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From: José Carlos Santos on 20 Jun 2010 12:36
On 20-06-2010 16:18, dushya wrote:

>>>>> suppose a space X be such that closure of every open set U (=! X) in X
>>>>> is compact. does it imply that the space X is compact ?
>>
>>>> If X contains at least one point _p_ such that the set {p} is closed
>>>> then, yes, it is true. Simply write:
>>
>>>> X = {p} U (X\{p}).
>>
>>>> Then
>>
>>>> X = (closure of {p}) U (closure of (X\{p}))
>>
>>>> = {p} U (closure of (X\{p})).
>>
>>>> It is the union of two compact sets, and therefore it is compact.
>>
>>> and we can replace {p} with any nonempty open set not equal to X.
>>
>> Why? Let A be such a set. Then X is the union of its closure and its
>> complement and the closure of A is compact. But X\A doesn't have to be
>> (or, at least, I don't see why it should be).
>
> i don't know but please let me know if anything is wrong with
> following argument --
>
> B=clr(A) is compact.(and we can assume that it is not equal to X fot
> if it is then X is compact)
> => X- clr(A) is open and is niether empty nor equal to X.
> => C=clr(X-clr(A))is compact.
> And X is union of B and C.
> so X is compact.

Yes, it is right. :-)

Best regards,

Jose Carlos Santos
From: William Elliot on 21 Jun 2010 01:36
On Sun, 20 Jun 2010, dushya wrote:

> suppose a space X be such that closure of every open set U (=! X) in X
> is compact. does it imply that the space X is compact ?

U = X - cl {a} for some a in X is a proper subset of X.
U is an open set and cl U = cl int X\a = X - int cl {a}.
Now show that int cl {a} is empty, so cl U = X.

BTW, In English and many other languages, the
beginning of a sentence is always capitalized.

What language is your native language that
doesn't capitalize the beginning of sentences?








From: dushya on 21 Jun 2010 02:37
On Jun 20, 10:36 pm, William Elliot <ma...(a)rdrop.remove.com> wrote:
> On Sun, 20 Jun 2010, dushya wrote:
> > suppose a space X be such that closure of every open set U (=! X) in X
> > is compact. does it imply that the space X is compact ?
>
> U = X - cl {a} for some a in X is a proper subset of X.
> U is an open set and cl U = cl int X\a = X - int cl {a}.
> Now show that int cl {a} is empty, so cl U = X.
>
> BTW, In English and many other languages, the
> beginning of a sentence is always capitalized.
>
> What language is your native language that
> doesn't capitalize the beginning of sentences?

Thanks :-)
Sorry for not using capital initial letter. I am not very good at
english and i am weak in topology too :-)

From: cwldoc on 21 Jun 2010 14:37
> On Sun, 20 Jun 2010, dushya wrote:
>
> > suppose a space X be such that closure of every
> open set U (=! X) in X
> > is compact. does it imply that the space X is
> compact ?
>
> U = X - cl {a} for some a in X is a proper subset of
> X.
> U is an open set and cl U = cl int X\a = X - int cl
> {a}.
> Now show that int cl {a} is empty, so cl U = X.

Do you have any hints about how to show that int cl {a} is empty?

Just because all closures of proper open subsets of X are compact does not guarantee that int cl {a} is empty.
(For example, take X = [0, 1] union {2} as a subspace of R. Then {a} is open and int cl {2} = {2}.)

If I can prove that int cl {a} is empty by making the additional assumption (for the sake of proof by contradiction) that X is noncompact, then this would lead to the contradiction that X is compact. But unfortunately I have not been successful in showing this.

>
> BTW, In English and many other languages, the
> beginning of a sentence is always capitalized.
>
> What language is your native language that
> doesn't capitalize the beginning of sentences?
>
>
>
>
>
>
>
>
From: William Elliot on 22 Jun 2010 01:01
On Mon, 21 Jun 2010, cwldoc wrote:
>> On Sun, 20 Jun 2010, dushya wrote:
>>
>>> suppose a space X be such that closure of every open set U (=! X) in X
>>> is compact. does it imply that the space X is compact ?
>>
>> U = X - cl {a} for some a in X is a proper subset of X.
>> U is an open set and cl U = cl int X\a = X - int cl {a}.
>> Now show that int cl {a} is empty, so cl U = X.
>
> Do you have any hints about how to show that int cl {a} is empty?
>
No. I've overstated my hand. The best answer was given in the
other subthread. Let cl {a} or nonnul cl A be compact. Then
K = cl (X - cl {a}) is compact by hypothesis and subsequently
X = K \/ cl {a} is compact.

> Just because all closures of proper open subsets of X are compact does
> not guarantee that int cl {a} is empty.

> (For example, take X = [0, 1] union {2} as a subspace of R. Then {a} is
> open and int cl {2} = {2}.)

> If I can prove that int cl {a} is empty by making the additional

One needs to assume {a} is not open. Still that's not enought.

> assumption (for the sake of proof by contradiction) that X is
> noncompact, then this would lead to the contradiction that X is compact.
> But unfortunately I have not been successful in showing this.

----
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