compact set's definition
oercim
definition( A subset of a topological space is compact if for every
open cover of there exists a finite subcover of S ) . If you help me
to understand it, I will be very glad. If you help me with an example,
I will be very glad. Thanks alot.
José Carlos Santos
Example of a non-compact set: the reals. Consider, for instance the set
of all intervals of the form (-a,a) (a > 0). They are all open and their
union is R itself; so, they form an open cover. However, if you take a
finite set of them, then its union is again an interval of the form
(-a,a), which is not R. This proves that R is not compact.
Example of a compact set: any finite set F. If you have an open cover of
such a set, then, for each _x_ in F, take an element of the cover to
which _x_ belongs. These elements form a finite sub-cover.
You might be interested in knowing that, for metric spaces, there is an
equivalent property: K is compact if and only if every sequence of
elements of K has a convergent subsequence.
I hope that this helps.
Best regards,
Jose Carlos Santos
Aatu Koskensilta
> Example of a compact set: any finite set F. If you have an open
> cover of such a set, then, for each _x_ in F, take an element of the
> cover to which _x_ belongs. These elements form a finite sub-cover.
Perhaps more helpfully we may also observe that many arguments that go
through for finite sets generalise with minimum fuss to compact sets.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
José Carlos Santos
>> Example of a compact set: any finite set F. If you have an open
>> cover of such a set, then, for each _x_ in F, take an element of the
>> cover to which _x_ belongs. These elements form a finite sub-cover.
>
> Perhaps more helpfully we may also observe that many arguments that go
> through for finite sets generalise with minimum fuss to compact sets.
Plus the fact that many arguments that go through for finite sets and
arbitrary functions generalize with minimum fuss to compact sets and
continuous functions.
Aatu Koskensilta
Gc
In R^n set is compact iff it is closed and bounded.
http://en.wikipedia.org/wiki/Heine-Borel_theorem
Gc
I have thought that this was the condition for weaker precompactness.
http://en.wikipedia.org/wiki/Relatively_compact_subspace
Gc
But of course you said that convergence happens in K, so never mind.
Jonathan Groves
K is compact if every open cover of K contains a finite subcover.
Here is what this definition is saying: Suppose I cover the set K with a collection of open sets (that means K is a subset of the union of all the open sets; or equivalently, that means every point in K belongs to at least one of these open sets). If it is possible to throw away all but finitely many of these open sets and still have an open cover of K (so every point of K belongs to one of these sets in the finite collection), regardless of what open cover I begin with, the set K is compact.
The first one to reply gave an excellent explanation of why R, the reals, is not compact.
Let us see from this definition why the open interval (0,1) is not compact. Let A_n=
(1/n, 1) where n=2,3,.... By choosing n sufficiently large, we can make the set (1/n,1) cover any point x in (0,1). So {A_n} is an open cover of (0,1). But, if we choose only finitely many of these sets A_n, the union of these sets, since they are nested (A_2 is contained in A_3, A_3 is contained in A_4, A_4 is contained in A_5, and so on), is some set A_m (m being the largest index of the sets in the finite subcollection). But A_m does not cover all of (0,1), so this open cover of (0,1) does not contain a finite subcover. Hence, (0,1) is not compact.
Dave L. Renfro
> Here is what the definition of compact set says:
>
> K is compact if every open cover of K contains a finite subcover.
A terrific historical survey of the historical development of
compactness is given in the paper below by Hildebrandt. This
paper also shows quite well how zoo-like the early history
of general topology was before our present terminology and
concepts were standardized.
T. H. Hildebrandt, "The Borel theorem and its generalizations",
Bulletin of the American Mathematical Society 32 (1926), 423-474.
<http://tinyurl.com/bw8f64> [from AMS pages; .pdf file]
<http://tinyurl.com/c8el3z> [from projecteuclid; .pdf file]
A more recent historical survey that is less technical than
Hildebrandt's paper is:
Manya Janaky Raman, "Understanding Compactness: A Historical
Perspective", M.A. Thesis (University of California at
Berkeley), Fall 1997, 39 pages.
http://www.matnv.umu.se/personal/untitled/manya_compactness.pdf
I've mentioned Raman's Thesis at least twice before in sci.math
(24 January 2003 and 15 June 2006), but the last time I looked
(probably 15 June 2006, since I mentioned this in my post and
don't think I've looked since), Raman's Thesis was no longer
available on the internet. Apparently that was temporary, since
it now seems to be available again (although at a different URL
than it used to be at).
Dave L. Renfro
FredJeffries
> A terrific historical survey of the historical development of
> compactness is given in the paper below by Hildebrandt. This
> paper also shows quite well how zoo-like the early history
> of general topology was before our present terminology and
> concepts were standardized.
>
> T. H. Hildebrandt, "The Borel theorem and its generalizations",
> Bulletin of the American Mathematical Society 32 (1926), 423-474.
> <http://tinyurl.com/bw8f64> [from AMS pages; .pdf file]
> <http://tinyurl.com/c8el3z> [from projecteuclid; .pdf file]
>
> A more recent historical survey that is less technical than
> Hildebrandt's paper is:
>
> Manya Janaky Raman, "Understanding Compactness: A Historical
> Perspective", M.A. Thesis (University of California at
>
> I've mentioned Raman's Thesis at least twice before in sci.math
> (24 January 2003 and 15 June 2006), but the last time I looked
> (probably 15 June 2006, since I mentioned this in my post and
> don't think I've looked since), Raman's Thesis was no longer
> available on the internet. Apparently that was temporary, since
> it now seems to be available again (although at a different URL
> than it used to be at).
>
> Dave L. Renfro
Dave, I'm going to have to buy another hard drive just to hold the
links to all of these great papers you post. When am I supposed to get
time to read them all? -- some of us have to work for a living, you
know.
Seriously, many thanks for the great references!
oercim
for the papers. They are great papers for me. Especially Raman's
thesis is a very good reference for my level. Again thanks a lot
rip pelletier
<b980d8c5-a8d4-46df...@g38g2000yqd.googlegroups.com>,
"Dave L. Renfro" <renf...@cmich.edu> wrote:
> Jonathan Groves wrote (in part):
>
> > Here is what the definition of compact set says:
> >
> > K is compact if every open cover of K contains a finite subcover.
>
> A terrific historical survey of the historical development of
> compactness is given in the paper below by Hildebrandt.
....
>
> T. H. Hildebrandt, "The Borel theorem and its generalizations",
> Bulletin of the American Mathematical Society 32 (1926), 423-474.
> <http://tinyurl.com/bw8f64> [from AMS pages; .pdf file]
> <http://tinyurl.com/c8el3z> [from projecteuclid; .pdf file]
>
> A more recent historical survey that is less technical than
> Hildebrandt's paper is:
>
> Manya Janaky Raman, "Understanding Compactness: A Historical
> Perspective", M.A. Thesis (University of California at
> Berkeley), Fall 1997, 39 pages.
> http://www.matnv.umu.se/personal/untitled/manya_compactness.pdf
>
....
> Dave L. Renfro
they both look interesting. thank you. and no, i'm not the OP.
vale,
rip
--
NB eddress is r i p 1 AT c o m c a s t DOT n e t
marc
The first exercise is to prove that a complete bounded set is compact.
Gc
> Intuitively compact sets are metric spaces that are complete and bounded. The idea of compactness originated with subsets of R_n.
>
> The first exercise is to prove that a complete bounded set is compact.
In R^n it is, but of course not in general. Compact set is always
complete and bounded in metric space, but in every infinite
dimensional Banach space Unit balls are not compact even if complete
and bounded.
oercim
statement in page 17: “Let En=[n,∞). Each En is closed becouse it
contains its limit points, and clearly E(n+1) is subset of En. But the
infinite intersection of these intervals is empty, so R is compact”.
I am confused with why the intersection of the intervals is
empty. I have problem with the infinitess concept.
Jonathan Groves
If R is the set of real numbers, then I think you meant to write,
"R is not compact."
I take it that n is any integer, positive or negative or zero.
Am I correct? I will assume so. To see that the intersection of
the intervals is infinite, pick any real number x. Choose n so that n>x.
Then the interval E_n=[n,\infinity)
does not contain x. But x is arbitrary, so the intersection
is empty (because there is no real number that belongs
to ALL these intervals).
A space is compact iff every collection of closed sets satisfying the
finite intersection property has nonempty intersection. To say that
a collection of sets having the finite intersection property means
that any finite subcollection of these sets is nonempty.
The intervals E_n have the finite intersection property:
Take a finite subcollection,
say E_{n_1},...,E_{n_m} where n_1<n_2<...<n_m.
Then any real number x>n_m belongs to all these intervals,
so the intersection of this finite subcollection is nonempty.
As I showed above, the intersection of all
these intervals is empty.
Thus, the set of real numbers R is not compact.
If it is not clear why this proves R is not compact, go back to the
finite intersection property of compactness. Since the statement is "iff,"
we may say that a space is not compact iff there is a collection of
closed sets having the finite intersection property whose intersection
is empty. The collection {E_n} is such a collection for R.
Zdislav V. Kovarik
Counterexample (not in R_n, of course): an infinite discrete space (e.g.
the integers) with zero-or-one metric. So easy...
Cheers, ZVK(Slavek).