answer 2

Let $X$ be a connected subspace of a metric space $S$ with metric $d$ on $S,$ and let a,b be distinct points of $X.$

1. Then $\forall r\in (0,d(a,b))\;\exists c\in X\; (d(a,c)=r).$

   Proof: If not then for some $r\in (0,d(a,b))$ the sets $V=X\cap \cup_{s>r}\cup \{B_d(p, (s+r)/2): s=d(a,p)\}$ and $U =X\cap B_d(a,r)$ are disjoint open subsets of $X$ such that $U\cup V=X.$

   But $a\in U$ and $b\in V,$ so $X$ is not connected.

2. By the Axiom of Choice, let $f: (0,d(a,b))\to X$ such that $d (a,f(r))=r.$ Then $f$ is injective so the cardinal of the image of $f$ is the cardinal of the reals. If this is also the cardinal of $S$ then by the Cantor-Schroeder-Berstein theorem, then this is the cardinal of $X$ too.