Re: a question of Lubarsky

[Thomas Streicher <streicher@mathematik.tu-darmstadt.de>]

Dear Peter,

> The answer to the general question is surely "no": the sentences
> which hold in Sh(X + X) are exactly those which hold in Sh(X).

Thanks for that explict argument. I told Bob a cardinality axiom. In the
languages under consideration there are just countably many formulas but
there are class many non-homeomorphic sober spaces.

> Whether one can separate Sh(R) from Sh([0,\infty)) in this way
> is a more interesting question. Does "Brouwer's continuity theorem"
> hold in Sh([0,\infty))? The proof that I know for Sh(R) doesn't
> work over [0,\infty), but that may be because it's not the best
> proof.

I don't know which proof you have in mind. Until recently I was just aware
of the argument in the book by MacLane and Moerdijk using a gros topos.
But just now I have found that in Troelstra & van Dalen Ch.15 Thm.3.24 says
that for any completely regular, first countable space T without isolated
points Sh(T) validates Brouwer's Theorem. They attribute it to Grayson.
Thus Brouwer's theorem holds in any of the four spaces mentioned.

Bob tells me that with his forcing with settling he can distinguish R and
[0,\infty) but does not know how to distinguish Q and Q \cap [),\infty).

Thomas