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mina_world science forum Guru Wannabe
Joined: 20 Jul 2005
Posts: 186

Posted: Wed Jul 19, 2006 4:55 am Post subject:
topology with connected.



hello sir~
if A is a connected subset of X,
int(A), bd(A) is connected ?

um... i think..
if A = (1,2)U[2,3) in usual topology of R.
A is connected.
and
int(A) = (1,2)U(2,3) is not connected.
bd(A) = {1,2,3} is not connected.
is this right ? 

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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Wed Jul 19, 2006 6:09 am Post subject:
Re: topology with connected.



On Wed, 19 Jul 2006, mina_world wrote:
Quote:  if A is a connected subset of X,
int(A), bd(A) is connected ?
if A = (1,2)U[2,3) in usual topology of R.
A is connected.
and
int(A) = (1,2)U(2,3) is not connected.
bd(A) = {1,2,3} is not connected.
is this right ?
No. A = (1,3) = int A 
bd A = { 1,3 }
Find a planar example for connected A, disconnected int A 

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The World Wide Wade science forum Guru
Joined: 24 Mar 2005
Posts: 790

Posted: Wed Jul 19, 2006 6:26 am Post subject:
Re: topology with connected.



In article <e9kds2$4is$1@news2.kornet.net>,
"mina_world" <mina_world@hanmail.net> wrote:
Quote:  hello sir~
if A is a connected subset of X,
int(A), bd(A) is connected ?

um... i think..
if A = (1,2)U[2,3) in usual topology of R.
A is connected.
and
int(A) = (1,2)U(2,3) is not connected.
bd(A) = {1,2,3} is not connected.
is this right ?

No. A = (1,3). So int(A) = A. bd(A) = {1,3}, so that's a step in
the right direction. Note that if A is a connected subset of R,
then int(A) is always connected. Proof: A is an interval.
To find a connected A such that int(A) is not connected, think
about subsets of R^2. (Try to find one so that both int(A) and
bd(A) are not connected.) 

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mina_world science forum Guru Wannabe
Joined: 20 Jul 2005
Posts: 186

Posted: Wed Jul 19, 2006 7:09 am Post subject:
Re: topology with connected.



"The World Wide Wade" <waderameyxiii@comcast.remove13.net> wrote in message
news:waderameyxiii7C1A4B.23264818072006@comcast.dca.giganews.com...
Quote:  In article <e9kds2$4is$1@news2.kornet.net>,
"mina_world" <mina_world@hanmail.net> wrote:
hello sir~
if A is a connected subset of X,
int(A), bd(A) is connected ?

um... i think..
if A = (1,2)U[2,3) in usual topology of R.
A is connected.
and
int(A) = (1,2)U(2,3) is not connected.
bd(A) = {1,2,3} is not connected.
is this right ?
No. A = (1,3). So int(A) = A. bd(A) = {1,3}, so that's a step in
the right direction. Note that if A is a connected subset of R,
then int(A) is always connected. Proof: A is an interval.
To find a connected A such that int(A) is not connected, think
about subsets of R^2. (Try to find one so that both int(A) and
bd(A) are not connected.)

um... i think...
X = {a,b,c,d,e}
T = {X,empty,{a,b,c},{a,b},{c}}
A = {a,b,c,d} is connected.
int(A) = {a,b,c} is not connected.
in the case R^2,
A = {R^2  (Q*{0})}
int A = R^2  (R*{0})} is not connected.
but i can't find the example with both int(A) and bd(A). 

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toni.lassila@gmail.com science forum beginner
Joined: 18 May 2006
Posts: 2

Posted: Wed Jul 19, 2006 9:44 am Post subject:
Re: topology with connected.



mina_world wrote:
Quote:  "The World Wide Wade" <waderameyxiii@comcast.remove13.net> wrote in message
news:waderameyxiii7C1A4B.23264818072006@comcast.dca.giganews.com...
In article <e9kds2$4is$1@news2.kornet.net>,
"mina_world" <mina_world@hanmail.net> wrote:
hello sir~
if A is a connected subset of X,
int(A), bd(A) is connected ?

um... i think..
if A = (1,2)U[2,3) in usual topology of R.
A is connected.
and
int(A) = (1,2)U(2,3) is not connected.
bd(A) = {1,2,3} is not connected.
is this right ?
No. A = (1,3). So int(A) = A. bd(A) = {1,3}, so that's a step in
the right direction. Note that if A is a connected subset of R,
then int(A) is always connected. Proof: A is an interval.
To find a connected A such that int(A) is not connected, think
about subsets of R^2. (Try to find one so that both int(A) and
bd(A) are not connected.)
um... i think...
X = {a,b,c,d,e}
T = {X,empty,{a,b,c},{a,b},{c}}
A = {a,b,c,d} is connected.
int(A) = {a,b,c} is not connected.
in the case R^2,
A = {R^2  (Q*{0})}
int A = R^2  (R*{0})} is not connected.
but i can't find the example with both int(A) and bd(A).

Take two disjoint infinite closed strips { (x,y) in R^2 : a <= x <= b,
y in R } and join them with a line segment in such a way that they
connect only in a boundary point. 

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mina_world science forum Guru Wannabe
Joined: 20 Jul 2005
Posts: 186

Posted: Wed Jul 19, 2006 10:06 am Post subject:
Re: topology with connected.



toni.lassila@gmail.com ìž‘ì„±:
Quote:  mina_world wrote:
"The World Wide Wade" <waderameyxiii@comcast.remove13.net> wrote in message
news:waderameyxiii7C1A4B.23264818072006@comcast.dca.giganews.com...
In article <e9kds2$4is$1@news2.kornet.net>,
"mina_world" <mina_world@hanmail.net> wrote:
hello sir~
if A is a connected subset of X,
int(A), bd(A) is connected ?

um... i think..
if A = (1,2)U[2,3) in usual topology of R.
A is connected.
and
int(A) = (1,2)U(2,3) is not connected.
bd(A) = {1,2,3} is not connected.
is this right ?
No. A = (1,3). So int(A) = A. bd(A) = {1,3}, so that's a step in
the right direction. Note that if A is a connected subset of R,
then int(A) is always connected. Proof: A is an interval.
To find a connected A such that int(A) is not connected, think
about subsets of R^2. (Try to find one so that both int(A) and
bd(A) are not connected.)
um... i think...
X = {a,b,c,d,e}
T = {X,empty,{a,b,c},{a,b},{c}}
A = {a,b,c,d} is connected.
int(A) = {a,b,c} is not connected.
in the case R^2,
A = {R^2  (Q*{0})}
int A = R^2  (R*{0})} is not connected.
but i can't find the example with both int(A) and bd(A).
Take two disjoint infinite closed strips { (x,y) in R^2 : a <= x <= b,
y in R } and join them with a line segment in such a way that they
connect only in a boundary point.

yes, maybe this is a form of   
thank you very much. 

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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Wed Jul 19, 2006 11:02 am Post subject:
Re: topology with connected.



On Wed, 19 Jul 2006 mina_world@hanmail.net wrote:
Quote:  toni.lassila@gmail.com ìž‘ì„±:
mina_world wrote:
"The World Wide Wade" <waderameyxiii@comcast.remove13.net> wrote in message
In article <e9kds2$4is$1@news2.kornet.net>,
if A is a connected subset of X,
int(A), bd(A) is connected ?
To find a connected A such that int(A) is not connected, think
about subsets of R^2. (Try to find one so that both int(A) and
bd(A) are not connected.)
but i can't find the example with both int(A) and bd(A).
Take two disjoint infinite closed strips { (x,y) in R^2 : a <= x <= b,
y in R } and join them with a line segment in such a way that they
connect only in a boundary point.
yes, maybe this is a form of   
Two closed tangent disks with an interior point removed from one. 


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Arturo Magidin science forum Guru
Joined: 25 Mar 2005
Posts: 1838

Posted: Wed Jul 19, 2006 4:12 pm Post subject:
Re: topology with connected.



In article <e9kds2$4is$1@news2.kornet.net>,
mina_world <mina_world@hanmail.net> wrote:
Quote:  hello sir~
if A is a connected subset of X,
int(A), bd(A) is connected ?

The interior, no. Take two tangent discs on the plane with the usual
R^2 topology; the interior is two disjoint open balls.
The boundary, no. Take a closed interval in R with the usual topology;
the boundary is two isolated points.
Quote:  
um... i think..
if A = (1,2)U[2,3) in usual topology of R.
A is connected.
and
int(A) = (1,2)U(2,3) is not connected.

This is wrong. If you take A=(1,2) U [2,3), then you are taken
A=(1,3). So the interior is A itself.
Quote:  bd(A) = {1,2,3} is not connected.

No, the boundary is {1,3}; 2 is an interior point. It is not
connected, however.
No on many counts, though two wrongs did make a right on the last
part.

======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
 Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin 

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mina_world science forum Guru Wannabe
Joined: 20 Jul 2005
Posts: 186

Posted: Thu Jul 20, 2006 5:06 am Post subject:
Re: topology with connected.



"William Elliot" <marsh@hevanet.remove.com> wrote in message
news:Pine.BSI.4.58.0607190359330.12173@vista.hevanet.com...
On Wed, 19 Jul 2006 mina_world@hanmail.net wrote:
Quote:  toni.lassila@gmail.com ÀÛ¼º:
mina_world wrote:
"The World Wide Wade" <waderameyxiii@comcast.remove13.net> wrote in
message
In article <e9kds2$4is$1@news2.kornet.net>,
if A is a connected subset of X,
int(A), bd(A) is connected ?
To find a connected A such that int(A) is not connected, think
about subsets of R^2. (Try to find one so that both int(A) and
bd(A) are not connected.)
but i can't find the example with both int(A) and bd(A).
Take two disjoint infinite closed strips { (x,y) in R^2 : a <= x <= b,
y in R } and join them with a line segment in such a way that they
connect only in a boundary point.
yes, maybe this is a form of   
Two closed tangent disks with an interior point removed from one. 
sorry. since i had a weak english, i can't understand it.
http://board2.blueweb.co.kr/user/math565/data/math/tangent.jpg
i need your advice one more. 

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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Thu Jul 20, 2006 10:38 am Post subject:
Re: topology with connected.



On Thu, 20 Jul 2006, mina_world wrote:
Find connected S with int S disconnected, bd S disconnected
S = ({ (x,y)  (x  1)^2 + y^2 <= 1 } \/
{ (x,y)  (x + 1)^2 + y^2 <= 1 })  { (1,0) }
Similar example.
S = { (x,y)  0 <= xy }  { (1,1) } 

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mina_world science forum Guru Wannabe
Joined: 20 Jul 2005
Posts: 186

Posted: Thu Jul 20, 2006 12:00 pm Post subject:
Re: topology with connected.



William Elliot ìž‘ì„±:
Quote:  On Thu, 20 Jul 2006, mina_world wrote:
Find connected S with int S disconnected, bd S disconnected
Two closed tangent disks with an interior point removed from one.
sorry. since i had a weak english, i can't understand it.
http://board2.blueweb.co.kr/user/math565/data/math/tangent.jpg
i need your advice one more.
S = ({ (x,y)  (x  1)^2 + y^2 <= 1 } \/
{ (x,y)  (x + 1)^2 + y^2 <= 1 })  { (1,0) }
Similar example.
S = { (x,y)  0 <= xy }  { (1,1) }

yes, i understanded. thank you very much. 

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