Prove that a function $f: \mathbb{R} \to \mathbb{R}$ that is continuous in lower and upper limit topologies is constant.

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Let $f: \mathbb{R} \to \mathbb{R}$ a function that is continuous in the topologies $\mathcal{T}_{u} \to \mathcal{T}_{[,)}$ and $\mathcal{T}_{u} \to \mathcal{T}_{(,]}$, where $\mathcal{T}_{u}$ is the usual topology on $\mathbb{R}$ and $\mathcal{T}_{[,)}$ and $\mathcal{T}_{(,]}$ are the lower and upper limit topologies on $\mathbb{R}$, respectively. Show that $f$ is constant.


I don't actually know how to begin the proof because I can't see an easy way of using some of the most common characterizations of continuity. Maybe it is useful to use the comparison between the limit topologies and the usual one... I've also about arguing by contradiction, but I cannot see an easy argument. Also, I know there is an easy way of proving this using connection concepts, but we are supposed to do it without using them. I would be pleased of any hints. Thanks in advance.

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Let $X\in\Bbb R$. Then $U=(\leftarrow,f(x)]$ is an open nbhd of $f(x)$ in the upper limit topology, and $V=[f(x),\to)$ is an open nbhd of $f(x)$ in the lower limit topology, so

$$f^{-1}\big[\{f(x)\}\big]=f^{-1}[U]\cap f^{-1}[V]$$

is open in the usual topology. On the other hand, $\{f(x)\}$ is closed in both the upper limit and the lower limit topology, so $f^{-1}\big[\{f(x)\}\big]$ is closed in the usual topology. What subsets of $\Bbb R$ are clopen (both closed and open) in the usual topology?