I was wrong about a countable set not being compact, now that I've decided to read Rudin's book a little bit closer. Chapter 2, Problem 12 is actually designed to prove this(and the proof is fairly easy). So yes...countable sets can be closed and they can also be compact. Hopefully this is all the hard-core math I'll need for quiz bowl this year! Joel Gluskin Wash U Academic Team --- In quizbowl_at_yahoogroups.com, arichman_at_b... wrote: > Quoting grapesmoker <no_reply_at_yahoogroups.com>: > > > And it seems we've lost the tack of the original remark in the > > process. The point made was that "compact" was ruled out by the use > > of "countable" so what I said in my previous post is wrong in that > > there are probably countable sets that are closed, but apparently > > not ones that are compact. Serves me right for not reading the post > > properly. I still stand by my statement that I don't believe in > > the "vacuously closed" proof, however. > > > In that case, you are wrong on two problems for the price of one. My copy of > Rudin is at my office, so it's not in front of me, but the vacuous proof of > being closed is absolutely correct. If it were not, then we would, for > example, say that the empty set or the set containing only a single point would > not be closed, which is certainly not standard usage. Anyway, if you don't > like that definition (and the proof that it generates), another definition used > by some (mostly topologists), and certainly a theorem in baby Rudin, is that > closed sets are those whose complements are open, a categorization which is > easy to see for the natural numbers (when viewed as a subset of the real > numbers). > > As for your second mistake, a countable set can easily be compact. The > simplest example is to take your own example {1, 1/2, 1/3, etc} and add the > single point {0} to the set. > > Alex
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