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If $C$ is a compact set and $(a,b)$ is any open interval, then $Csetminus (a,b)$ is a compact set.
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Is this proof fine or does it lack rigor? Please help me spot mistakes or improve on it. Thanks!
Proof Attempt:
Let $C$ be compact, so it must be closed and bounded due to the Heine-Borel Theorem. Consider the open interval $(a,b)$ and, w.l.o.g., suppose that $Ccap(a,b)=[a,x]$. Then $(a,x]nsubseteq Csetminus (a,b)$, but $ain Csetminus (a,b)$. For the sake of contradiction, suppose $Csetminus (a,b)$ is not closed, then $Csetminus (a,b)$ does not contain all its points of closure; that is $exists$ a point $p$ of $Csetminus (a,b)$ such that $in(a,x]$, then $a<pleq x$ which implies $existsdelta>0$ such that $(p-delta, p+delta)cap Csetminus (a,b)=emptyset$. So $forall yin(a,x]$, $y$ cannot be a point of closure of $Csetminus (a,b)$ ~ a contradiction.
EDIT
Attempt(Direct Approach): Let $U$ be an open cover of $Csetminus (a,b)$ such that $C$ is closed and bounded. Since $Ucup(a,b)$ is an open cover of $C$. But, $C$ is compact so $Ucup(a,b)$ has a finite subcover. Knowing that $(a,b)$ does not cover $Csetminus (a,b)$ then $U$ must be finite.
real-analysis proof-verification compactness
asked Jul 15 at 15:32
TheLast Cipher
538414
 |Â
show 7 more comments
up vote
2
down vote
favorite
Is this proof fine or does it lack rigor? Please help me spot mistakes or improve on it. Thanks!
Proof Attempt:
Let $C$ be compact, so it must be closed and bounded due to the Heine-Borel Theorem. Consider the open interval $(a,b)$ and, w.l.o.g., suppose that $Ccap(a,b)=[a,x]$. Then $(a,x]nsubseteq Csetminus (a,b)$, but $ain Csetminus (a,b)$. For the sake of contradiction, suppose $Csetminus (a,b)$ is not closed, then $Csetminus (a,b)$ does not contain all its points of closure; that is $exists$ a point $p$ of $Csetminus (a,b)$ such that $in(a,x]$, then $a<pleq x$ which implies $existsdelta>0$ such that $(p-delta, p+delta)cap Csetminus (a,b)=emptyset$. So $forall yin(a,x]$, $y$ cannot be a point of closure of $Csetminus (a,b)$ ~ a contradiction.
EDIT
Attempt(Direct Approach): Let $U$ be an open cover of $Csetminus (a,b)$ such that $C$ is closed and bounded. Since $Ucup(a,b)$ is an open cover of $C$. But, $C$ is compact so $Ucup(a,b)$ has a finite subcover. Knowing that $(a,b)$ does not cover $Csetminus (a,b)$ then $U$ must be finite.
real-analysis proof-verification compactness
asked Jul 15 at 15:32
TheLast Cipher
538414
1
Why can we suppose that $Ccap(a,b)$ is an interval? What if, say, $C=leftfrac1n,middle$?
– José Carlos Santos
Jul 15 at 15:36
I was thinking of considering a compact set $C$ such that $Csetminus (a,b)$ is not empty
– TheLast Cipher
Jul 15 at 15:38
1
Don't use Heine-Borel. The definition of $Csetminus(a,b)$ makes it perfectly suited for a direct argument with open covers
– Hagen von Eitzen
Jul 15 at 15:49
2
$C$ is closed (because it's compact). $Csetminus (a, b)$ is closed, it's intersection of $(a, b)^c$ and $C$, two closed sets. Now, we have a closed subset $Csetminus (a, b)$ of $C$, a compact set. So $Csetminus (a, b)$ is compact
– Rumpelstiltskin
Jul 15 at 16:10
1
@TheLastCipher As Adam's argument shows, in general removing an open set $U$ from a closed set $F$ gives a closed set, as $F setminus U = F cap U^mathsfc$ is an intersection of closed sets.
– Hayden
Jul 15 at 16:57
 |Â
show 7 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is this proof fine or does it lack rigor? Please help me spot mistakes or improve on it. Thanks!
Proof Attempt:
Let $C$ be compact, so it must be closed and bounded due to the Heine-Borel Theorem. Consider the open interval $(a,b)$ and, w.l.o.g., suppose that $Ccap(a,b)=[a,x]$. Then $(a,x]nsubseteq Csetminus (a,b)$, but $ain Csetminus (a,b)$. For the sake of contradiction, suppose $Csetminus (a,b)$ is not closed, then $Csetminus (a,b)$ does not contain all its points of closure; that is $exists$ a point $p$ of $Csetminus (a,b)$ such that $in(a,x]$, then $a<pleq x$ which implies $existsdelta>0$ such that $(p-delta, p+delta)cap Csetminus (a,b)=emptyset$. So $forall yin(a,x]$, $y$ cannot be a point of closure of $Csetminus (a,b)$ ~ a contradiction.
EDIT
Attempt(Direct Approach): Let $U$ be an open cover of $Csetminus (a,b)$ such that $C$ is closed and bounded. Since $Ucup(a,b)$ is an open cover of $C$. But, $C$ is compact so $Ucup(a,b)$ has a finite subcover. Knowing that $(a,b)$ does not cover $Csetminus (a,b)$ then $U$ must be finite.
real-analysis proof-verification compactness
asked Jul 15 at 15:32
TheLast Cipher
538414
Is this proof fine or does it lack rigor? Please help me spot mistakes or improve on it. Thanks!
Proof Attempt:
Let $C$ be compact, so it must be closed and bounded due to the Heine-Borel Theorem. Consider the open interval $(a,b)$ and, w.l.o.g., suppose that $Ccap(a,b)=[a,x]$. Then $(a,x]nsubseteq Csetminus (a,b)$, but $ain Csetminus (a,b)$. For the sake of contradiction, suppose $Csetminus (a,b)$ is not closed, then $Csetminus (a,b)$ does not contain all its points of closure; that is $exists$ a point $p$ of $Csetminus (a,b)$ such that $in(a,x]$, then $a<pleq x$ which implies $existsdelta>0$ such that $(p-delta, p+delta)cap Csetminus (a,b)=emptyset$. So $forall yin(a,x]$, $y$ cannot be a point of closure of $Csetminus (a,b)$ ~ a contradiction.
EDIT
Attempt(Direct Approach): Let $U$ be an open cover of $Csetminus (a,b)$ such that $C$ is closed and bounded. Since $Ucup(a,b)$ is an open cover of $C$. But, $C$ is compact so $Ucup(a,b)$ has a finite subcover. Knowing that $(a,b)$ does not cover $Csetminus (a,b)$ then $U$ must be finite.
real-analysis proof-verification compactness
asked Jul 15 at 15:32
TheLast Cipher
538414
edited Jul 16 at 2:23
asked Jul 15 at 15:32
TheLast Cipher
538414
asked Jul 15 at 15:32
TheLast Cipher
538414
asked Jul 15 at 15:32
TheLast Cipher
538414
538414
1
Why can we suppose that $Ccap(a,b)$ is an interval? What if, say, $C=leftfrac1n,middle$?
– José Carlos Santos
Jul 15 at 15:36
I was thinking of considering a compact set $C$ such that $Csetminus (a,b)$ is not empty
– TheLast Cipher
Jul 15 at 15:38
1
Don't use Heine-Borel. The definition of $Csetminus(a,b)$ makes it perfectly suited for a direct argument with open covers
– Hagen von Eitzen
Jul 15 at 15:49
2
$C$ is closed (because it's compact). $Csetminus (a, b)$ is closed, it's intersection of $(a, b)^c$ and $C$, two closed sets. Now, we have a closed subset $Csetminus (a, b)$ of $C$, a compact set. So $Csetminus (a, b)$ is compact
– Rumpelstiltskin
Jul 15 at 16:10
1
@TheLastCipher As Adam's argument shows, in general removing an open set $U$ from a closed set $F$ gives a closed set, as $F setminus U = F cap U^mathsfc$ is an intersection of closed sets.
– Hayden
Jul 15 at 16:57
 |Â
show 7 more comments
1
Why can we suppose that $Ccap(a,b)$ is an interval? What if, say, $C=leftfrac1n,middle$?
– José Carlos Santos
Jul 15 at 15:36
I was thinking of considering a compact set $C$ such that $Csetminus (a,b)$ is not empty
– TheLast Cipher
Jul 15 at 15:38
1
Don't use Heine-Borel. The definition of $Csetminus(a,b)$ makes it perfectly suited for a direct argument with open covers
– Hagen von Eitzen
Jul 15 at 15:49
2
$C$ is closed (because it's compact). $Csetminus (a, b)$ is closed, it's intersection of $(a, b)^c$ and $C$, two closed sets. Now, we have a closed subset $Csetminus (a, b)$ of $C$, a compact set. So $Csetminus (a, b)$ is compact
– Rumpelstiltskin
Jul 15 at 16:10
1
@TheLastCipher As Adam's argument shows, in general removing an open set $U$ from a closed set $F$ gives a closed set, as $F setminus U = F cap U^mathsfc$ is an intersection of closed sets.
– Hayden
Jul 15 at 16:57
1
1
Why can we suppose that $Ccap(a,b)$ is an interval? What if, say, $C=leftfrac1n,middle$?
– José Carlos Santos
Jul 15 at 15:36
Why can we suppose that $Ccap(a,b)$ is an interval? What if, say, $C=leftfrac1n,middle$?
– José Carlos Santos
Jul 15 at 15:36
I was thinking of considering a compact set $C$ such that $Csetminus (a,b)$ is not empty
– TheLast Cipher
Jul 15 at 15:38
I was thinking of considering a compact set $C$ such that $Csetminus (a,b)$ is not empty
– TheLast Cipher
Jul 15 at 15:38
1
1
Don't use Heine-Borel. The definition of $Csetminus(a,b)$ makes it perfectly suited for a direct argument with open covers
– Hagen von Eitzen
Jul 15 at 15:49
Don't use Heine-Borel. The definition of $Csetminus(a,b)$ makes it perfectly suited for a direct argument with open covers
– Hagen von Eitzen
Jul 15 at 15:49
2
2
$C$ is closed (because it's compact). $Csetminus (a, b)$ is closed, it's intersection of $(a, b)^c$ and $C$, two closed sets. Now, we have a closed subset $Csetminus (a, b)$ of $C$, a compact set. So $Csetminus (a, b)$ is compact
– Rumpelstiltskin
Jul 15 at 16:10
$C$ is closed (because it's compact). $Csetminus (a, b)$ is closed, it's intersection of $(a, b)^c$ and $C$, two closed sets. Now, we have a closed subset $Csetminus (a, b)$ of $C$, a compact set. So $Csetminus (a, b)$ is compact
– Rumpelstiltskin
Jul 15 at 16:10
1
1
@TheLastCipher As Adam's argument shows, in general removing an open set $U$ from a closed set $F$ gives a closed set, as $F setminus U = F cap U^mathsfc$ is an intersection of closed sets.
– Hayden
Jul 15 at 16:57
@TheLastCipher As Adam's argument shows, in general removing an open set $U$ from a closed set $F$ gives a closed set, as $F setminus U = F cap U^mathsfc$ is an intersection of closed sets.
– Hayden
Jul 15 at 16:57
 |Â
show 7 more comments
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
You seem to be assuming that $Ccap (a,b)$ is a closed interval. That is not at all the case. For instance, what if $C=1,2,3,ldots, n$ for some $n$? Or $C=0,frac11, frac12,frac13,ldots$? Or the Cantor set? As you can see, it doesn't even have to be a union of closed intervals in any nice way. As a general intuition, open sets look "nice", but closed sets can be quite "ugly".
However, if $C$ is closed and bounded, what can you say about $Csetminus (a,b)$? Is it closed? Is it bounded?
You can do it a lot more directly from the definition of compact as well. Let $mathcal U$ be an open cover of $Csetminus (a,b)$. Then $mathcal Ucup (a,b)$ is an open cover of $C$. Can you do the rest?
answered Jul 15 at 15:38
Arthur
99k793175
+1 Started typing then saw your answer. Lol.
– SinTan1729
Jul 15 at 15:38
I want to say that $Csetminus (a,b)$ is closed before I've written this proof attempt. But was unsure how to show.
– TheLast Cipher
Jul 15 at 15:45
@TheLastCipher It's easier to show that $Bbb Rsetminus (Csetminus(a,b))$ is open, using that $(a,b)$ and $Bbb Rsetminus C$ are both open.
– Arthur
Jul 15 at 15:48
I'll also give the direct approach a try. I'll comment if I have any more problems. Thank you.
– TheLast Cipher
Jul 15 at 15:49
To the proposer: The point of the last paragraph of this Answer is that if $ C$ is a compact subset of any space $X$ and if $U$ is an open subset of $X$ then $C$ $U$ is compact.
– DanielWainfleet
Jul 16 at 0:42
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
You seem to be assuming that $Ccap (a,b)$ is a closed interval. That is not at all the case. For instance, what if $C=1,2,3,ldots, n$ for some $n$? Or $C=0,frac11, frac12,frac13,ldots$? Or the Cantor set? As you can see, it doesn't even have to be a union of closed intervals in any nice way. As a general intuition, open sets look "nice", but closed sets can be quite "ugly".
However, if $C$ is closed and bounded, what can you say about $Csetminus (a,b)$? Is it closed? Is it bounded?
You can do it a lot more directly from the definition of compact as well. Let $mathcal U$ be an open cover of $Csetminus (a,b)$. Then $mathcal Ucup (a,b)$ is an open cover of $C$. Can you do the rest?
answered Jul 15 at 15:38
Arthur
99k793175
+1 Started typing then saw your answer. Lol.
– SinTan1729
Jul 15 at 15:38
I want to say that $Csetminus (a,b)$ is closed before I've written this proof attempt. But was unsure how to show.
– TheLast Cipher
Jul 15 at 15:45
@TheLastCipher It's easier to show that $Bbb Rsetminus (Csetminus(a,b))$ is open, using that $(a,b)$ and $Bbb Rsetminus C$ are both open.
– Arthur
Jul 15 at 15:48
I'll also give the direct approach a try. I'll comment if I have any more problems. Thank you.
– TheLast Cipher
Jul 15 at 15:49
To the proposer: The point of the last paragraph of this Answer is that if $ C$ is a compact subset of any space $X$ and if $U$ is an open subset of $X$ then $C$ $U$ is compact.
– DanielWainfleet
Jul 16 at 0:42
 |Â
show 1 more comment
up vote
7
down vote
accepted
You seem to be assuming that $Ccap (a,b)$ is a closed interval. That is not at all the case. For instance, what if $C=1,2,3,ldots, n$ for some $n$? Or $C=0,frac11, frac12,frac13,ldots$? Or the Cantor set? As you can see, it doesn't even have to be a union of closed intervals in any nice way. As a general intuition, open sets look "nice", but closed sets can be quite "ugly".
However, if $C$ is closed and bounded, what can you say about $Csetminus (a,b)$? Is it closed? Is it bounded?
You can do it a lot more directly from the definition of compact as well. Let $mathcal U$ be an open cover of $Csetminus (a,b)$. Then $mathcal Ucup (a,b)$ is an open cover of $C$. Can you do the rest?
answered Jul 15 at 15:38
Arthur
99k793175
+1 Started typing then saw your answer. Lol.
– SinTan1729
Jul 15 at 15:38
I want to say that $Csetminus (a,b)$ is closed before I've written this proof attempt. But was unsure how to show.
– TheLast Cipher
Jul 15 at 15:45
@TheLastCipher It's easier to show that $Bbb Rsetminus (Csetminus(a,b))$ is open, using that $(a,b)$ and $Bbb Rsetminus C$ are both open.
– Arthur
Jul 15 at 15:48
I'll also give the direct approach a try. I'll comment if I have any more problems. Thank you.
– TheLast Cipher
Jul 15 at 15:49
To the proposer: The point of the last paragraph of this Answer is that if $ C$ is a compact subset of any space $X$ and if $U$ is an open subset of $X$ then $C$ $U$ is compact.
– DanielWainfleet
Jul 16 at 0:42
 |Â
show 1 more comment
up vote
7
down vote
accepted
up vote
7
down vote
accepted
You seem to be assuming that $Ccap (a,b)$ is a closed interval. That is not at all the case. For instance, what if $C=1,2,3,ldots, n$ for some $n$? Or $C=0,frac11, frac12,frac13,ldots$? Or the Cantor set? As you can see, it doesn't even have to be a union of closed intervals in any nice way. As a general intuition, open sets look "nice", but closed sets can be quite "ugly".
However, if $C$ is closed and bounded, what can you say about $Csetminus (a,b)$? Is it closed? Is it bounded?
You can do it a lot more directly from the definition of compact as well. Let $mathcal U$ be an open cover of $Csetminus (a,b)$. Then $mathcal Ucup (a,b)$ is an open cover of $C$. Can you do the rest?
answered Jul 15 at 15:38
Arthur
99k793175
You seem to be assuming that $Ccap (a,b)$ is a closed interval. That is not at all the case. For instance, what if $C=1,2,3,ldots, n$ for some $n$? Or $C=0,frac11, frac12,frac13,ldots$? Or the Cantor set? As you can see, it doesn't even have to be a union of closed intervals in any nice way. As a general intuition, open sets look "nice", but closed sets can be quite "ugly".
However, if $C$ is closed and bounded, what can you say about $Csetminus (a,b)$? Is it closed? Is it bounded?
You can do it a lot more directly from the definition of compact as well. Let $mathcal U$ be an open cover of $Csetminus (a,b)$. Then $mathcal Ucup (a,b)$ is an open cover of $C$. Can you do the rest?
answered Jul 15 at 15:38
Arthur
99k793175
edited Jul 15 at 15:46
answered Jul 15 at 15:38
Arthur
99k793175
answered Jul 15 at 15:38
Arthur
99k793175
answered Jul 15 at 15:38
Arthur
99k793175
99k793175
+1 Started typing then saw your answer. Lol.
– SinTan1729
Jul 15 at 15:38
I want to say that $Csetminus (a,b)$ is closed before I've written this proof attempt. But was unsure how to show.
– TheLast Cipher
Jul 15 at 15:45
@TheLastCipher It's easier to show that $Bbb Rsetminus (Csetminus(a,b))$ is open, using that $(a,b)$ and $Bbb Rsetminus C$ are both open.
– Arthur
Jul 15 at 15:48
I'll also give the direct approach a try. I'll comment if I have any more problems. Thank you.
– TheLast Cipher
Jul 15 at 15:49
To the proposer: The point of the last paragraph of this Answer is that if $ C$ is a compact subset of any space $X$ and if $U$ is an open subset of $X$ then $C$ $U$ is compact.
– DanielWainfleet
Jul 16 at 0:42
 |Â
show 1 more comment
+1 Started typing then saw your answer. Lol.
– SinTan1729
Jul 15 at 15:38
I want to say that $Csetminus (a,b)$ is closed before I've written this proof attempt. But was unsure how to show.
– TheLast Cipher
Jul 15 at 15:45
@TheLastCipher It's easier to show that $Bbb Rsetminus (Csetminus(a,b))$ is open, using that $(a,b)$ and $Bbb Rsetminus C$ are both open.
– Arthur
Jul 15 at 15:48
I'll also give the direct approach a try. I'll comment if I have any more problems. Thank you.
– TheLast Cipher
Jul 15 at 15:49
To the proposer: The point of the last paragraph of this Answer is that if $ C$ is a compact subset of any space $X$ and if $U$ is an open subset of $X$ then $C$ $U$ is compact.
– DanielWainfleet
Jul 16 at 0:42
+1 Started typing then saw your answer. Lol.
– SinTan1729
Jul 15 at 15:38
+1 Started typing then saw your answer. Lol.
– SinTan1729
Jul 15 at 15:38
I want to say that $Csetminus (a,b)$ is closed before I've written this proof attempt. But was unsure how to show.
– TheLast Cipher
Jul 15 at 15:45
I want to say that $Csetminus (a,b)$ is closed before I've written this proof attempt. But was unsure how to show.
– TheLast Cipher
Jul 15 at 15:45
@TheLastCipher It's easier to show that $Bbb Rsetminus (Csetminus(a,b))$ is open, using that $(a,b)$ and $Bbb Rsetminus C$ are both open.
– Arthur
Jul 15 at 15:48
@TheLastCipher It's easier to show that $Bbb Rsetminus (Csetminus(a,b))$ is open, using that $(a,b)$ and $Bbb Rsetminus C$ are both open.
– Arthur
Jul 15 at 15:48
I'll also give the direct approach a try. I'll comment if I have any more problems. Thank you.
– TheLast Cipher
Jul 15 at 15:49
I'll also give the direct approach a try. I'll comment if I have any more problems. Thank you.
– TheLast Cipher
Jul 15 at 15:49
To the proposer: The point of the last paragraph of this Answer is that if $ C$ is a compact subset of any space $X$ and if $U$ is an open subset of $X$ then $C$ $U$ is compact.
– DanielWainfleet
Jul 16 at 0:42
To the proposer: The point of the last paragraph of this Answer is that if $ C$ is a compact subset of any space $X$ and if $U$ is an open subset of $X$ then $C$ $U$ is compact.
– DanielWainfleet
Jul 16 at 0:42
 |Â
show 1 more comment
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1
Why can we suppose that $Ccap(a,b)$ is an interval? What if, say, $C=leftfrac1n,middle$?
– José Carlos Santos
Jul 15 at 15:36
I was thinking of considering a compact set $C$ such that $Csetminus (a,b)$ is not empty
– TheLast Cipher
Jul 15 at 15:38
1
Don't use Heine-Borel. The definition of $Csetminus(a,b)$ makes it perfectly suited for a direct argument with open covers
– Hagen von Eitzen
Jul 15 at 15:49
2
$C$ is closed (because it's compact). $Csetminus (a, b)$ is closed, it's intersection of $(a, b)^c$ and $C$, two closed sets. Now, we have a closed subset $Csetminus (a, b)$ of $C$, a compact set. So $Csetminus (a, b)$ is compact
– Rumpelstiltskin
Jul 15 at 16:10
1
@TheLastCipher As Adam's argument shows, in general removing an open set $U$ from a closed set $F$ gives a closed set, as $F setminus U = F cap U^mathsfc$ is an intersection of closed sets.
– Hayden
Jul 15 at 16:57