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

In article <dnc23e$7fi$1@xxxxxxxxxxxxxxxxxx>, Bob <Bob@xxxxxxxxxx> wrote:
>i decided to start the problem from scratch
>
>co-fiinite topology on R.
>this means UcR is open iff U =empty set or R-U is finite
>
>WTS want to show R is compact (with this topology)

I thought you wanted to show taht ANY subset of R is compact with this
topology, but fine. Let's start by showing that R itself is compact in
this topology.

>Let V be an open cover,
>look at any non empty subset U. Complement of U is finite.
>Complement can be written {x_1, x_2,.....x_n}
>pick any U_x_i from V containing x_i 'sand these together with U cover R (I
>think this is all correct)

I would prefer if you be more precise in your wording, but yes: for
each i, we know that there is a U_{x_i} in V which contains x_i; so
now we have U, U_{x_1}, ..., and U_{x_n}.

Now, you ->think<- the union of them is all of R. Why can't you
->prove<- that the union is of all R?

Well, the union contains U, contains x_1 (since x_1 is in U_1),
contains x_2 (because it is in U_2), ..., and contains x_n. Since U
union {x_1,...,x_n} is equal to R, the union contains R, and therefore
is equal to R.

>Now moving on to showing that ANY subset of R is compact
>
>If X is empty, then pick any U open set from the open cover V (as stated in
>the beginning V is an open cover of R)

No, no, no, no, no, no, no, no, no.

Let X be a subset of R.

Let V be an open cover ->of X<-, NOT of R. You want to show that any
open cover ->of X<- has a finite subcover.

But yes, since V is an open cover for X, if X is empty then we can
pick any U in V, and U will cover X.


>X is empty, =>empty set element of U (as the empty set is subset of every
>set)

No.

The empty set is not an element of any U. It is a SUBSET.

>this is the part of the proof im struggling with, im unsure what i have to
>show.

> To show compactness i have to show FOR ALL families of open covers,

of X,

>there always exists finite subcover

of that family.

> As i have assumed V is any open cover,

You didn't. You assumed V was an open of R. But assume it is an open
cover of X.

>and U is any open set in V, is it enough just to state X is
>contained in any U. as U is one set, this means it is a subcover
>which is finite.

Yes. So the empty set is compact (this is true in ANY topological
space).

> Now X is non empty.

Now, we ->assume<- X is nonempty.

>and x is an element of X. this
>means there exists at least 1 U_i s.t x element U_i, will call this
>U_{i_1} As this is open (U's are a collection of OPEN subsets, so
>U_{i_1} is obviously open, it cannot be empty. as it contains the
>element x, this means that R\U_{i_1} is finite) As complement of
>U_{i_1} is finite, this means there are only a finite number of real
>numbers that are not in U_{i_1}, that is also the maximum number of
>elements that X can have that are not in U_{i_1} So suppose
>{x_2,....x_m} do not lie in U_i, but in X,

you mean, in U_{i_1}.

>But we know there exist
>{U_i} which are are a collection of open sets which cover R,

No, they cover X.

>so there
>must exist U_{i_2}s.t x_2 is an element of it...

Yes, because they cover X.

>carry on this
>argument until x_m Now U_i_1\/U_i_2......\/U_i_m cover X (and more
>importantly are finite)

Now, the union is not finite. The collection U_{i_1},..., U_{i_n} is
finite, and their union cover X.

You should PROVE they cover X. If you were not sure before that you
had a cover for R, you are probably not sure now that you have a cover
for X. So you should make sure to PROVE that X is indeed contained in
that union.

>U_i_1\/U_i_2......\/U_i_m are a finite
>collection of the {U_i}

No, this is not a finite collection, it is a union.

>(which cover R)

which cover X.

> so U_i_1\/U_i_2......\/U_i_m
>is a finite subcover.

Techincally, it is the collections of sets {U_{i_1},....,U_{i_n}}
which is the finite subcover, NOT the union of them.

>i do see the general way this proof works, it
>starts with the fact is that i dont have control over the open cover
>which covers R,

Of X.

> but i do have control over the subcover, and if i can
>show (as i hope i have) that this subcover is finite, then it shows
>it is compact. The only thing that worries me is the empty set
>part. i think the answer is staring me in the face!

Yes. Any one open set of an open cover for the empty st is a finite
subcover.


--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

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