Situation where a non-$T_0$ path-connected space is injectively path-connected

Question

Suppose in all that follows that $X$ is a path connected topological space. That is, between any two points $a$ and $b$ there is a path $f:[0,1]\to X$ joining $a$ to $b$.

It is a classical result (see here) that if $X$ is Hausdorff, one can find an injective path between any two distinct points of $X$. (Steen & Seebach's Counterexamples in topology calls this "arc connected").

Now suppose $X$ is not even $T_0$ but in an extreme way: for every $x\in X$, there are at least $\mathfrak c=|\mathbb R|$ points that are topologically indistinguishable from $x$. Can distinct points of $X$ be joined by an injective path?

Answer 1

Let $f: [0,1] \to X$ be a path from $a$ to $b$. We shall now modify it so that it becomes an arc from $a$ to $b$. For each $x \in X$, let $I_x = f^{-1}(\{x\})$. Since $|I_x| \leq \mathfrak{c}$, we may choose an injective map $\pi_x: I_x \to [x]$ where $[x]$ is the collection of all points topologically indistinguishable from $x$. Furthermore, we shall require:

1. $\pi_a(0) = a$, $\pi_b(1) = b$;
2. $\pi_x(I_x) \cap \pi_y(I_y) = \varnothing$ whenever $x \neq y$. This can be arranged because there are at most $\mathfrak{c}$ many $z$ s.t. $I_z$ is nonempty.

Now define,

$$f': [0,1] \to X, f'(t) = \pi_{f(t)}(t)$$

By the conditions on $\pi_x$, it is not hard to check $f'$ is injective and $f'(0) = a$, $f'(1) = b$. Furthermore, $f(t)$ and $f'(t)$ are topologically indistinguishable for any $t$, from which it is easy to conclude that $f'$ must be continuous as $f$ is continuous.