Given two domains, $D_1$, $D_2$, already equipped with Scott (or Lawson) topology, the product domain $D=D_1\times D_2$ has the Tychonoff product topology, e.g., Mathematical Theory of Domains, page 124.

I can't seem to find a similar discussion of the function space topology for $D=[D_1\rightarrow D_2]$, though I assume it's well-known. Can someone explain it and/or give a reference which discusses it?

More generally, if you have a general domain equation $D=F(D_1,\ldots,D_n)$, where $F(\cdot)$ is built up from $D_1,\ldots,D_n$ by usual operations like product and function space, what can you say about $D$'s Scott/Lawson topology? In particular, how can you construct the open sets of $D$ from the open sets of $D_1,\ldots,D_n$?

  • $\begingroup$ @Tom, domains are part of programming language semantics. Dana Scott invented them to provide a semantic for lambda calculus. $\endgroup$
    – Kaveh
    Jan 4, 2016 at 7:10
  • $\begingroup$ The references I use when I want to look up something is G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. Mislove, D. S. Scott, Continuous Lattices and Domains and Samson Abramsky and Achim Jung Domain Theory. For topology vs. order theory S. Vickers Topology via Logic is very good. $\endgroup$
    – Kaveh
    Jan 4, 2016 at 7:19
  • $\begingroup$ It has been years since I have looked at these but generally when thinking about domains I always find it useful to think about them in terms of information: so intuitively the topology of $[D \to E]$ should be similar to the topology of $D^{op} \times E$. I will check Vicker's book when I have time (or you can do it yourself if you don't want to wait :). $\endgroup$
    – Kaveh
    Jan 4, 2016 at 7:29
  • $\begingroup$ @Kaveh Thanks for the hint to consider $D^{op}\times E$, which likely solves the problem (of specifying the open sets of $[D\rightarrow E]$ in terms of those of $D$ and $E$). I already have print copies of all three references you suggested (Handbook Volume 3 containing the Abramsky and Jung extended article, although your online version is an "expanded and extended" version). So I'll check them all myself -- certainly a worthwhile exercise regardless of reason:). P.S. and thanks for adding the tags (funny, I couldn't find a domain theory tag when I looked). $\endgroup$ Jan 4, 2016 at 7:56
  • $\begingroup$ I'd thought I'd just missed something obvious, and maybe it is, but I'm still not seeing how $D^{op}\times E\sim [D\rightarrow E]$, where I'm taking it you mean the usual (usual to me, anyway) $\sqsubseteq\Leftrightarrow\sqsupseteq$ for $D^{op}$. At first blush, doesn't seem cardinality-wise right, so I guess I'm still missing what you're saying. $\endgroup$ Jan 4, 2016 at 8:32


Your Answer

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

Browse other questions tagged or ask your own question.