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Suppose we have $I \subseteq R$. We few definition is required of this problem:

A family $\{V_i\}_{i \in I}$ of subsets of a set $X$ is an $I$-scale if

\begin{align*} V_i \subseteq V_j \quad \mathrm{whenever} \quad i \leq j \quad \mathrm{and} \quad \displaystyle\bigcup_{i \in I}V_i = X \end{align*}

Associate every $I$-scale subsets of $X$, the function

$$f(x):=\inf\{i \in I | x \in V_i\} \tag{1}$$

Question: Suppose $I$ is a dense subset of $[0,1]$. Show that for any $I$-scale of open subsets of $X$, $(1)$ is upper semi-continuous.

For simplicity, the definition I am using for upper semi-continuous is that a $f: X \to \mathbb{R}$ from a topological space to the real numbers is uppersemi-continuous if for any $x \in X$, $\epsilon > 0$, there is a neighborhood $N$ of $x$ such that $f(x') - f(x) < \epsilon$ for all $x' \in N$.

First, my first thought in approaching this problem was to parse the definition first. $I$ being dense subset of $[0,1]$ meant that every open set of $[0,1]$ (non-empty) intersects $I$. I was looking at Rudin's definition of dense in a set. Every point of $[0,1]$ is a limit point of $I$. I start with picking an $I$-scale of open subsets of $X$. One observation I was able to make was that given that if $I$ was finite, then it would contradict the definition of $I$ being dense in $[0,1]$. Also, I reasoned that $I$ can't be at most countable, so it has to be uncountable (not sure if this observation is relevant).

After gathering facts, I just thought about showing that $f$ had to be upper-semicontinuous using the definition I gave above. But implementing $f(x')-f(x)$ gave me a horrible expression involving infima. So I thought that if I can show that $f^{-1}(-\infty,a)$ is open for any $a \in \mathbb{R}$, I would be done. So I got that $f^{-1}(-\infty,a) = f^{-1}(-\infty,0) \cup f^{-1}[0,1]$. The first preimage is the emptyset. So we have that $f^{-1}[0,1]$ has to be open or closed. If it is open, we're done. However, if $f^{-1}[0,1]$ is closed, then there's a theorem I came across: $f^{-1}[a, \infty)$ is closed for any $a \in \mathbb{R}$ iff $f$ is upper semicontinuous. I don't think this is right, but any feedback is appreciated.

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1 Answer 1

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Suppose $x\in X$ and assume $\varepsilon > 0$. Since $f(x)=\inf\{i\in I:x\in V_i\}$, we may choose $k\in I$ such that $x\in V_k$ and such that $f(x) \le k < f(x)+\varepsilon$. Now consider $V_k$ and you should be able to finish the proof on your own.

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