Is it false that the complement of an open set is closed?
$begingroup$
Let $f:mathbb{C}rightarrow mathbb{C}$ be a continuous function.
Let $Z(f)$ be the zero of $f$.
Prove that $Z(f)$ is closed.
This is one of problems in my mid-term exam. I have used $epsilon-delta$ argument and showed that the complement of $Z(f)$ is open. Hence, $Z(f)$ is closed.
The problem is my professor didn't marked my answer sheet, but a teaching assistant marked it and he marked it wrong. He said, "there is a counter example that a complement of an open set is not closed".
I have no words. As far as i know, that is the definition of closed sets. Am i wrong?
general-topology definition
edited May 15 '14 at 2:54
Jack M
18.7k33880
asked May 15 '14 at 2:48
John. pJohn. p
684513
$endgroup$
|
show 5 more comments
$begingroup$
Let $f:mathbb{C}rightarrow mathbb{C}$ be a continuous function.
Let $Z(f)$ be the zero of $f$.
Prove that $Z(f)$ is closed.
This is one of problems in my mid-term exam. I have used $epsilon-delta$ argument and showed that the complement of $Z(f)$ is open. Hence, $Z(f)$ is closed.
The problem is my professor didn't marked my answer sheet, but a teaching assistant marked it and he marked it wrong. He said, "there is a counter example that a complement of an open set is not closed".
I have no words. As far as i know, that is the definition of closed sets. Am i wrong?
general-topology definition
edited May 15 '14 at 2:54
Jack M
18.7k33880
asked May 15 '14 at 2:48
John. pJohn. p
684513
$endgroup$
1
$begingroup$
Yes, a set is closed iff its complement is open.
$endgroup$
– André Nicolas
May 15 '14 at 2:52
$begingroup$
I edited your question title and tags to something that I think better reflects the real heart of your question. Feel free to revert the edit if you disagree.
$endgroup$
– Jack M
May 15 '14 at 2:54
8
$begingroup$
I believe that the teaching assistant is wrong. You should (politely and calmly!) speak to your professor about this.
$endgroup$
– David
May 15 '14 at 2:56
2
$begingroup$
There are other possible definitions of a closed set. For example, I think Rudin's analysis textbook defines a closed set as one which contains all its limit points. But in all cases, the fact that the complement of an open set is closed should be either a definition or a theorem (or else someone is using very unusual terminology).
$endgroup$
– Nate Eldredge
May 15 '14 at 2:59
1
$begingroup$
A confusion in some students arise when they learn that "$A$ is open" and "$A$ is closed" are not contradictory, and hence the implication "$A$ is open $iff$ $A$ is not closed" is false. One should not confuse that with "$A$ is open $iff$ $A_c$ is closed", which is true.
$endgroup$
– leonbloy
May 15 '14 at 3:37
|
show 5 more comments
$begingroup$
Let $f:mathbb{C}rightarrow mathbb{C}$ be a continuous function.
Let $Z(f)$ be the zero of $f$.
Prove that $Z(f)$ is closed.
This is one of problems in my mid-term exam. I have used $epsilon-delta$ argument and showed that the complement of $Z(f)$ is open. Hence, $Z(f)$ is closed.
The problem is my professor didn't marked my answer sheet, but a teaching assistant marked it and he marked it wrong. He said, "there is a counter example that a complement of an open set is not closed".
I have no words. As far as i know, that is the definition of closed sets. Am i wrong?
general-topology definition
edited May 15 '14 at 2:54
Jack M
18.7k33880
asked May 15 '14 at 2:48
John. pJohn. p
684513
$endgroup$
Let $f:mathbb{C}rightarrow mathbb{C}$ be a continuous function.
Let $Z(f)$ be the zero of $f$.
Prove that $Z(f)$ is closed.
This is one of problems in my mid-term exam. I have used $epsilon-delta$ argument and showed that the complement of $Z(f)$ is open. Hence, $Z(f)$ is closed.
The problem is my professor didn't marked my answer sheet, but a teaching assistant marked it and he marked it wrong. He said, "there is a counter example that a complement of an open set is not closed".
I have no words. As far as i know, that is the definition of closed sets. Am i wrong?
general-topology definition
general-topology definition
edited May 15 '14 at 2:54
Jack M
18.7k33880
asked May 15 '14 at 2:48
John. pJohn. p
684513
edited May 15 '14 at 2:54
Jack M
18.7k33880
asked May 15 '14 at 2:48
John. pJohn. p
684513
edited May 15 '14 at 2:54
Jack M
18.7k33880
edited May 15 '14 at 2:54
Jack M
18.7k33880
edited May 15 '14 at 2:54
Jack M
18.7k33880
18.7k33880
asked May 15 '14 at 2:48
John. pJohn. p
684513
asked May 15 '14 at 2:48
John. pJohn. p
684513
asked May 15 '14 at 2:48
John. pJohn. p
684513
684513
1
$begingroup$
Yes, a set is closed iff its complement is open.
$endgroup$
– André Nicolas
May 15 '14 at 2:52
$begingroup$
I edited your question title and tags to something that I think better reflects the real heart of your question. Feel free to revert the edit if you disagree.
$endgroup$
– Jack M
May 15 '14 at 2:54
8
$begingroup$
I believe that the teaching assistant is wrong. You should (politely and calmly!) speak to your professor about this.
$endgroup$
– David
May 15 '14 at 2:56
2
$begingroup$
There are other possible definitions of a closed set. For example, I think Rudin's analysis textbook defines a closed set as one which contains all its limit points. But in all cases, the fact that the complement of an open set is closed should be either a definition or a theorem (or else someone is using very unusual terminology).
$endgroup$
– Nate Eldredge
May 15 '14 at 2:59
1
$begingroup$
A confusion in some students arise when they learn that "$A$ is open" and "$A$ is closed" are not contradictory, and hence the implication "$A$ is open $iff$ $A$ is not closed" is false. One should not confuse that with "$A$ is open $iff$ $A_c$ is closed", which is true.
$endgroup$
– leonbloy
May 15 '14 at 3:37
|
show 5 more comments
1
$begingroup$
Yes, a set is closed iff its complement is open.
$endgroup$
– André Nicolas
May 15 '14 at 2:52
$begingroup$
I edited your question title and tags to something that I think better reflects the real heart of your question. Feel free to revert the edit if you disagree.
$endgroup$
– Jack M
May 15 '14 at 2:54
8
$begingroup$
I believe that the teaching assistant is wrong. You should (politely and calmly!) speak to your professor about this.
$endgroup$
– David
May 15 '14 at 2:56
2
$begingroup$
There are other possible definitions of a closed set. For example, I think Rudin's analysis textbook defines a closed set as one which contains all its limit points. But in all cases, the fact that the complement of an open set is closed should be either a definition or a theorem (or else someone is using very unusual terminology).
$endgroup$
– Nate Eldredge
May 15 '14 at 2:59
1
$begingroup$
A confusion in some students arise when they learn that "$A$ is open" and "$A$ is closed" are not contradictory, and hence the implication "$A$ is open $iff$ $A$ is not closed" is false. One should not confuse that with "$A$ is open $iff$ $A_c$ is closed", which is true.
$endgroup$
– leonbloy
May 15 '14 at 3:37
1
1
$begingroup$
Yes, a set is closed iff its complement is open.
$endgroup$
– André Nicolas
May 15 '14 at 2:52
$begingroup$
Yes, a set is closed iff its complement is open.
$endgroup$
– André Nicolas
May 15 '14 at 2:52
$begingroup$
I edited your question title and tags to something that I think better reflects the real heart of your question. Feel free to revert the edit if you disagree.
$endgroup$
– Jack M
May 15 '14 at 2:54
$begingroup$
I edited your question title and tags to something that I think better reflects the real heart of your question. Feel free to revert the edit if you disagree.
$endgroup$
– Jack M
May 15 '14 at 2:54
8
8
$begingroup$
I believe that the teaching assistant is wrong. You should (politely and calmly!) speak to your professor about this.
$endgroup$
– David
May 15 '14 at 2:56
$begingroup$
I believe that the teaching assistant is wrong. You should (politely and calmly!) speak to your professor about this.
$endgroup$
– David
May 15 '14 at 2:56
2
2
$begingroup$
There are other possible definitions of a closed set. For example, I think Rudin's analysis textbook defines a closed set as one which contains all its limit points. But in all cases, the fact that the complement of an open set is closed should be either a definition or a theorem (or else someone is using very unusual terminology).
$endgroup$
– Nate Eldredge
May 15 '14 at 2:59
$begingroup$
There are other possible definitions of a closed set. For example, I think Rudin's analysis textbook defines a closed set as one which contains all its limit points. But in all cases, the fact that the complement of an open set is closed should be either a definition or a theorem (or else someone is using very unusual terminology).
$endgroup$
– Nate Eldredge
May 15 '14 at 2:59
1
1
$begingroup$
A confusion in some students arise when they learn that "$A$ is open" and "$A$ is closed" are not contradictory, and hence the implication "$A$ is open $iff$ $A$ is not closed" is false. One should not confuse that with "$A$ is open $iff$ $A_c$ is closed", which is true.
$endgroup$
– leonbloy
May 15 '14 at 3:37
$begingroup$
A confusion in some students arise when they learn that "$A$ is open" and "$A$ is closed" are not contradictory, and hence the implication "$A$ is open $iff$ $A$ is not closed" is false. One should not confuse that with "$A$ is open $iff$ $A_c$ is closed", which is true.
$endgroup$
– leonbloy
May 15 '14 at 3:37
|
show 5 more comments
1 Answer
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$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
The complement of an open set is closed. This is one definition of being closed.
Alternatively, notice that $mathbb C$ is Hausdorff and so singletons are closed, in particular ${0}$ is closed. One definition of a function being continuous is that inverse images of closed sets are closed, thus $Z(f) = f^{-1}({0})$ is closed.
$endgroup$
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$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
The complement of an open set is closed. This is one definition of being closed.
Alternatively, notice that $mathbb C$ is Hausdorff and so singletons are closed, in particular ${0}$ is closed. One definition of a function being continuous is that inverse images of closed sets are closed, thus $Z(f) = f^{-1}({0})$ is closed.
$endgroup$
add a comment |
$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
The complement of an open set is closed. This is one definition of being closed.
Alternatively, notice that $mathbb C$ is Hausdorff and so singletons are closed, in particular ${0}$ is closed. One definition of a function being continuous is that inverse images of closed sets are closed, thus $Z(f) = f^{-1}({0})$ is closed.
$endgroup$
add a comment |
$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
The complement of an open set is closed. This is one definition of being closed.
Alternatively, notice that $mathbb C$ is Hausdorff and so singletons are closed, in particular ${0}$ is closed. One definition of a function being continuous is that inverse images of closed sets are closed, thus $Z(f) = f^{-1}({0})$ is closed.
$endgroup$
This community wiki solution is intended to clear the question from the unanswered queue.
The complement of an open set is closed. This is one definition of being closed.
Alternatively, notice that $mathbb C$ is Hausdorff and so singletons are closed, in particular ${0}$ is closed. One definition of a function being continuous is that inverse images of closed sets are closed, thus $Z(f) = f^{-1}({0})$ is closed.
answered Dec 10 '18 at 21:03
community wiki
Robert Cardona
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1
$begingroup$
Yes, a set is closed iff its complement is open.
$endgroup$
– André Nicolas
May 15 '14 at 2:52
$begingroup$
I edited your question title and tags to something that I think better reflects the real heart of your question. Feel free to revert the edit if you disagree.
$endgroup$
– Jack M
May 15 '14 at 2:54
8
$begingroup$
I believe that the teaching assistant is wrong. You should (politely and calmly!) speak to your professor about this.
$endgroup$
– David
May 15 '14 at 2:56
2
$begingroup$
There are other possible definitions of a closed set. For example, I think Rudin's analysis textbook defines a closed set as one which contains all its limit points. But in all cases, the fact that the complement of an open set is closed should be either a definition or a theorem (or else someone is using very unusual terminology).
$endgroup$
– Nate Eldredge
May 15 '14 at 2:59
1
$begingroup$
A confusion in some students arise when they learn that "$A$ is open" and "$A$ is closed" are not contradictory, and hence the implication "$A$ is open $iff$ $A$ is not closed" is false. One should not confuse that with "$A$ is open $iff$ $A_c$ is closed", which is true.
$endgroup$
– leonbloy
May 15 '14 at 3:37