Mixing
Avoiding choice in proving “Sequential compactness implies Lebesgue Number Lemma”
$begingroup$
The standard proof can be found in ProofWiki. From what it is shown on that, it uses the Axiom of Countable Choice when choosing the subsequence ${x_n}$ to produce a contradiction. And normally, as I discover, when some instances of AC is used on ProofWiki, it will be remarked at the bottom of the page. So this leads me to question whether AC is indeed necessary here. Please provide some clarification. Thanks in advance.
Context: This is a step in proving "sequential compactness implies compactness for metric spaces". I know it must use AC at a certain point, but is it here?
general-topology set-theory axiom-of-choice
asked Jan 19 at 16:11
YuiTo ChengYuiTo Cheng
1,9452633
$endgroup$
add a comment |
$begingroup$
The standard proof can be found in ProofWiki. From what it is shown on that, it uses the Axiom of Countable Choice when choosing the subsequence ${x_n}$ to produce a contradiction. And normally, as I discover, when some instances of AC is used on ProofWiki, it will be remarked at the bottom of the page. So this leads me to question whether AC is indeed necessary here. Please provide some clarification. Thanks in advance.
Context: This is a step in proving "sequential compactness implies compactness for metric spaces". I know it must use AC at a certain point, but is it here?
general-topology set-theory axiom-of-choice
asked Jan 19 at 16:11
YuiTo ChengYuiTo Cheng
1,9452633
$endgroup$
add a comment |
$begingroup$
The standard proof can be found in ProofWiki. From what it is shown on that, it uses the Axiom of Countable Choice when choosing the subsequence ${x_n}$ to produce a contradiction. And normally, as I discover, when some instances of AC is used on ProofWiki, it will be remarked at the bottom of the page. So this leads me to question whether AC is indeed necessary here. Please provide some clarification. Thanks in advance.
Context: This is a step in proving "sequential compactness implies compactness for metric spaces". I know it must use AC at a certain point, but is it here?
general-topology set-theory axiom-of-choice
asked Jan 19 at 16:11
YuiTo ChengYuiTo Cheng
1,9452633
$endgroup$
The standard proof can be found in ProofWiki. From what it is shown on that, it uses the Axiom of Countable Choice when choosing the subsequence ${x_n}$ to produce a contradiction. And normally, as I discover, when some instances of AC is used on ProofWiki, it will be remarked at the bottom of the page. So this leads me to question whether AC is indeed necessary here. Please provide some clarification. Thanks in advance.
Context: This is a step in proving "sequential compactness implies compactness for metric spaces". I know it must use AC at a certain point, but is it here?
general-topology set-theory axiom-of-choice
general-topology set-theory axiom-of-choice
asked Jan 19 at 16:11
YuiTo ChengYuiTo Cheng
1,9452633
asked Jan 19 at 16:11
YuiTo ChengYuiTo Cheng
1,9452633
edited Jan 19 at 16:27
YuiTo Cheng
asked Jan 19 at 16:11
YuiTo ChengYuiTo Cheng
1,9452633
asked Jan 19 at 16:11
YuiTo ChengYuiTo Cheng
1,9452633
asked Jan 19 at 16:11
YuiTo ChengYuiTo Cheng
1,9452633
1,9452633
add a comment |
add a comment |
1 Answer
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$begingroup$
Not really, if you insist on sequential compactness.
Suppose that $A$ is an infinite Dedekind-finite set of real numbers. It is easy to show that $A$ is not closed, as therefore not compact. Alas, $A$ is sequentially compact, since every sequence can only have finitely many values, and thus has a convergent subsequence.
Let's assume that $A$ is a dense subset of $(0,1)$. Since $(0,1)$ can be covered by almost disjoint intervals of arbitrarily small size (i.e., an open cover without a Lebesgue number), this defines a cover of open intervals of $A$ without a Lebesgue number. Indeed, this can even be a countable sequence of intervals (the crux here is that the midpoints of the intervals are not in $A$, so you cannot use them to define a countable subset of $A$).
The interesting question, to which I don't have an answer off-hand, is what happens when we assume compactness, rather than sequential compactness. On the surface, it seems like it should work. But it is also seemingly requiring that the countable union of finite sets is countable, which itself needs a bit choice.
answered Jan 19 at 21:35
Asaf Karagila♦Asaf Karagila
305k33435766
$endgroup$
$begingroup$
I don't think "Compactness implies Lebesgue Number Lemma" requires AC at all.
$endgroup$
– YuiTo Cheng
Jan 20 at 1:28
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Not really, if you insist on sequential compactness.
Suppose that $A$ is an infinite Dedekind-finite set of real numbers. It is easy to show that $A$ is not closed, as therefore not compact. Alas, $A$ is sequentially compact, since every sequence can only have finitely many values, and thus has a convergent subsequence.
Let's assume that $A$ is a dense subset of $(0,1)$. Since $(0,1)$ can be covered by almost disjoint intervals of arbitrarily small size (i.e., an open cover without a Lebesgue number), this defines a cover of open intervals of $A$ without a Lebesgue number. Indeed, this can even be a countable sequence of intervals (the crux here is that the midpoints of the intervals are not in $A$, so you cannot use them to define a countable subset of $A$).
The interesting question, to which I don't have an answer off-hand, is what happens when we assume compactness, rather than sequential compactness. On the surface, it seems like it should work. But it is also seemingly requiring that the countable union of finite sets is countable, which itself needs a bit choice.
answered Jan 19 at 21:35
Asaf Karagila♦Asaf Karagila
305k33435766
$endgroup$
$begingroup$
I don't think "Compactness implies Lebesgue Number Lemma" requires AC at all.
$endgroup$
– YuiTo Cheng
Jan 20 at 1:28
add a comment |
$begingroup$
Not really, if you insist on sequential compactness.
Suppose that $A$ is an infinite Dedekind-finite set of real numbers. It is easy to show that $A$ is not closed, as therefore not compact. Alas, $A$ is sequentially compact, since every sequence can only have finitely many values, and thus has a convergent subsequence.
Let's assume that $A$ is a dense subset of $(0,1)$. Since $(0,1)$ can be covered by almost disjoint intervals of arbitrarily small size (i.e., an open cover without a Lebesgue number), this defines a cover of open intervals of $A$ without a Lebesgue number. Indeed, this can even be a countable sequence of intervals (the crux here is that the midpoints of the intervals are not in $A$, so you cannot use them to define a countable subset of $A$).
The interesting question, to which I don't have an answer off-hand, is what happens when we assume compactness, rather than sequential compactness. On the surface, it seems like it should work. But it is also seemingly requiring that the countable union of finite sets is countable, which itself needs a bit choice.
answered Jan 19 at 21:35
Asaf Karagila♦Asaf Karagila
305k33435766
$endgroup$
$begingroup$
I don't think "Compactness implies Lebesgue Number Lemma" requires AC at all.
$endgroup$
– YuiTo Cheng
Jan 20 at 1:28
add a comment |
$begingroup$
Not really, if you insist on sequential compactness.
Suppose that $A$ is an infinite Dedekind-finite set of real numbers. It is easy to show that $A$ is not closed, as therefore not compact. Alas, $A$ is sequentially compact, since every sequence can only have finitely many values, and thus has a convergent subsequence.
Let's assume that $A$ is a dense subset of $(0,1)$. Since $(0,1)$ can be covered by almost disjoint intervals of arbitrarily small size (i.e., an open cover without a Lebesgue number), this defines a cover of open intervals of $A$ without a Lebesgue number. Indeed, this can even be a countable sequence of intervals (the crux here is that the midpoints of the intervals are not in $A$, so you cannot use them to define a countable subset of $A$).
The interesting question, to which I don't have an answer off-hand, is what happens when we assume compactness, rather than sequential compactness. On the surface, it seems like it should work. But it is also seemingly requiring that the countable union of finite sets is countable, which itself needs a bit choice.
answered Jan 19 at 21:35
Asaf Karagila♦Asaf Karagila
305k33435766
$endgroup$
Not really, if you insist on sequential compactness.
Suppose that $A$ is an infinite Dedekind-finite set of real numbers. It is easy to show that $A$ is not closed, as therefore not compact. Alas, $A$ is sequentially compact, since every sequence can only have finitely many values, and thus has a convergent subsequence.
Let's assume that $A$ is a dense subset of $(0,1)$. Since $(0,1)$ can be covered by almost disjoint intervals of arbitrarily small size (i.e., an open cover without a Lebesgue number), this defines a cover of open intervals of $A$ without a Lebesgue number. Indeed, this can even be a countable sequence of intervals (the crux here is that the midpoints of the intervals are not in $A$, so you cannot use them to define a countable subset of $A$).
The interesting question, to which I don't have an answer off-hand, is what happens when we assume compactness, rather than sequential compactness. On the surface, it seems like it should work. But it is also seemingly requiring that the countable union of finite sets is countable, which itself needs a bit choice.
answered Jan 19 at 21:35
Asaf Karagila♦Asaf Karagila
305k33435766
answered Jan 19 at 21:35
Asaf Karagila♦Asaf Karagila
305k33435766
answered Jan 19 at 21:35
Asaf Karagila♦Asaf Karagila
305k33435766
answered Jan 19 at 21:35
Asaf Karagila♦Asaf Karagila
305k33435766
305k33435766
$begingroup$
I don't think "Compactness implies Lebesgue Number Lemma" requires AC at all.
$endgroup$
– YuiTo Cheng
Jan 20 at 1:28
add a comment |
$begingroup$
I don't think "Compactness implies Lebesgue Number Lemma" requires AC at all.
$endgroup$
– YuiTo Cheng
Jan 20 at 1:28
$begingroup$
I don't think "Compactness implies Lebesgue Number Lemma" requires AC at all.
$endgroup$
– YuiTo Cheng
Jan 20 at 1:28
$begingroup$
I don't think "Compactness implies Lebesgue Number Lemma" requires AC at all.
$endgroup$
– YuiTo Cheng
Jan 20 at 1:28
add a comment |
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