Isn't this proof of a theorem about the closedness of a set wrong?
$begingroup$
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 Jan 8 at 23:21
Asaf Karagila♦
302k32429760
asked Jan 8 at 22:41
cppcodercppcoder
1484
$endgroup$
add a comment |
$begingroup$
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 Jan 8 at 23:21
Asaf Karagila♦
302k32429760
asked Jan 8 at 22:41
cppcodercppcoder
1484
$endgroup$
5
$begingroup$
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.
$endgroup$
– angryavian
Jan 8 at 22:43
1
$begingroup$
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
$endgroup$
– copper.hat
Jan 8 at 22:47
2
$begingroup$
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.
$endgroup$
– fleablood
Jan 8 at 22:48
1
$begingroup$
As one of my professors used to say: “sets are not doors.”
$endgroup$
– Jim
Jan 8 at 23:11
add a comment |
$begingroup$
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 Jan 8 at 23:21
Asaf Karagila♦
302k32429760
asked Jan 8 at 22:41
cppcodercppcoder
1484
$endgroup$
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 Jan 8 at 23:21
Asaf Karagila♦
302k32429760
asked Jan 8 at 22:41
cppcodercppcoder
1484
edited Jan 8 at 23:21
Asaf Karagila♦
302k32429760
asked Jan 8 at 22:41
cppcodercppcoder
1484
edited Jan 8 at 23:21
Asaf Karagila♦
302k32429760
edited Jan 8 at 23:21
Asaf Karagila♦
302k32429760
edited Jan 8 at 23:21
Asaf Karagila♦
302k32429760
302k32429760
asked Jan 8 at 22:41
cppcodercppcoder
1484
asked Jan 8 at 22:41
cppcodercppcoder
1484
asked Jan 8 at 22:41
cppcodercppcoder
1484
1484
5
$begingroup$
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.
$endgroup$
– angryavian
Jan 8 at 22:43
1
$begingroup$
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
$endgroup$
– copper.hat
Jan 8 at 22:47
2
$begingroup$
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.
$endgroup$
– fleablood
Jan 8 at 22:48
1
$begingroup$
As one of my professors used to say: “sets are not doors.”
$endgroup$
– Jim
Jan 8 at 23:11
add a comment |
5
$begingroup$
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.
$endgroup$
– angryavian
Jan 8 at 22:43
1
$begingroup$
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
$endgroup$
– copper.hat
Jan 8 at 22:47
2
$begingroup$
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.
$endgroup$
– fleablood
Jan 8 at 22:48
1
$begingroup$
As one of my professors used to say: “sets are not doors.”
$endgroup$
– Jim
Jan 8 at 23:11
5
5
$begingroup$
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.
$endgroup$
– angryavian
Jan 8 at 22:43
$begingroup$
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.
$endgroup$
– angryavian
Jan 8 at 22:43
1
1
$begingroup$
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
$endgroup$
– copper.hat
Jan 8 at 22:47
$begingroup$
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
$endgroup$
– copper.hat
Jan 8 at 22:47
2
2
$begingroup$
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.
$endgroup$
– fleablood
Jan 8 at 22:48
$begingroup$
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.
$endgroup$
– fleablood
Jan 8 at 22:48
1
1
$begingroup$
As one of my professors used to say: “sets are not doors.”
$endgroup$
– Jim
Jan 8 at 23:11
$begingroup$
As one of my professors used to say: “sets are not doors.”
$endgroup$
– Jim
Jan 8 at 23:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 Jan 8 at 22:47
ArthurArthur
112k7109191
$endgroup$
$begingroup$
Yes, you're right. I'm such an idiot. Thank you!
$endgroup$
– cppcoder
Jan 8 at 22:57
add a comment |
$begingroup$
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 Jan 8 at 23:02
fleabloodfleablood
68.9k22685
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
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 Jan 8 at 22:47
ArthurArthur
112k7109191
$endgroup$
$begingroup$
Yes, you're right. I'm such an idiot. Thank you!
$endgroup$
– cppcoder
Jan 8 at 22:57
add a comment |
$begingroup$
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 Jan 8 at 22:47
ArthurArthur
112k7109191
$endgroup$
$begingroup$
Yes, you're right. I'm such an idiot. Thank you!
$endgroup$
– cppcoder
Jan 8 at 22:57
add a comment |
$begingroup$
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 Jan 8 at 22:47
ArthurArthur
112k7109191
$endgroup$
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 Jan 8 at 22:47
ArthurArthur
112k7109191
answered Jan 8 at 22:47
ArthurArthur
112k7109191
answered Jan 8 at 22:47
ArthurArthur
112k7109191
answered Jan 8 at 22:47
ArthurArthur
112k7109191
112k7109191
$begingroup$
Yes, you're right. I'm such an idiot. Thank you!
$endgroup$
– cppcoder
Jan 8 at 22:57
add a comment |
$begingroup$
Yes, you're right. I'm such an idiot. Thank you!
$endgroup$
– cppcoder
Jan 8 at 22:57
$begingroup$
Yes, you're right. I'm such an idiot. Thank you!
$endgroup$
– cppcoder
Jan 8 at 22:57
$begingroup$
Yes, you're right. I'm such an idiot. Thank you!
$endgroup$
– cppcoder
Jan 8 at 22:57
add a comment |
$begingroup$
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 Jan 8 at 23:02
fleabloodfleablood
68.9k22685
$endgroup$
add a comment |
$begingroup$
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 Jan 8 at 23:02
fleabloodfleablood
68.9k22685
$endgroup$
add a comment |
$begingroup$
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 Jan 8 at 23:02
fleabloodfleablood
68.9k22685
$endgroup$
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 Jan 8 at 23:02
fleabloodfleablood
68.9k22685
answered Jan 8 at 23:02
fleabloodfleablood
68.9k22685
answered Jan 8 at 23:02
fleabloodfleablood
68.9k22685
answered Jan 8 at 23:02
fleabloodfleablood
68.9k22685
68.9k22685
add a comment |
add a comment |
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$begingroup$
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.
$endgroup$
– angryavian
Jan 8 at 22:43
1
$begingroup$
To show that $P,Q$ are equivalent, you can prove $P implies Q$ and $lnot P implies lnot Q$.
$endgroup$
– copper.hat
Jan 8 at 22:47
2
$begingroup$
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.
$endgroup$
– fleablood
Jan 8 at 22:48
1
$begingroup$
As one of my professors used to say: “sets are not doors.”
$endgroup$
– Jim
Jan 8 at 23:11