Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.