Mixing
Prove that complement of the interior equals closure of the complement
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I tried as follows:
If x is not in the interior of set E, then no neighborhood of x is contained in E. So for every neighborhood of x there is a point y which is not in E. That is, every neighborhood of x contains a point y which lies in complement of E. So x is a limit point of complement of E.
-
On the other hand, if y is in closure of complement of E, then y is in E complement or set of all limit points of it. In the first case we are done since Int(E) is contained in E.
In the second case, y is the limit point of complement of E. So every neighborhood of y contains a point which is not in E. Hence no neighborhood of y completely lies in E. Thus y is not an interior point of E.
Is this correct??
Thanks in advance.
general-topology
asked Jan 12 at 4:24
USKUSK
335
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add a comment |
$begingroup$
I tried as follows:
If x is not in the interior of set E, then no neighborhood of x is contained in E. So for every neighborhood of x there is a point y which is not in E. That is, every neighborhood of x contains a point y which lies in complement of E. So x is a limit point of complement of E.
-
On the other hand, if y is in closure of complement of E, then y is in E complement or set of all limit points of it. In the first case we are done since Int(E) is contained in E.
In the second case, y is the limit point of complement of E. So every neighborhood of y contains a point which is not in E. Hence no neighborhood of y completely lies in E. Thus y is not an interior point of E.
Is this correct??
Thanks in advance.
general-topology
asked Jan 12 at 4:24
USKUSK
335
$endgroup$
1
$begingroup$
Your proof is kind of hard to read. It seems like it could use some paragraph breaks or something.
$endgroup$
– Chris Custer
Jan 12 at 4:32
1
$begingroup$
I think it's correct.
$endgroup$
– Chris Custer
Jan 12 at 4:36
add a comment |
$begingroup$
I tried as follows:
If x is not in the interior of set E, then no neighborhood of x is contained in E. So for every neighborhood of x there is a point y which is not in E. That is, every neighborhood of x contains a point y which lies in complement of E. So x is a limit point of complement of E.
-
On the other hand, if y is in closure of complement of E, then y is in E complement or set of all limit points of it. In the first case we are done since Int(E) is contained in E.
In the second case, y is the limit point of complement of E. So every neighborhood of y contains a point which is not in E. Hence no neighborhood of y completely lies in E. Thus y is not an interior point of E.
Is this correct??
Thanks in advance.
general-topology
asked Jan 12 at 4:24
USKUSK
335
$endgroup$
I tried as follows:
If x is not in the interior of set E, then no neighborhood of x is contained in E. So for every neighborhood of x there is a point y which is not in E. That is, every neighborhood of x contains a point y which lies in complement of E. So x is a limit point of complement of E.
-
On the other hand, if y is in closure of complement of E, then y is in E complement or set of all limit points of it. In the first case we are done since Int(E) is contained in E.
In the second case, y is the limit point of complement of E. So every neighborhood of y contains a point which is not in E. Hence no neighborhood of y completely lies in E. Thus y is not an interior point of E.
Is this correct??
Thanks in advance.
general-topology
general-topology
asked Jan 12 at 4:24
USKUSK
335
asked Jan 12 at 4:24
USKUSK
335
edited Jan 12 at 19:17
USK
asked Jan 12 at 4:24
USKUSK
335
asked Jan 12 at 4:24
USKUSK
335
asked Jan 12 at 4:24
USKUSK
335
335
1
$begingroup$
Your proof is kind of hard to read. It seems like it could use some paragraph breaks or something.
$endgroup$
– Chris Custer
Jan 12 at 4:32
1
$begingroup$
I think it's correct.
$endgroup$
– Chris Custer
Jan 12 at 4:36
add a comment |
1
$begingroup$
Your proof is kind of hard to read. It seems like it could use some paragraph breaks or something.
$endgroup$
– Chris Custer
Jan 12 at 4:32
1
$begingroup$
I think it's correct.
$endgroup$
– Chris Custer
Jan 12 at 4:36
1
1
$begingroup$
Your proof is kind of hard to read. It seems like it could use some paragraph breaks or something.
$endgroup$
– Chris Custer
Jan 12 at 4:32
$begingroup$
Your proof is kind of hard to read. It seems like it could use some paragraph breaks or something.
$endgroup$
– Chris Custer
Jan 12 at 4:32
1
1
$begingroup$
I think it's correct.
$endgroup$
– Chris Custer
Jan 12 at 4:36
$begingroup$
I think it's correct.
$endgroup$
– Chris Custer
Jan 12 at 4:36
add a comment |
1 Answer
1
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$begingroup$
Let X be discrete, E = X - {a}. Assume x in interior E.
In your 1st proof, you conclude x is a limit point of the
complement of E, namely {a} which has no limit points.
That example shows your proof is wrong.
You do however, have the right idea.
Use the right definition of closure:
the set of all adherence points.
a is an adherance point of A when for all
open U nhood a, U $cap$ A is not empty.
answered Jan 12 at 6:35
William ElliotWilliam Elliot
8,0712720
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1 Answer
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1 Answer
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$begingroup$
Let X be discrete, E = X - {a}. Assume x in interior E.
In your 1st proof, you conclude x is a limit point of the
complement of E, namely {a} which has no limit points.
That example shows your proof is wrong.
You do however, have the right idea.
Use the right definition of closure:
the set of all adherence points.
a is an adherance point of A when for all
open U nhood a, U $cap$ A is not empty.
answered Jan 12 at 6:35
William ElliotWilliam Elliot
8,0712720
$endgroup$
add a comment |
$begingroup$
Let X be discrete, E = X - {a}. Assume x in interior E.
In your 1st proof, you conclude x is a limit point of the
complement of E, namely {a} which has no limit points.
That example shows your proof is wrong.
You do however, have the right idea.
Use the right definition of closure:
the set of all adherence points.
a is an adherance point of A when for all
open U nhood a, U $cap$ A is not empty.
answered Jan 12 at 6:35
William ElliotWilliam Elliot
8,0712720
$endgroup$
add a comment |
$begingroup$
Let X be discrete, E = X - {a}. Assume x in interior E.
In your 1st proof, you conclude x is a limit point of the
complement of E, namely {a} which has no limit points.
That example shows your proof is wrong.
You do however, have the right idea.
Use the right definition of closure:
the set of all adherence points.
a is an adherance point of A when for all
open U nhood a, U $cap$ A is not empty.
answered Jan 12 at 6:35
William ElliotWilliam Elliot
8,0712720
$endgroup$
Let X be discrete, E = X - {a}. Assume x in interior E.
In your 1st proof, you conclude x is a limit point of the
complement of E, namely {a} which has no limit points.
That example shows your proof is wrong.
You do however, have the right idea.
Use the right definition of closure:
the set of all adherence points.
a is an adherance point of A when for all
open U nhood a, U $cap$ A is not empty.
answered Jan 12 at 6:35
William ElliotWilliam Elliot
8,0712720
answered Jan 12 at 6:35
William ElliotWilliam Elliot
8,0712720
answered Jan 12 at 6:35
William ElliotWilliam Elliot
8,0712720
answered Jan 12 at 6:35
William ElliotWilliam Elliot
8,0712720
8,0712720
add a comment |
add a comment |
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1
$begingroup$
Your proof is kind of hard to read. It seems like it could use some paragraph breaks or something.
$endgroup$
– Chris Custer
Jan 12 at 4:32
1
$begingroup$
I think it's correct.
$endgroup$
– Chris Custer
Jan 12 at 4:36