# Showing that a function $f:(X, \mathcal{T}) \to \mathbb{R}$ is right-continuous if and only if $f$ is continuous to a given topology on $\mathbb{R}$

What I'm asking for is some tips/hints on the following problem. As of writing I have no idea how to even begin proving the claim:

Let $$\mathcal{T}' = \{\varnothing, \mathbb{R}\} \cup \{(t, \infty)\mid t \in \mathbb{R}\}$$ be a topology on $$\mathbb{R}$$. I have to show that a function $$f:(X, \mathcal{T})\to \mathbb{R}$$ is right-continuous if and only if $$f$$ is continuous w.r.t. the topology $$\mathcal{T}'$$ of $$\mathbb{R}$$, where $$(X, \mathcal{T})$$ is an arbitrary topological space. The given description of a right-continuity is that $$f$$ is right-continuous if for every $$x \in X, \epsilon > 0$$ there exists a neighborhood $$U$$ of $$x$$ such that $$f(x) - \epsilon < f(y), \forall y \in U$$. To show that $$f$$ is cont. w.r.t. the topology $$\mathcal{T}'$$, we'd have to show that for every $$V \in \mathcal{T}'$$, $$f^{-1}[V]$$ is open in $$(X, \mathcal{T})$$. By right-continuity we know that for every $$x \in f^{-1}[V]$$ and $$\epsilon > 0$$ there exists $$U \in \mathcal{T}$$ s.t. $$f(x) - \epsilon < f(y), \forall y \in U$$. But how do we extend this fact to imply that $$f^{-1}[V]$$ is open? Surely we can find a cover for $$f^{-1}[V]$$ by taking the union over all neighborhoods of all elements of $$f^{-1}[V]$$. But then we only know that $$f^{-1}[V]$$ is a subset to the union of some elements of $$\mathcal{T}$$.

• That's not the definition of right-continuity. You should look into that. Commented Mar 6, 2022 at 14:52
• @jjagmath My source material is somewhat old and not in English, so translation error is possible. The direct translation of the given definition is downward half-continuous, which I took to refer to right-continuity. Commented Mar 6, 2022 at 15:33

## 1 Answer

Two thing. First, as hinted in the comments you have the wrong translation. This is lower semicontinuity, not right-continuity.

Second, your proof is on the right track. You are correct to pick, for each $$x\in f^{-1}[V]$$, an open neighborhood $$U$$ of $$x$$, and then take the union of these $$U$$ to get a cover of $$f^{-1}[V]$$. The trick is to choose $$\varepsilon$$ in the definition of lower semicontinuity correctly so that each $$U$$ is contained in $$f^{-1}[V]$$ and hence the union is exactly $$f^{-1}[V]$$.

So here is my hint: Suppose $$V=(t,\infty)$$ for some $$t\in\mathbb R$$. Given $$x\in f^{-1}[V]$$, you know that $$f(x)>t$$. What can $$\varepsilon$$ be (depending on $$f(x)$$ and $$t$$) so that the $$U$$ given by the definition is necessarily a subset of $$f^{-1}[V]$$?