Equivalence of Path-Connectedness and Arc-Connectedness for Hausdorff Spaces

I have a classical sort of question. If we define a path in a space $X$ from points $a$ to $b$ to be a continuous image $f: I \rightarrow X$ from the unit interval to $X$ such that $f(0) = a$ and $f(1) = b$, then we can define an arc from $a$ to $b$ to be the same except we require that $f$ is a topological embedding, i.e. $f$ is a homeomorphism between $I$ and its image in $X$.

We can then define path-connectedness as per usual, and define arc-connectedness to mean that for every $a, b \in X$ there is an arc from $a$ to $b$ in $X$. It's a well-known theorem that for Hausdorff spaces these two properties are equivalent. However, the only proof I've seen uses the Hahn-Mazurkiewicz Theorem and the Moore/'cut point' characterization of the arc, neither of which are very fun to prove and which require an exposition of the order topology.

One of the main applications of these theorems is that every compact, connected, locally connected metric space (the definition of a Peano Continuum) is locally arc-connected. Then the equivalence of path-connectedness and arc-connectedness is established as a corollary of this.

What I want to do is prove the theorems in the following order:

(1a) Path-connectedness and arc-connectedness are equivalent for Hausdorff spaces.

(1b) If $X$ is a Peano continuum and $x \in X$ then $x$ has a local basis of Peano (sub)continua.

(2) Every Peano continuum is path-connected (more precisely, there is a surjection of $I$ onto $X$, which is the Hahn-Mazurkiewicz Theorem).

(3) Immediate corollary: Every Peano continuum is locally arc-connected.

The proof of (1b) is standard and involves a concept called Property $S$. I have developed a short, self-contained proof of (2) using an esoteric theorem called the Order Arc Theorem (which I need for other reasons, anyway) and (1b). Then (3) follows immediately.

So the only task is to prove (1a), the equivalence of arc-connectedness and path-connectedness, in a self-contained manner. Does anyone know of a proof of this fact that doesn't involve the Hahn-Mazurkiewicz Theorem/Moore classification/order topology? These theorems are very fundamental and attractive (particularly (1a)), so it would be nice to have a shorter proof of them so that they could be included in topology courses. I have sketched a proof; could someone look it over and let me know if there is an error? If there is, it's probably of the glaring sort haha

I assume familiarity with the devil's staircase function $d$ from the standard Cantor ternary set onto $I$. It can be shown that this function is continuous, and is 1-1 everywhere except the end points of the deleted middle-third intervals, where it is 2-1. Using the Moore classification of the Cantor set (or by explicit construction), one can show that the set created by recursively removing intervals of sizes other than $\frac{1}{3^n}$ and/or moving their centers also results in a Cantor set that admits an order-preserving homeomorphism $\phi$ with the standard Cantor set. For a Cantor set $D$ constructed as such, we call the map $d \circ \phi: D \rightarrow I$ a pinching map for $D$.

So suppose $X$ is a path-connected Hausdorff space and let $a, b \in X$. Let $f$ be a path from $a$ to $b$; we need to show there is a bona-fide arc from $a$ to $b$. After discarding and reparametrizing if need be, since $X$ is Hausdorff we can assume that $f^{-1}(a)$ and $f^{-1}(b)$ are $\lbrace 0 \rbrace$ and $\lbrace 1 \rbrace$ respectively.

Let $E_1$ be a (not necessarily unique) largest non-trivial closed interval in $I$ such that its end points have the same image under $f$, i.e. $f(E_1)$ has a 'loop' (existence uses the Hausdorff property again). If none exists then $f$ is a bijection so we're done. Assuming otherwise, consider $J_1 = I \setminus E_1^\circ$, where $E_1^\circ$ is the interior of $E_1$. This chops $I$ into two closed intervals, and we can then repeat inductively, cutting away successive intervals from the interiors (and possibly leaving some intervals intact if $f$ is bijective on them) of remaining intervals. We end up with a closed set $Z = \cap J_n$ consisting of closed intervals and Cantor sets.

Then form the 'pinching map' $g$ of $Z$ by just having the map be linear on the non-trivial intervals in $Z$. This will be a continuous map - in fact a quotient map - from $Z$ onto $I$ that is 1-1 everywhere except the end-points of the removed intervals $E_k$. However, $f$ restricted to $Z$ agrees precisely with the identifications of $g$ so by a standard quotient map argument will lift to a continuous map from $I$ to $f(Z)$.

But this will be a bijection on $I$ if we can show it's 1-1 on $Z$ everywhere except the end points of the sets $E_n$, and thus will be our desired arc from $a$ to $b$ since continuous bijections on a compact $I$ into a Hausdorff $X$ are homeomorphisms. Suppose not, i.e. $f(x) = f(y)$ with $x \neq y$. Then $|x - y| > 0$ but length$(E_n) \rightarrow 0$ since, for example, it's an infinite collection of disjoint intervals in $[0,1]$, impossible.

Does anyone see an error here and/or want to help me out with the fine details?

• It looks fine to me. – Moishe Kohan Dec 7 '17 at 7:59