Irreducible closed subset
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The dimension of a topological space $X$ is defined as the supremum of all integers $nge0$ for which there is a strictly increasing chain of $n+1$ irreducible closed subsets of $X$. Thus is the dimension of the empty topological space $-infty$? Is the dimension $>-infty$ for all nonempty topological spaces?
general-topology
asked Dec 8 '18 at 8:32
saisai
1376
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add a comment |
$begingroup$
The dimension of a topological space $X$ is defined as the supremum of all integers $nge0$ for which there is a strictly increasing chain of $n+1$ irreducible closed subsets of $X$. Thus is the dimension of the empty topological space $-infty$? Is the dimension $>-infty$ for all nonempty topological spaces?
general-topology
asked Dec 8 '18 at 8:32
saisai
1376
$endgroup$
$begingroup$
Why would the dimension of the empty set not be $-1$ by your definition?
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– Matt
Dec 8 '18 at 8:39
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@Matt: Because I consider it to be the supremum of a set of extended real numbers, and so the supremum of the empty set is $-infty$.
$endgroup$
– sai
Dec 8 '18 at 8:58
add a comment |
$begingroup$
The dimension of a topological space $X$ is defined as the supremum of all integers $nge0$ for which there is a strictly increasing chain of $n+1$ irreducible closed subsets of $X$. Thus is the dimension of the empty topological space $-infty$? Is the dimension $>-infty$ for all nonempty topological spaces?
general-topology
asked Dec 8 '18 at 8:32
saisai
1376
$endgroup$
The dimension of a topological space $X$ is defined as the supremum of all integers $nge0$ for which there is a strictly increasing chain of $n+1$ irreducible closed subsets of $X$. Thus is the dimension of the empty topological space $-infty$? Is the dimension $>-infty$ for all nonempty topological spaces?
general-topology
general-topology
asked Dec 8 '18 at 8:32
saisai
1376
asked Dec 8 '18 at 8:32
saisai
1376
asked Dec 8 '18 at 8:32
saisai
1376
asked Dec 8 '18 at 8:32
saisai
1376
asked Dec 8 '18 at 8:32
saisai
1376
1376
$begingroup$
Why would the dimension of the empty set not be $-1$ by your definition?
$endgroup$
– Matt
Dec 8 '18 at 8:39
$begingroup$
@Matt: Because I consider it to be the supremum of a set of extended real numbers, and so the supremum of the empty set is $-infty$.
$endgroup$
– sai
Dec 8 '18 at 8:58
add a comment |
$begingroup$
Why would the dimension of the empty set not be $-1$ by your definition?
$endgroup$
– Matt
Dec 8 '18 at 8:39
$begingroup$
@Matt: Because I consider it to be the supremum of a set of extended real numbers, and so the supremum of the empty set is $-infty$.
$endgroup$
– sai
Dec 8 '18 at 8:58
$begingroup$
Why would the dimension of the empty set not be $-1$ by your definition?
$endgroup$
– Matt
Dec 8 '18 at 8:39
$begingroup$
Why would the dimension of the empty set not be $-1$ by your definition?
$endgroup$
– Matt
Dec 8 '18 at 8:39
$begingroup$
@Matt: Because I consider it to be the supremum of a set of extended real numbers, and so the supremum of the empty set is $-infty$.
$endgroup$
– sai
Dec 8 '18 at 8:58
$begingroup$
@Matt: Because I consider it to be the supremum of a set of extended real numbers, and so the supremum of the empty set is $-infty$.
$endgroup$
– sai
Dec 8 '18 at 8:58
add a comment |
1 Answer
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Yes and yes. For your first question, since there's no such $n ge 1$, we have $sup emptyset = -infty$. So you're right. For your second question, it suffices to show that any nonempty space $X$ has at least one irreducible closed subset, so that $dim X = sup {0, dots} ge 0$
Since $X$ is nonempty, we can pick some $x in X$. By lemma 8.3 (1), the closure of any irreducible subset is irreducible. Since ${x}$ is irreducible, $overline{{x}}$ is an irreducible closed subset of $X$. So we're done.
Note: I refer to definition 8.1 (1) for the definition of irreducible subset.
answered Dec 8 '18 at 13:01
Alex VongAlex Vong
1,284819
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Thank you for your answer.
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– sai
Dec 10 '18 at 13:18
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@sai You can accept it if you want to.
$endgroup$
– Alex Vong
Dec 10 '18 at 13:42
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Is this argument plausible? There is a minimal nonempty closed subset of $X$ by Zorn's lemma, and then this must necessarily be irreducible.
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– sai
Dec 11 '18 at 2:26
$begingroup$
@sai I suppose you mean to partially order all nonempty closed subsets of $X$ by reversed inclusion $supseteq$ and then apply Zorn's lemma. I think you would run into problem in showing any nonempty chain has an upper bound since the intersection of a chain of nonempty closed sets can be empty.
$endgroup$
– Alex Vong
Dec 11 '18 at 6:08
$begingroup$
Okay, I accepted this answer. Thank you.
$endgroup$
– sai
Dec 12 '18 at 4:10
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Yes and yes. For your first question, since there's no such $n ge 1$, we have $sup emptyset = -infty$. So you're right. For your second question, it suffices to show that any nonempty space $X$ has at least one irreducible closed subset, so that $dim X = sup {0, dots} ge 0$
Since $X$ is nonempty, we can pick some $x in X$. By lemma 8.3 (1), the closure of any irreducible subset is irreducible. Since ${x}$ is irreducible, $overline{{x}}$ is an irreducible closed subset of $X$. So we're done.
Note: I refer to definition 8.1 (1) for the definition of irreducible subset.
answered Dec 8 '18 at 13:01
Alex VongAlex Vong
1,284819
$endgroup$
$begingroup$
Thank you for your answer.
$endgroup$
– sai
Dec 10 '18 at 13:18
$begingroup$
@sai You can accept it if you want to.
$endgroup$
– Alex Vong
Dec 10 '18 at 13:42
$begingroup$
Is this argument plausible? There is a minimal nonempty closed subset of $X$ by Zorn's lemma, and then this must necessarily be irreducible.
$endgroup$
– sai
Dec 11 '18 at 2:26
$begingroup$
@sai I suppose you mean to partially order all nonempty closed subsets of $X$ by reversed inclusion $supseteq$ and then apply Zorn's lemma. I think you would run into problem in showing any nonempty chain has an upper bound since the intersection of a chain of nonempty closed sets can be empty.
$endgroup$
– Alex Vong
Dec 11 '18 at 6:08
$begingroup$
Okay, I accepted this answer. Thank you.
$endgroup$
– sai
Dec 12 '18 at 4:10
add a comment |
$begingroup$
Yes and yes. For your first question, since there's no such $n ge 1$, we have $sup emptyset = -infty$. So you're right. For your second question, it suffices to show that any nonempty space $X$ has at least one irreducible closed subset, so that $dim X = sup {0, dots} ge 0$
Since $X$ is nonempty, we can pick some $x in X$. By lemma 8.3 (1), the closure of any irreducible subset is irreducible. Since ${x}$ is irreducible, $overline{{x}}$ is an irreducible closed subset of $X$. So we're done.
Note: I refer to definition 8.1 (1) for the definition of irreducible subset.
answered Dec 8 '18 at 13:01
Alex VongAlex Vong
1,284819
$endgroup$
$begingroup$
Thank you for your answer.
$endgroup$
– sai
Dec 10 '18 at 13:18
$begingroup$
@sai You can accept it if you want to.
$endgroup$
– Alex Vong
Dec 10 '18 at 13:42
$begingroup$
Is this argument plausible? There is a minimal nonempty closed subset of $X$ by Zorn's lemma, and then this must necessarily be irreducible.
$endgroup$
– sai
Dec 11 '18 at 2:26
$begingroup$
@sai I suppose you mean to partially order all nonempty closed subsets of $X$ by reversed inclusion $supseteq$ and then apply Zorn's lemma. I think you would run into problem in showing any nonempty chain has an upper bound since the intersection of a chain of nonempty closed sets can be empty.
$endgroup$
– Alex Vong
Dec 11 '18 at 6:08
$begingroup$
Okay, I accepted this answer. Thank you.
$endgroup$
– sai
Dec 12 '18 at 4:10
add a comment |
$begingroup$
Yes and yes. For your first question, since there's no such $n ge 1$, we have $sup emptyset = -infty$. So you're right. For your second question, it suffices to show that any nonempty space $X$ has at least one irreducible closed subset, so that $dim X = sup {0, dots} ge 0$
Since $X$ is nonempty, we can pick some $x in X$. By lemma 8.3 (1), the closure of any irreducible subset is irreducible. Since ${x}$ is irreducible, $overline{{x}}$ is an irreducible closed subset of $X$. So we're done.
Note: I refer to definition 8.1 (1) for the definition of irreducible subset.
answered Dec 8 '18 at 13:01
Alex VongAlex Vong
1,284819
$endgroup$
Yes and yes. For your first question, since there's no such $n ge 1$, we have $sup emptyset = -infty$. So you're right. For your second question, it suffices to show that any nonempty space $X$ has at least one irreducible closed subset, so that $dim X = sup {0, dots} ge 0$
Since $X$ is nonempty, we can pick some $x in X$. By lemma 8.3 (1), the closure of any irreducible subset is irreducible. Since ${x}$ is irreducible, $overline{{x}}$ is an irreducible closed subset of $X$. So we're done.
Note: I refer to definition 8.1 (1) for the definition of irreducible subset.
answered Dec 8 '18 at 13:01
Alex VongAlex Vong
1,284819
edited Dec 8 '18 at 16:08
answered Dec 8 '18 at 13:01
Alex VongAlex Vong
1,284819
answered Dec 8 '18 at 13:01
Alex VongAlex Vong
1,284819
answered Dec 8 '18 at 13:01
Alex VongAlex Vong
1,284819
1,284819
$begingroup$
Thank you for your answer.
$endgroup$
– sai
Dec 10 '18 at 13:18
$begingroup$
@sai You can accept it if you want to.
$endgroup$
– Alex Vong
Dec 10 '18 at 13:42
$begingroup$
Is this argument plausible? There is a minimal nonempty closed subset of $X$ by Zorn's lemma, and then this must necessarily be irreducible.
$endgroup$
– sai
Dec 11 '18 at 2:26
$begingroup$
@sai I suppose you mean to partially order all nonempty closed subsets of $X$ by reversed inclusion $supseteq$ and then apply Zorn's lemma. I think you would run into problem in showing any nonempty chain has an upper bound since the intersection of a chain of nonempty closed sets can be empty.
$endgroup$
– Alex Vong
Dec 11 '18 at 6:08
$begingroup$
Okay, I accepted this answer. Thank you.
$endgroup$
– sai
Dec 12 '18 at 4:10
add a comment |
$begingroup$
Thank you for your answer.
$endgroup$
– sai
Dec 10 '18 at 13:18
$begingroup$
@sai You can accept it if you want to.
$endgroup$
– Alex Vong
Dec 10 '18 at 13:42
$begingroup$
Is this argument plausible? There is a minimal nonempty closed subset of $X$ by Zorn's lemma, and then this must necessarily be irreducible.
$endgroup$
– sai
Dec 11 '18 at 2:26
$begingroup$
@sai I suppose you mean to partially order all nonempty closed subsets of $X$ by reversed inclusion $supseteq$ and then apply Zorn's lemma. I think you would run into problem in showing any nonempty chain has an upper bound since the intersection of a chain of nonempty closed sets can be empty.
$endgroup$
– Alex Vong
Dec 11 '18 at 6:08
$begingroup$
Okay, I accepted this answer. Thank you.
$endgroup$
– sai
Dec 12 '18 at 4:10
$begingroup$
Thank you for your answer.
$endgroup$
– sai
Dec 10 '18 at 13:18
$begingroup$
Thank you for your answer.
$endgroup$
– sai
Dec 10 '18 at 13:18
$begingroup$
@sai You can accept it if you want to.
$endgroup$
– Alex Vong
Dec 10 '18 at 13:42
$begingroup$
@sai You can accept it if you want to.
$endgroup$
– Alex Vong
Dec 10 '18 at 13:42
$begingroup$
Is this argument plausible? There is a minimal nonempty closed subset of $X$ by Zorn's lemma, and then this must necessarily be irreducible.
$endgroup$
– sai
Dec 11 '18 at 2:26
$begingroup$
Is this argument plausible? There is a minimal nonempty closed subset of $X$ by Zorn's lemma, and then this must necessarily be irreducible.
$endgroup$
– sai
Dec 11 '18 at 2:26
$begingroup$
@sai I suppose you mean to partially order all nonempty closed subsets of $X$ by reversed inclusion $supseteq$ and then apply Zorn's lemma. I think you would run into problem in showing any nonempty chain has an upper bound since the intersection of a chain of nonempty closed sets can be empty.
$endgroup$
– Alex Vong
Dec 11 '18 at 6:08
$begingroup$
@sai I suppose you mean to partially order all nonempty closed subsets of $X$ by reversed inclusion $supseteq$ and then apply Zorn's lemma. I think you would run into problem in showing any nonempty chain has an upper bound since the intersection of a chain of nonempty closed sets can be empty.
$endgroup$
– Alex Vong
Dec 11 '18 at 6:08
$begingroup$
Okay, I accepted this answer. Thank you.
$endgroup$
– sai
Dec 12 '18 at 4:10
$begingroup$
Okay, I accepted this answer. Thank you.
$endgroup$
– sai
Dec 12 '18 at 4:10
add a comment |
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$begingroup$
Why would the dimension of the empty set not be $-1$ by your definition?
$endgroup$
– Matt
Dec 8 '18 at 8:39
$begingroup$
@Matt: Because I consider it to be the supremum of a set of extended real numbers, and so the supremum of the empty set is $-infty$.
$endgroup$
– sai
Dec 8 '18 at 8:58