Re: is this as easy as i think? finite complement topology

In article <dna72i$vfv$1@xxxxxxxxxxxxxxxxxx>, Bob <Bob@xxxxxxxxxx> wrote:

Problem: prove that every subset of R is compact in the cofinite topology.


>Thank you for helping me with the big flaw in my understanding, mind if i
>ask how i should problem this problem ?

I don't know how one goes about probleming a problem. But if you want
to know how to go about solving the problem... (-;

Let X be ANY subset of R. We are NOT going to assume ->anything<-
about X, other than the fact that it is a subset of X.

Let {U_i} i in I be a collection of open subsets of R, such that X is
contained in the union of the U_i. We do NOT assume ANYTHING about the
U_i, except that they are cofinite (have finite complement) or empty.

We want to show that there must be a finite collection U_{i_1},...,
U_{i_n} of open sets in our collection such that X is contained in the
union of them.

So, first, if X is empty then this is easy: just pick any one U_i and
that will do (can you see why?)

So we may assume that X is not empty. Let x be some element of X. Then
there has to exist at least on U_i such that x is in U_i. Let's call
that U_{i_1}.

What do we know about U_{i_1}? It is an open set, so it is either
empty, or else R\U_{i_1} is finite. It is most certainly not empty
(can you see why?), so we know that there are only a finite number of
real numbers which are not in U{i_1}.

How many elements can be in X but not in U_{i_1}? Well, at most a
finite number! (Can you explain why?). So suppose that x_2, ..., x_m
are in X but not in U_{i_1}. Since X is contained in the union of ALL
U_i, there must be some U_{i_2} such that x_2 is in U_{i_2}. So
U_{i_1}\/U_{i_2} covers more of X than U_{i_1} above.

Can you see how to continue and finish the argument to come up with a
finite subcover of X?

--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
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Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

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