For the following question, I think the answer is the lower limit topology as the weak topology. I am not sure how to go about showing it. also, do I have to consider the fact that the lower limit topology is finer than the usual topology, because the weak topology is suppose to be the smallest topology which makes a family of functions not only continuous but a particular type of continuiuty, being right continuous. I don't think the usual topology would satisfy it.
The question is as follows: a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is said to be continuous from the right or right continuous for every $a\in\mathbb{R}$, $\lim_{x \to a^{+}} f(x) = f(a).$ Find the weak topology on the domain set $\mathbb{R}$, determined by the collection of all right continuous functions on it.
Thank you in advance.
Best Answer
I define the continuous from the right at $a \in \Bbb R$ condition for $f:\Bbb R \to \Bbb R$ as follows:
$$\forall \varepsilon>0: \exists \delta > 0: \forall x \in \Bbb R: (x>a \land |x-a| < \delta) \to |f(x)-f(a)| < \varepsilon\tag{1}$$
which is I think pretty standard. Let the smallest topology that makes all the functions that are continuous from the right at every point, continuous, be called $\mathcal{T}_w$
Then if $f: (\Bbb R, \mathcal{T}_l) \to (\Bbb R, \mathcal{T}_e)$ (from lower limit topology to Euclidean topology) is continuous, then $f$ fulfills $(1)$ at all points:
Having $a$ and $\varepsilon >0$ arbitrary, $f^{-1}[(f(a)-\varepsilon, f(a)+\varepsilon)]$ is open in $\mathcal{T}_l$ (by continuity of $f$, and the fact that open intervals are open in $\mathcal{T}_e$) and contains $a$, so there is some $\delta>0$ such that $$[a,a+\delta) \subseteq f^{-1}[(f(a)-\varepsilon, f(a)+\varepsilon)]$$ which immediately implies $(1)$ by definition.
On the other hand if $f:\Bbb R \to \Bbb R$ is continuous on the right at every point, we have that $f$ seen as a function from $(\Bbb R, \mathcal{T}_l)$ to $(\Bbb R, \mathcal{T}_e)$ is continuous: let $O$ be open in the Euclidean topology and let $a \in f^{-1}[O]$; we need to show that $a$ is an interior point of $f^{-1}[O]$. First note that $f(a) \in O$ and $O$ is Euclidean open so there is some $\varepsilon>0$ such that $(f(a)-\varepsilon, f(a)+\varepsilon) \subseteq O$. Apply $(1)$ to $a$ and this $\varepsilon$ and the condition gives that $x \in [a,a+\delta)$ implies that $f(x) \in (f(a)-\varepsilon, f(a)+\varepsilon)$, or equivalently $$[a,a+\delta) \subseteq f^{-1}[(f(a)-\varepsilon, f(a)+\varepsilon)] \subseteq f^{-1}[O]$$ showing that indeed $a$ is an interior point of $f^{-1}[O]$, as required. So $f^{-1}[O]$ is lower-limit open and $f$ is continuous.
Now $\mathcal{T}_l$ is one topology on $\Bbb R$ that makes all right-continuous $f$ continuous, and $\mathcal{T}_w$ is the minimal such, so $$\mathcal{T}_w \subseteq \mathcal{T}_l$$ And if $\mathcal{T}$ is any topology such that all right continuous $f$ are continuous when seen as $f(\Bbb R,\mathcal{T}) \to (\Bbb R, \mathcal{T}_e)$ then $\mathcal{T}_l \subseteq \mathcal{T}$: let $[a,b)$ be a basic open set of the lower limit topology, then the function $\chi: \Bbb R \to \Bbb R$ defined by $\chi(x)=1$ if $x \in [a,b)$ and $\chi(x)=0$ otherwise, is continuous from the right at any point. This is easy to check by the definition $(1)$. Now, as $\chi$ thus is continuous with domain $(\Bbb R, \mathcal{T})$ we know that $[a,b) = \chi^{-1}(\frac12, \rightarrow)] \in \mathcal{T}$ and so $\mathcal{T}$ contains all basic open sets of $\mathcal{T}_l$, so $\mathcal{T}_l \subseteq \mathcal{T}$. This shows directly that $\mathcal{T}_l$ is the minimal topology that makes all right continuous functions continuous.