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From: Niels Diepeveen on 24 Jul 2010 09:21 William Elliot wrote: > 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. Oops. I see now that that's the same idea. By the way, it's easy to do a tiny bit better. Since X is not the exterior of any nonempty set |T| >= |X| + 1. Assuming a well-ordering on T it's also easy to show that |T| = |X| + 1 is attainable. -- Niels Diepeveen |