Irreducible closed subset












0












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











Your Answer





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1 Answer
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1 Answer
1






active

oldest

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active

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active

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0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 16:08

























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


















  • $begingroup$
    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


















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