Isn't this proof of a theorem about the closedness of a set wrong?
I was reading a proof of the following theorem in my textbook:
A set $A$ is closed iff $A' subseteq A$.
Proof: Suppose $A$ is closed and $x in A'$. If $x notin A$, then $xin A^c$, an open set. Thus $mathcal{N}(x, delta)subseteq A^c$ for some positive $delta$. But then $mathcal{N}(x, delta)$ can contain no points of $A$. Thus $x$ is not an accumulation point of $A$ and so $xnotin A'$, a contradiction. We conclude that $xin A$. Therefore $A'subseteq A$.
Now suppose $A'subseteq A$. To show that $A$ is closed, we show $A^c$ is open. If $A^c$ is not open, there is $xin A^c$ that is not an interior point of $A^c$. Therefore, no $delta$-neighborhood of $x$ is a subset of $A^c$; that is, each $delta$-neighborhood of $x$ contains a point of $A$. This point must be different from $x$, since $xin A^c$. Thus $xin A'$. But $A'subseteq A$, so $xin A$. This is a contradiction. We conclude that $A$ is closed.
Somewhere along the proof, the complement of $A$ is proven to be open. In order to achieve this, the author first assumes that it's not open and produces a contradiction. But isn't that wrong since it's possible for a set to be neither open nor closed?
general-topology
edited yesterday
Asaf Karagila♦
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asked yesterday
cppcodercppcoder
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I was reading a proof of the following theorem in my textbook:
A set $A$ is closed iff $A' subseteq A$.
Proof: Suppose $A$ is closed and $x in A'$. If $x notin A$, then $xin A^c$, an open set. Thus $mathcal{N}(x, delta)subseteq A^c$ for some positive $delta$. But then $mathcal{N}(x, delta)$ can contain no points of $A$. Thus $x$ is not an accumulation point of $A$ and so $xnotin A'$, a contradiction. We conclude that $xin A$. Therefore $A'subseteq A$.
Now suppose $A'subseteq A$. To show that $A$ is closed, we show $A^c$ is open. If $A^c$ is not open, there is $xin A^c$ that is not an interior point of $A^c$. Therefore, no $delta$-neighborhood of $x$ is a subset of $A^c$; that is, each $delta$-neighborhood of $x$ contains a point of $A$. This point must be different from $x$, since $xin A^c$. Thus $xin A'$. But $A'subseteq A$, so $xin A$. This is a contradiction. We conclude that $A$ is closed.
Somewhere along the proof, the complement of $A$ is proven to be open. In order to achieve this, the author first assumes that it's not open and produces a contradiction. But isn't that wrong since it's possible for a set to be neither open nor closed?
general-topology
edited yesterday
Asaf Karagila♦
302k32427757
asked yesterday
cppcodercppcoder
1484
New contributor
cppcoder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
What exactly is wrong? The proof starts with the assumption that $A^c$ is not open; it is not assuming that $A^c$ is closed.
– angryavian
yesterday
1
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
– copper.hat
yesterday
2
He makes absolutely no assumptions about whether $A^c$ is closed. He only assume $A^c$ is not open. He gets a contradiction which means..... It is false that $A^c$ is not open. That means $A^c$ is open. That's all.
– fleablood
yesterday
1
As one of my professors used to say: “sets are not doors.”
– Jim
yesterday
add a comment |
I was reading a proof of the following theorem in my textbook:
A set $A$ is closed iff $A' subseteq A$.
Proof: Suppose $A$ is closed and $x in A'$. If $x notin A$, then $xin A^c$, an open set. Thus $mathcal{N}(x, delta)subseteq A^c$ for some positive $delta$. But then $mathcal{N}(x, delta)$ can contain no points of $A$. Thus $x$ is not an accumulation point of $A$ and so $xnotin A'$, a contradiction. We conclude that $xin A$. Therefore $A'subseteq A$.
Now suppose $A'subseteq A$. To show that $A$ is closed, we show $A^c$ is open. If $A^c$ is not open, there is $xin A^c$ that is not an interior point of $A^c$. Therefore, no $delta$-neighborhood of $x$ is a subset of $A^c$; that is, each $delta$-neighborhood of $x$ contains a point of $A$. This point must be different from $x$, since $xin A^c$. Thus $xin A'$. But $A'subseteq A$, so $xin A$. This is a contradiction. We conclude that $A$ is closed.
Somewhere along the proof, the complement of $A$ is proven to be open. In order to achieve this, the author first assumes that it's not open and produces a contradiction. But isn't that wrong since it's possible for a set to be neither open nor closed?
general-topology
edited yesterday
Asaf Karagila♦
302k32427757
asked yesterday
cppcodercppcoder
1484
New contributor
cppcoder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I was reading a proof of the following theorem in my textbook:
A set $A$ is closed iff $A' subseteq A$.
Proof: Suppose $A$ is closed and $x in A'$. If $x notin A$, then $xin A^c$, an open set. Thus $mathcal{N}(x, delta)subseteq A^c$ for some positive $delta$. But then $mathcal{N}(x, delta)$ can contain no points of $A$. Thus $x$ is not an accumulation point of $A$ and so $xnotin A'$, a contradiction. We conclude that $xin A$. Therefore $A'subseteq A$.
Now suppose $A'subseteq A$. To show that $A$ is closed, we show $A^c$ is open. If $A^c$ is not open, there is $xin A^c$ that is not an interior point of $A^c$. Therefore, no $delta$-neighborhood of $x$ is a subset of $A^c$; that is, each $delta$-neighborhood of $x$ contains a point of $A$. This point must be different from $x$, since $xin A^c$. Thus $xin A'$. But $A'subseteq A$, so $xin A$. This is a contradiction. We conclude that $A$ is closed.
Somewhere along the proof, the complement of $A$ is proven to be open. In order to achieve this, the author first assumes that it's not open and produces a contradiction. But isn't that wrong since it's possible for a set to be neither open nor closed?
general-topology
general-topology
edited yesterday
Asaf Karagila♦
302k32427757
asked yesterday
cppcodercppcoder
1484
New contributor
cppcoder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Asaf Karagila♦
302k32427757
asked yesterday
cppcodercppcoder
1484
New contributor
cppcoder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Asaf Karagila♦
302k32427757
edited yesterday
Asaf Karagila♦
302k32427757
edited yesterday
Asaf Karagila♦
302k32427757
302k32427757
asked yesterday
cppcodercppcoder
1484
New contributor
cppcoder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
cppcodercppcoder
1484
asked yesterday
cppcodercppcoder
1484
1484
New contributor
cppcoder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
cppcoder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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5
What exactly is wrong? The proof starts with the assumption that $A^c$ is not open; it is not assuming that $A^c$ is closed.
– angryavian
yesterday
1
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
– copper.hat
yesterday
2
He makes absolutely no assumptions about whether $A^c$ is closed. He only assume $A^c$ is not open. He gets a contradiction which means..... It is false that $A^c$ is not open. That means $A^c$ is open. That's all.
– fleablood
yesterday
1
As one of my professors used to say: “sets are not doors.”
– Jim
yesterday
add a comment |
5
What exactly is wrong? The proof starts with the assumption that $A^c$ is not open; it is not assuming that $A^c$ is closed.
– angryavian
yesterday
1
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
– copper.hat
yesterday
2
He makes absolutely no assumptions about whether $A^c$ is closed. He only assume $A^c$ is not open. He gets a contradiction which means..... It is false that $A^c$ is not open. That means $A^c$ is open. That's all.
– fleablood
yesterday
1
As one of my professors used to say: “sets are not doors.”
– Jim
yesterday
5
5
What exactly is wrong? The proof starts with the assumption that $A^c$ is not open; it is not assuming that $A^c$ is closed.
– angryavian
yesterday
What exactly is wrong? The proof starts with the assumption that $A^c$ is not open; it is not assuming that $A^c$ is closed.
– angryavian
yesterday
1
1
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
– copper.hat
yesterday
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
– copper.hat
yesterday
2
2
He makes absolutely no assumptions about whether $A^c$ is closed. He only assume $A^c$ is not open. He gets a contradiction which means..... It is false that $A^c$ is not open. That means $A^c$ is open. That's all.
– fleablood
yesterday
He makes absolutely no assumptions about whether $A^c$ is closed. He only assume $A^c$ is not open. He gets a contradiction which means..... It is false that $A^c$ is not open. That means $A^c$ is open. That's all.
– fleablood
yesterday
1
1
As one of my professors used to say: “sets are not doors.”
– Jim
yesterday
As one of my professors used to say: “sets are not doors.”
– Jim
yesterday
add a comment |
2 Answers
2
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oldest
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No problems here. Note that the author doesn't assume $A^c$ to be closed. Just that it isn't open. And since "open" means "for any point in the set, we have a certain property", we get that "not open" means "for at least one point in the set, we don't have that property". The point $x$ they choose is one such point guaranteed to exist by that assumption.
answered yesterday
ArthurArthur
111k7107189
Yes, you're right. I'm such an idiot. Thank you!
– cppcoder
yesterday
add a comment |
There are four possiblities
1) $A^c$ is open and closed.
2) $A^c$ is open and not closed.
3) $A^c$ is not open and closed.
4) $A^c$ is not open and not closed.
He assumed: $A^c$ is not open. That means either $3$ or $4$. He got a contradiction. That shows that both $3$ and $4$ are both false. Thus the only possibilities are $1$ or $2$. So he concludes either $1$ or $2$ is true. Which means.... $A^c$ is open.
However as he didn't give a carnivore's fig whether $A^c$ was closed or not he made no mention of it at all.
So it comes down to:
I) $A^c$ is open. or
II) $A^c$ is not open.
He assumes II) gets a contradiction, and concludes I). $A^c$ is open. Now maybe $A^c$ is closed. Or maybe it isn't. He doesn't care and made no assumptions to the fact.
But he does know, and correctly so. That $A^c$ is open....
answered yesterday
fleabloodfleablood
68.7k22685
add a comment |
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2 Answers
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No problems here. Note that the author doesn't assume $A^c$ to be closed. Just that it isn't open. And since "open" means "for any point in the set, we have a certain property", we get that "not open" means "for at least one point in the set, we don't have that property". The point $x$ they choose is one such point guaranteed to exist by that assumption.
answered yesterday
ArthurArthur
111k7107189
Yes, you're right. I'm such an idiot. Thank you!
– cppcoder
yesterday
add a comment |
No problems here. Note that the author doesn't assume $A^c$ to be closed. Just that it isn't open. And since "open" means "for any point in the set, we have a certain property", we get that "not open" means "for at least one point in the set, we don't have that property". The point $x$ they choose is one such point guaranteed to exist by that assumption.
answered yesterday
ArthurArthur
111k7107189
Yes, you're right. I'm such an idiot. Thank you!
– cppcoder
yesterday
add a comment |
No problems here. Note that the author doesn't assume $A^c$ to be closed. Just that it isn't open. And since "open" means "for any point in the set, we have a certain property", we get that "not open" means "for at least one point in the set, we don't have that property". The point $x$ they choose is one such point guaranteed to exist by that assumption.
answered yesterday
ArthurArthur
111k7107189
No problems here. Note that the author doesn't assume $A^c$ to be closed. Just that it isn't open. And since "open" means "for any point in the set, we have a certain property", we get that "not open" means "for at least one point in the set, we don't have that property". The point $x$ they choose is one such point guaranteed to exist by that assumption.
answered yesterday
ArthurArthur
111k7107189
answered yesterday
ArthurArthur
111k7107189
answered yesterday
ArthurArthur
111k7107189
answered yesterday
ArthurArthur
111k7107189
111k7107189
Yes, you're right. I'm such an idiot. Thank you!
– cppcoder
yesterday
add a comment |
Yes, you're right. I'm such an idiot. Thank you!
– cppcoder
yesterday
Yes, you're right. I'm such an idiot. Thank you!
– cppcoder
yesterday
Yes, you're right. I'm such an idiot. Thank you!
– cppcoder
yesterday
add a comment |
There are four possiblities
1) $A^c$ is open and closed.
2) $A^c$ is open and not closed.
3) $A^c$ is not open and closed.
4) $A^c$ is not open and not closed.
He assumed: $A^c$ is not open. That means either $3$ or $4$. He got a contradiction. That shows that both $3$ and $4$ are both false. Thus the only possibilities are $1$ or $2$. So he concludes either $1$ or $2$ is true. Which means.... $A^c$ is open.
However as he didn't give a carnivore's fig whether $A^c$ was closed or not he made no mention of it at all.
So it comes down to:
I) $A^c$ is open. or
II) $A^c$ is not open.
He assumes II) gets a contradiction, and concludes I). $A^c$ is open. Now maybe $A^c$ is closed. Or maybe it isn't. He doesn't care and made no assumptions to the fact.
But he does know, and correctly so. That $A^c$ is open....
answered yesterday
fleabloodfleablood
68.7k22685
add a comment |
There are four possiblities
1) $A^c$ is open and closed.
2) $A^c$ is open and not closed.
3) $A^c$ is not open and closed.
4) $A^c$ is not open and not closed.
He assumed: $A^c$ is not open. That means either $3$ or $4$. He got a contradiction. That shows that both $3$ and $4$ are both false. Thus the only possibilities are $1$ or $2$. So he concludes either $1$ or $2$ is true. Which means.... $A^c$ is open.
However as he didn't give a carnivore's fig whether $A^c$ was closed or not he made no mention of it at all.
So it comes down to:
I) $A^c$ is open. or
II) $A^c$ is not open.
He assumes II) gets a contradiction, and concludes I). $A^c$ is open. Now maybe $A^c$ is closed. Or maybe it isn't. He doesn't care and made no assumptions to the fact.
But he does know, and correctly so. That $A^c$ is open....
answered yesterday
fleabloodfleablood
68.7k22685
add a comment |
There are four possiblities
1) $A^c$ is open and closed.
2) $A^c$ is open and not closed.
3) $A^c$ is not open and closed.
4) $A^c$ is not open and not closed.
He assumed: $A^c$ is not open. That means either $3$ or $4$. He got a contradiction. That shows that both $3$ and $4$ are both false. Thus the only possibilities are $1$ or $2$. So he concludes either $1$ or $2$ is true. Which means.... $A^c$ is open.
However as he didn't give a carnivore's fig whether $A^c$ was closed or not he made no mention of it at all.
So it comes down to:
I) $A^c$ is open. or
II) $A^c$ is not open.
He assumes II) gets a contradiction, and concludes I). $A^c$ is open. Now maybe $A^c$ is closed. Or maybe it isn't. He doesn't care and made no assumptions to the fact.
But he does know, and correctly so. That $A^c$ is open....
answered yesterday
fleabloodfleablood
68.7k22685
There are four possiblities
1) $A^c$ is open and closed.
2) $A^c$ is open and not closed.
3) $A^c$ is not open and closed.
4) $A^c$ is not open and not closed.
He assumed: $A^c$ is not open. That means either $3$ or $4$. He got a contradiction. That shows that both $3$ and $4$ are both false. Thus the only possibilities are $1$ or $2$. So he concludes either $1$ or $2$ is true. Which means.... $A^c$ is open.
However as he didn't give a carnivore's fig whether $A^c$ was closed or not he made no mention of it at all.
So it comes down to:
I) $A^c$ is open. or
II) $A^c$ is not open.
He assumes II) gets a contradiction, and concludes I). $A^c$ is open. Now maybe $A^c$ is closed. Or maybe it isn't. He doesn't care and made no assumptions to the fact.
But he does know, and correctly so. That $A^c$ is open....
answered yesterday
fleabloodfleablood
68.7k22685
answered yesterday
fleabloodfleablood
68.7k22685
answered yesterday
fleabloodfleablood
68.7k22685
answered yesterday
fleabloodfleablood
68.7k22685
68.7k22685
add a comment |
add a comment |
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5
What exactly is wrong? The proof starts with the assumption that $A^c$ is not open; it is not assuming that $A^c$ is closed.
– angryavian
yesterday
1
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
– copper.hat
yesterday
2
He makes absolutely no assumptions about whether $A^c$ is closed. He only assume $A^c$ is not open. He gets a contradiction which means..... It is false that $A^c$ is not open. That means $A^c$ is open. That's all.
– fleablood
yesterday
1
As one of my professors used to say: “sets are not doors.”
– Jim
yesterday