In "A domain theoretic account of Picard's theorem" (http://www.doc.ic.ac.uk/~dirk/Publications/icalp2004.pdf), the authors assert the following.

Let $\mathbb{IR}$ be the interval domain $\lbrace [a^-,a⁺] | a^- \leq a^+, a^-, a^+ \in \mathbb{R} \rbrace \cup \lbrace \mathbb{R} \rbrace$.

Suppose $f = [f^-,f^+] : [-a,a] \rightarrow \mathbb{IR}$ is Scott continuous. Then $f^-$ and $f^+$ are measurable.

Their proof consists in asserting that $f⁻$ and $f⁺$ are lower (resp. upper) semi-continuous, which I have tried without success to establish. Can anyone help with either an explanation or a reference that proves it?

  • $\begingroup$ Interesting question. Am I right to assume that if $f(x) = [z_1, z_2] \in \mathbb{IR}$ then $f^-(x) = z_1$ and $f^+(x) = z_2$? (I didn't find the definition of $f^-, f^+$ in the paper you linked to.) $\endgroup$ – Dominic van der Zypen Jul 24 '15 at 7:38
  • $\begingroup$ Searching for "scott continuous" on google only gives me link after link of air "scott continuous air freshener" sites... $\endgroup$ – Trevor J Richards Oct 20 '17 at 14:02

First, we note that $\mathbb{IR}$ is ordered by $[a,b] \sqsubseteq [c,d]$ iff $[c,d] \subseteq [a,b]$, which makes $\mathbb{IR}$ into a domain. Now suppose $f: [-a,a]\to \mathbb{IR}$ is Scott-continuous (that is, $f$ preserves all directed suprema).

We want to prove that $f^-$ as defined in the comment above is lower semi-continuous. Let $(x_n)$ be an increasing sequence in $[-a, a]$. So $(x_n) \to s$ where $s = \sup\{x_n:n\in\mathbb{N}\} \in [-a, a]$. We want to show that $\lim_{n\to\infty} f^-(x_n) = f^-(s)$.

The set $$D:=\{x_n: n\in \mathbb{N}\}$$ is a directed subset of the domain $[-a,a]$ (with the ordering inherited from $\mathbb{R}$), and we have $s=\bigsqcup D$.

For $n\in \mathbb{N}$ we have $f(x_n)\supseteq f(x_{n+1})$ since $f$ is order-preserving (which is implied by $f$ being Scott-continuous). The fact that $f$ is Scott-continuous means that $$[f^-(s), f^+(s)] = f(s)=f(\bigsqcup D) = \bigsqcup f(D) = \bigcap \{ [f^-(x_n), f^+(x_n)]:n\in\mathbb{N}\}.$$

So $(f^-(x_n))_{n\in\mathbb{N}}$ is an increasing sequence in $\mathbb{R}$ and it is bounded by $f^+(x_1)$, so it converges. The equations above imply that $\lim_{n\to\infty} f^-(x_n) = f^-(s)$, so $f^-: [-a,a]\to \mathbb{R}$ is lower semi-continuous. A similar argument shows that $f^+$ is upper semi-continuous.

| cite | improve this answer | |
  • $\begingroup$ You got an anonymous edit suggestion to change "directed set" to "domain" in the first paragraph. $\endgroup$ – Joonas Ilmavirta Jul 24 '15 at 19:57
  • $\begingroup$ Thank you so much for this detailed answer. There is something that seems strange to me: if $f$ being Scott-continuous truly means that for all $x \leq y$, $f(x) \supseteq f(y)$, then in particular $f(-a) \supseteq f([-a,a])$? So this function can't afford to be very "thin" (if we think of it as an approximation to a real function, as is the goal of the paper). $\endgroup$ – user3078439 Jul 25 '15 at 0:14
  • $\begingroup$ I'll answer your question in an answer below as it doesn't fit in a comment. $\endgroup$ – Dominic van der Zypen Jul 25 '15 at 11:22

This answers @user3078439's question in his comment to my original answer.

First, a short argument to show that if $P,Q$ are domains and $f:P\to Q$ is Scott-continuous, then $f$ is order preserving.

Let $a\leq b \in P$. So $D:= \{a,b\}$ is a directed set with $b=\bigsqcup D$. Since $f$ is Scott-continuous, $f(D) = \{f(a), f(b)\}$ is also directed, so at least one of the relations $f(a) \leq f(b)$ or $f(b) \leq f(a)$ holds in $Q$. Now $f$ being Scott-continuous means $$f(b) = f(\bigsqcup D) = \bigsqcup f(D) = \sup\{f(a), f(b)\},$$ which directly implies $f(a)\leq f(b)$. So if $f:P\to Q$ is Scott-continuous, then it is order-preserving. (The converse of this statement does not hold.)

Since in your question the ordering on $Q:= \mathbb{IR}$ is given by $\supseteq$, I hope that I have answered your implicit question "if $f$ being Scott-continuous truly means that for all $x\leq y, f(x)\supseteq f(y)$" in the positive.

As to the question whether $f(-a)\supseteq f([-a,a])$ there is a negative answer: $f(-a)$ is just a real interval, namely $[f^-(-a), f^+(-a)]$. But $f([-a,a])$ is a set of intervals.

What is true, though, is that $f(-a)=[f^-(-a), f^+(-a)]$ is an interval that is a superset of every interval $f(x)=[f^-(x), f^+(x)]$ for all $x\in[-a,a]$. If that was the intent of your question in the comment, then there is a positive answer.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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