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From: William Elliot on 18 Jul 2010 03:33 On Sat, 17 Jul 2010, Prof Lobster wrote: > William Elliot <marsh(a)rdrop.remove.com> wrote in > >> Let (S,tau) be a T0 space with a finite topology. Show S is finite. >> How's this for a proof? Any suggestions for a more direct proof? >> > The map > x |-> {U : U is open and x is an element of U} > injects S into a finite set. > Ok, I get the idea. Let (S,tau) be a T0 space. For all x,y, (cl {x} = cl {y} ==> x = y). Thus the map f:S -> tau, x -> S\cl {x} is an injection and |S| <= |tau|. -- How are you doing Professor Lobster? Has the gulf oil spill spared you? ----
From: David C. Ullrich on 18 Jul 2010 06:53 On Sun, 18 Jul 2010 00:33:06 -0700, William Elliot <marsh(a)rdrop.remove.com> wrote: >On Sat, 17 Jul 2010, Prof Lobster wrote: >> William Elliot <marsh(a)rdrop.remove.com> wrote in >> >>> Let (S,tau) be a T0 space with a finite topology. Show S is finite. >>> How's this for a proof? Any suggestions for a more direct proof? >>> >> The map >> x |-> {U : U is open and x is an element of U} >> injects S into a finite set. >> >Ok, I get the idea. ??? He gave a _proof_, not just an idea. The definition of T0 says _precisely_ that if x <> y then the class of open sets containing x is not the same as the class of open sets containing y. No need for any lemmas, that's exactly what the definition says. And now _that_ says precisely that x |-> {U : U is open and x is an element of U} defines a 1-1 mapping of S into the power set of tau. This is a much more "direct" proof, exactly as you requested. > Let (S,tau) be a T0 space. >For all x,y, (cl {x} = cl {y} ==> x = y). >Thus the map > f:S -> tau, x -> S\cl {x} >is an injection and > |S| <= |tau|.
From: Niels Diepeveen on 18 Jul 2010 19:57 William Elliot wrote: > Let (S,tau) be a T0 space with a finite topology. Show S is finite. > > How's this for a proof? Any suggestions for a more direct proof? > > Since S has a finite topology, it has isolated points, > only finite many. Remove from S, the isolated points > and from tau, the singletons. The new space has fewer > open sets. Iteration of this construction will end in > a finite number of steps, with the empty space. > > Lemma. A T0 space with a finite topology has an isolated point. > By the Hausdorff maximality theorm, there's a maximal > chain C of nonnul open sets. /\C is a nonnul minimal > open set. By T0, /\C is a open singleton. > Isn't it more generally the case that if (X, T) is a T0 space, then |T| >= |X| ? It seems to me that f: X -> T with f(x) = Ext {x} must be injective. -- Niels Diepeveen
From: William Elliot on 19 Jul 2010 00:30 On Sun, 18 Jul 2010, David C. Ullrich wrote: > <marsh(a)rdrop.remove.com> wrote: >> On Sat, 17 Jul 2010, Prof Lobster wrote: >>> William Elliot <marsh(a)rdrop.remove.com> wrote in >>> >>>> Let (S,tau) be a T0 space with a finite topology. Show S is finite. >>>> How's this for a proof? Any suggestions for a more direct proof? >>>> >>> The map >>> x |-> {U : U is open and x is an element of U} >>> injects S into a finite set. >>> >> Ok, I get the idea. > > ??? He gave a _proof_, not just an idea. I used the idea to prove a stronger theorem. > The definition of T0 says _precisely_ that if x <> y then the class of > open sets containing x is not the same as the class of open sets > containing y. No need for any lemmas, that's exactly what the definition > says. It's equivalent to for all x,y, (cl {x} = cl {y} implies x = y). > And now _that_ says precisely that > x |-> {U : U is open and x is an element of U} > defines a 1-1 mapping of S into the power set > of tau. The injection x -> S - cl {x} is an injection into tau instead of the larger P(tau). > This is a much more "direct" proof, exactly as > you requested. > It was and with it, I generalized the theorem. Any T0 space has more open sets than points. >> Let (S,tau) be a T0 space. >> For all x,y, (cl {x} = cl {y} ==> x = y). >> Thus the map >> f:S -> tau, x -> S\cl {x} >> is an injection and >> |S| <= |tau|. ----
From: William Elliot on 19 Jul 2010 00:33
On Sun, 18 Jul 2010, Niels Diepeveen wrote: > William Elliot wrote: > >> Let (S,tau) be a T0 space with a finite topology. Show S is finite. >> > > Isn't it more generally the case that if (X, T) is a T0 space, > then |T| >= |X| ? > It seems to me that f: X -> T with f(x) = Ext {x} must be injective. > Yes. See my reply to Professor Lobster. |