Equivalence of continuity on sorgenfrey line and a subspace

Question

Suppose $f: \mathbb{R} \mapsto \mathbb{R}$ and $a \in \mathbb{R}$. Are the following statements equivalent?

i. $f:R_s \mapsto \mathbb{R}$ is continuous in $a$ where $R_s$ is the Sorgenfrey line

ii. $f|_{(-\infty,a]}: (-\infty,a] \mapsto \mathbb{R}$ the restriction of $f$ in $(-\infty,a]$ is continuous in $a$, where $(-\infty, a]$ is equipped with the relative topology as a subspace of $\mathbb{R}$.

So far I showed that $i \Rightarrow ii$ using the fact that for every $V$ neighborhood of $f(a)$ there exists an open set $G \in \mathcal{T}_S$ so that $a \in G \subset f^{-1}(V)$ and then $(-\infty,a] \cap G \in \mathcal{T}_a$ and so $(f|_{(-\infty,a]})^{-1}(V)$ is a neighborhood of $a$, hence the continuity of the restricted $f$ in $a$.

For the inverse I have for every $V$ neighborhood of $f(a)$ there is $G \in \mathcal{T}_a$ such that $a\in G \subset (f|_{(-\infty,a]})^{-1}(V)$, while it is needed that $G \in \mathcal{T}_S$.

Any ideas, for the second statement?

Answer 1

If I understood you ideas right, it remains to pick $G$ in the form $(b;a]$ for some $b<a$ to finish the proof oft the inverse.

Answer 2

Two general facts :

If $f: X \to Y$ is continuous at $p\in X$ then for any $A \subseteq X$ with $p \in A$: $f|_A: A \to Y$ is continuous at $p$.

I.e. we cannot kill continuity at a point by going to a subspace.

Proof: Let $V \subseteq Y$ be a(n open) neighbourhood of $f(p)$. By continuity of the original $f$ at $p$ there is a neighbourhood $U$ of $p$ such that $f[U] \subseteq V$. Then by definition of the subspace topology $U \cap A$ is an open neighbourhood of $p$ in $A$ and $(f|_A)[U \cap A] = f[U \cap A] \subseteq f[U] \subseteq V$, so that indeed $f|A$ is continuous at $p$. This is also essentially your argument for i. implies ii.

For the reverse we need another general fact:

Let $f: X \to Y$ be a function betweeen spaces $X$ and $Y$, and let $p \in X$. If $f|_O \to Y$ is continuous for some open $O\subseteq X$ with $p \in O$, then $f$ is also continuous at $p$.

Let $V$ be a(n open) neighbourhood of $f(p)$ again. By continuity of $f|O$ we have some neighbourhood $N$ of $p$ in $O$ such that $(f|_O)[N] \subseteq V$. Because $N$ is a neighbourhood of $p$ in $O$, this means that or some open $U$ in $X$ we have $p \in U \cap O \subseteq N$. But $O \cap O$ is open in $X$ (finite intersection) and contains $p$, and $f[O \cap U] \subseteq f[N] = (f_O)[N] \subseteq V$, so that we cna use $O \cap U$ as witness that $f$ is continuous at $p$.

Thsis fact applies directly to (ii) to (i) as the set $(-\infty,a]$ is open in the Sorgenfrey line and contains $a$.