Mixing
Can we prove that a point is either in the interior, exterior or boundary of a set?
$begingroup$
So, I am reading Michael Spivak’s book Calculus on Manifolds and when he starts speaking of sets, he states something like this:
If we take an arbitrary set $A$ and an arbitrary point $x$, then either one of three has to be true:
- There exists an open set $B$ such that $xin Bsubset A$
- There exists an open set $B$ such that $xin B subset overline A$
- For all open sets $B$ such that $xin B$, $B$ contains elements of $A$ and $overline A$
My question is: Can we prove this?
Thank you in advance.
real-analysis elementary-set-theory
asked Jan 26 at 13:07
DaàvidDaàvid
505
$endgroup$
add a comment |
$begingroup$
So, I am reading Michael Spivak’s book Calculus on Manifolds and when he starts speaking of sets, he states something like this:
If we take an arbitrary set $A$ and an arbitrary point $x$, then either one of three has to be true:
- There exists an open set $B$ such that $xin Bsubset A$
- There exists an open set $B$ such that $xin B subset overline A$
- For all open sets $B$ such that $xin B$, $B$ contains elements of $A$ and $overline A$
My question is: Can we prove this?
Thank you in advance.
real-analysis elementary-set-theory
asked Jan 26 at 13:07
DaàvidDaàvid
505
$endgroup$
add a comment |
$begingroup$
So, I am reading Michael Spivak’s book Calculus on Manifolds and when he starts speaking of sets, he states something like this:
If we take an arbitrary set $A$ and an arbitrary point $x$, then either one of three has to be true:
- There exists an open set $B$ such that $xin Bsubset A$
- There exists an open set $B$ such that $xin B subset overline A$
- For all open sets $B$ such that $xin B$, $B$ contains elements of $A$ and $overline A$
My question is: Can we prove this?
Thank you in advance.
real-analysis elementary-set-theory
asked Jan 26 at 13:07
DaàvidDaàvid
505
$endgroup$
So, I am reading Michael Spivak’s book Calculus on Manifolds and when he starts speaking of sets, he states something like this:
If we take an arbitrary set $A$ and an arbitrary point $x$, then either one of three has to be true:
- There exists an open set $B$ such that $xin Bsubset A$
- There exists an open set $B$ such that $xin B subset overline A$
- For all open sets $B$ such that $xin B$, $B$ contains elements of $A$ and $overline A$
My question is: Can we prove this?
Thank you in advance.
real-analysis elementary-set-theory
real-analysis elementary-set-theory
asked Jan 26 at 13:07
DaàvidDaàvid
505
asked Jan 26 at 13:07
DaàvidDaàvid
505
asked Jan 26 at 13:07
DaàvidDaàvid
505
asked Jan 26 at 13:07
DaàvidDaàvid
505
asked Jan 26 at 13:07
DaàvidDaàvid
505
505
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Call the three conditions a, b, c.
For any set $A$, either c holds, and we're done, or c does not hold.
In the latter case (c does not hold), there's some open set $B$ containing $x$ that either contains only elements of $A$ (which is "a") or contains only elements of $bar{A}$ (which is case "b").
answered Jan 26 at 13:14
John HughesJohn Hughes
64.8k24191
$endgroup$
$begingroup$
Thank you for the answer. But how do we even know that there are points satisfying c?
$endgroup$
– Daàvid
Jan 26 at 13:20
$begingroup$
We don't know that, and Spivak doesn't claim that, which is a good thing, because it's false. If, for instance, we're working in $Bbb R$, and $A = Bbb R$, then there are no points satisfying either b or c, but the statement that every point satisfies exactly one of the three is still true: they all satisfy the first one!
$endgroup$
– John Hughes
Jan 26 at 13:22
$begingroup$
Hmm ok, I got it. Thank you
$endgroup$
– Daàvid
Jan 26 at 14:18
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Call the three conditions a, b, c.
For any set $A$, either c holds, and we're done, or c does not hold.
In the latter case (c does not hold), there's some open set $B$ containing $x$ that either contains only elements of $A$ (which is "a") or contains only elements of $bar{A}$ (which is case "b").
answered Jan 26 at 13:14
John HughesJohn Hughes
64.8k24191
$endgroup$
$begingroup$
Thank you for the answer. But how do we even know that there are points satisfying c?
$endgroup$
– Daàvid
Jan 26 at 13:20
$begingroup$
We don't know that, and Spivak doesn't claim that, which is a good thing, because it's false. If, for instance, we're working in $Bbb R$, and $A = Bbb R$, then there are no points satisfying either b or c, but the statement that every point satisfies exactly one of the three is still true: they all satisfy the first one!
$endgroup$
– John Hughes
Jan 26 at 13:22
$begingroup$
Hmm ok, I got it. Thank you
$endgroup$
– Daàvid
Jan 26 at 14:18
add a comment |
$begingroup$
Call the three conditions a, b, c.
For any set $A$, either c holds, and we're done, or c does not hold.
In the latter case (c does not hold), there's some open set $B$ containing $x$ that either contains only elements of $A$ (which is "a") or contains only elements of $bar{A}$ (which is case "b").
answered Jan 26 at 13:14
John HughesJohn Hughes
64.8k24191
$endgroup$
$begingroup$
Thank you for the answer. But how do we even know that there are points satisfying c?
$endgroup$
– Daàvid
Jan 26 at 13:20
$begingroup$
We don't know that, and Spivak doesn't claim that, which is a good thing, because it's false. If, for instance, we're working in $Bbb R$, and $A = Bbb R$, then there are no points satisfying either b or c, but the statement that every point satisfies exactly one of the three is still true: they all satisfy the first one!
$endgroup$
– John Hughes
Jan 26 at 13:22
$begingroup$
Hmm ok, I got it. Thank you
$endgroup$
– Daàvid
Jan 26 at 14:18
add a comment |
$begingroup$
Call the three conditions a, b, c.
For any set $A$, either c holds, and we're done, or c does not hold.
In the latter case (c does not hold), there's some open set $B$ containing $x$ that either contains only elements of $A$ (which is "a") or contains only elements of $bar{A}$ (which is case "b").
answered Jan 26 at 13:14
John HughesJohn Hughes
64.8k24191
$endgroup$
Call the three conditions a, b, c.
For any set $A$, either c holds, and we're done, or c does not hold.
In the latter case (c does not hold), there's some open set $B$ containing $x$ that either contains only elements of $A$ (which is "a") or contains only elements of $bar{A}$ (which is case "b").
answered Jan 26 at 13:14
John HughesJohn Hughes
64.8k24191
answered Jan 26 at 13:14
John HughesJohn Hughes
64.8k24191
answered Jan 26 at 13:14
John HughesJohn Hughes
64.8k24191
answered Jan 26 at 13:14
John HughesJohn Hughes
64.8k24191
64.8k24191
$begingroup$
Thank you for the answer. But how do we even know that there are points satisfying c?
$endgroup$
– Daàvid
Jan 26 at 13:20
$begingroup$
We don't know that, and Spivak doesn't claim that, which is a good thing, because it's false. If, for instance, we're working in $Bbb R$, and $A = Bbb R$, then there are no points satisfying either b or c, but the statement that every point satisfies exactly one of the three is still true: they all satisfy the first one!
$endgroup$
– John Hughes
Jan 26 at 13:22
$begingroup$
Hmm ok, I got it. Thank you
$endgroup$
– Daàvid
Jan 26 at 14:18
add a comment |
$begingroup$
Thank you for the answer. But how do we even know that there are points satisfying c?
$endgroup$
– Daàvid
Jan 26 at 13:20
$begingroup$
We don't know that, and Spivak doesn't claim that, which is a good thing, because it's false. If, for instance, we're working in $Bbb R$, and $A = Bbb R$, then there are no points satisfying either b or c, but the statement that every point satisfies exactly one of the three is still true: they all satisfy the first one!
$endgroup$
– John Hughes
Jan 26 at 13:22
$begingroup$
Hmm ok, I got it. Thank you
$endgroup$
– Daàvid
Jan 26 at 14:18
$begingroup$
Thank you for the answer. But how do we even know that there are points satisfying c?
$endgroup$
– Daàvid
Jan 26 at 13:20
$begingroup$
Thank you for the answer. But how do we even know that there are points satisfying c?
$endgroup$
– Daàvid
Jan 26 at 13:20
$begingroup$
We don't know that, and Spivak doesn't claim that, which is a good thing, because it's false. If, for instance, we're working in $Bbb R$, and $A = Bbb R$, then there are no points satisfying either b or c, but the statement that every point satisfies exactly one of the three is still true: they all satisfy the first one!
$endgroup$
– John Hughes
Jan 26 at 13:22
$begingroup$
We don't know that, and Spivak doesn't claim that, which is a good thing, because it's false. If, for instance, we're working in $Bbb R$, and $A = Bbb R$, then there are no points satisfying either b or c, but the statement that every point satisfies exactly one of the three is still true: they all satisfy the first one!
$endgroup$
– John Hughes
Jan 26 at 13:22
$begingroup$
Hmm ok, I got it. Thank you
$endgroup$
– Daàvid
Jan 26 at 14:18
$begingroup$
Hmm ok, I got it. Thank you
$endgroup$
– Daàvid
Jan 26 at 14:18
add a comment |
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