If f: $\mathbb{R}_{l} \rightarrow S_{\Omega}$ is continuous, then f is not injective.

Question

If f: $\mathbb{R}_{l} \rightarrow S_{\Omega}$ is continuous, then f is not injective.

I've been trying to solve this problem for a few days, but I haven't been able to see how can I do it. First, $\mathbb{R}_{l}$ refers to $\mathbb{R}$ with the lower limit topology (generated by intervals of the form $[a,b)$ ) and $S_{\Omega}$ refers to the uncountable well-ordered set with the order topology (generated by (a,b) and rays). Until now, I've tried to use the fact that both of those topologic spaces are Hausdorff and suppose that f is injective to get a contradiction, but I don't see where can I get it.

Answer 1

HINT: Let $y=\min S_\Omega$; then $\{y\}$ is open in $S_\Omega$, so if $f:\Bbb R_\ell\to S_\Omega$ is continuous, then $f^{-1}[\{y\}]$ is open in $\Bbb R_\ell$. Does $\Bbb R_\ell$ have any isolated points?

Answer 2

If $f$ is injective, then you have an imbedding of $[a,b]$ onto $f([a,b])$. But it is easy to see that an uncountable set with the order topology cannot have the same topology as the lower limit topology. See here.

Answer 3

Let $f: \Bbb R_l \to S_\Omega$ be an injective continuous function. We strive for a contradiction:

Let $f[\Bbb R_l]$ be the image in $S_\Omega$, which is a non-empty set so $m = \min f[\Bbb R_l]$ exists, so there is a unique (by injectivity of $f$) $x_0 \in \Bbb R_l$ with $f(x_0)=m$. In the order topology on $S_\Omega$, the set $O= \{x\mid x < m+1\}$ is open so that $f^{-1}[O] = \{x_0\}$ (identity clear from minimality of $m$ and uncity of $x_0$) is open in $\Bbb R_l$, contradicting that all non-empty open subsets of $\Bbb R_l$ are uncountable (they contain a set of the form $[a,b), a < b$).