Is $mathcal Bleqmathcal B'$ $iff $ $B'in mathcal B'implies B'in mathcal B$ an order on topology basis? The 2019 Stack Overflow Developer Survey Results Are InOrder topology and partial orderordered topologyClarification regarding basis for a topologyBasis in Topology confusionQuestion about a basis for a topology vs the topology generated by a basis?Homeomorphism, subspace topology, and basis of topologyIf $mathcalT$ is a topology and $B$ is a basis, does $mathcalT_1 subset mathcalT_2 implies mathcalB_1 subset mathcalB_2$?Show basis for a topologyShow $mathcalB$ is a topology basis on $X$$coprod_i in I X_i$ is second countable $implies |I| leq |mathbbN|$
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Is $mathcal Bleqmathcal B'$ $iff $ $B'in mathcal B'implies B'in mathcal B$ an order on topology basis?
The 2019 Stack Overflow Developer Survey Results Are InOrder topology and partial orderordered topologyClarification regarding basis for a topologyBasis in Topology confusionQuestion about a basis for a topology vs the topology generated by a basis?Homeomorphism, subspace topology, and basis of topologyIf $mathcalT$ is a topology and $B$ is a basis, does $mathcalT_1 subset mathcalT_2 implies mathcalB_1 subset mathcalB_2$?Show basis for a topologyShow $mathcalB$ is a topology basis on $X$$coprod_i in I X_i$ is second countable $implies |I| leq |mathbbN|$
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I was wondering, Let $(X,mathcal T)$ a topological space. Is $$mathcal Bleqmathcal B'iff (B'in mathcal B'implies B'in mathcal B)$$ an order on topology basis ? (i.e. $mathcal B$ is thinner tan $mathcal B'$). Or we prefer to define such a relation as $mathcal B'leq mathcal B$ ?
general-topology
asked Mar 30 at 10:23
user657324user657324
5119
$endgroup$
add a comment |
$begingroup$
I was wondering, Let $(X,mathcal T)$ a topological space. Is $$mathcal Bleqmathcal B'iff (B'in mathcal B'implies B'in mathcal B)$$ an order on topology basis ? (i.e. $mathcal B$ is thinner tan $mathcal B'$). Or we prefer to define such a relation as $mathcal B'leq mathcal B$ ?
general-topology
asked Mar 30 at 10:23
user657324user657324
5119
$endgroup$
$begingroup$
Yes, sure you can define a partial order $mathcalB le mathcalB'$ iff $mathcalB' subseteq mathcalB$ on the set of all bases of the topology $(X, mathcalT)$. What's your purpose with it, Zorn?
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– Henno Brandsma
Mar 30 at 13:41
add a comment |
$begingroup$
I was wondering, Let $(X,mathcal T)$ a topological space. Is $$mathcal Bleqmathcal B'iff (B'in mathcal B'implies B'in mathcal B)$$ an order on topology basis ? (i.e. $mathcal B$ is thinner tan $mathcal B'$). Or we prefer to define such a relation as $mathcal B'leq mathcal B$ ?
general-topology
asked Mar 30 at 10:23
user657324user657324
5119
$endgroup$
I was wondering, Let $(X,mathcal T)$ a topological space. Is $$mathcal Bleqmathcal B'iff (B'in mathcal B'implies B'in mathcal B)$$ an order on topology basis ? (i.e. $mathcal B$ is thinner tan $mathcal B'$). Or we prefer to define such a relation as $mathcal B'leq mathcal B$ ?
general-topology
general-topology
asked Mar 30 at 10:23
user657324user657324
5119
asked Mar 30 at 10:23
user657324user657324
5119
asked Mar 30 at 10:23
user657324user657324
5119
asked Mar 30 at 10:23
user657324user657324
5119
asked Mar 30 at 10:23
user657324user657324
5119
5119
$begingroup$
Yes, sure you can define a partial order $mathcalB le mathcalB'$ iff $mathcalB' subseteq mathcalB$ on the set of all bases of the topology $(X, mathcalT)$. What's your purpose with it, Zorn?
$endgroup$
– Henno Brandsma
Mar 30 at 13:41
add a comment |
$begingroup$
Yes, sure you can define a partial order $mathcalB le mathcalB'$ iff $mathcalB' subseteq mathcalB$ on the set of all bases of the topology $(X, mathcalT)$. What's your purpose with it, Zorn?
$endgroup$
– Henno Brandsma
Mar 30 at 13:41
$begingroup$
Yes, sure you can define a partial order $mathcalB le mathcalB'$ iff $mathcalB' subseteq mathcalB$ on the set of all bases of the topology $(X, mathcalT)$. What's your purpose with it, Zorn?
$endgroup$
– Henno Brandsma
Mar 30 at 13:41
$begingroup$
Yes, sure you can define a partial order $mathcalB le mathcalB'$ iff $mathcalB' subseteq mathcalB$ on the set of all bases of the topology $(X, mathcalT)$. What's your purpose with it, Zorn?
$endgroup$
– Henno Brandsma
Mar 30 at 13:41
add a comment |
1 Answer
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What you have described is just inclusion (in the reversed order), and inclusion forms a partial order on the set of all basis for a topological space $(X,mathcal T)$. In fact, if you are given any set $S$, and $Tsubset P(S)$ a collection of subsets of $S$, then $T$ is naturally equipped with the partial order given by set inclusion.
answered Mar 30 at 10:48
lEmlEm
3,4321921
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$begingroup$
What you have described is just inclusion (in the reversed order), and inclusion forms a partial order on the set of all basis for a topological space $(X,mathcal T)$. In fact, if you are given any set $S$, and $Tsubset P(S)$ a collection of subsets of $S$, then $T$ is naturally equipped with the partial order given by set inclusion.
answered Mar 30 at 10:48
lEmlEm
3,4321921
$endgroup$
add a comment |
$begingroup$
What you have described is just inclusion (in the reversed order), and inclusion forms a partial order on the set of all basis for a topological space $(X,mathcal T)$. In fact, if you are given any set $S$, and $Tsubset P(S)$ a collection of subsets of $S$, then $T$ is naturally equipped with the partial order given by set inclusion.
answered Mar 30 at 10:48
lEmlEm
3,4321921
$endgroup$
add a comment |
$begingroup$
What you have described is just inclusion (in the reversed order), and inclusion forms a partial order on the set of all basis for a topological space $(X,mathcal T)$. In fact, if you are given any set $S$, and $Tsubset P(S)$ a collection of subsets of $S$, then $T$ is naturally equipped with the partial order given by set inclusion.
answered Mar 30 at 10:48
lEmlEm
3,4321921
$endgroup$
What you have described is just inclusion (in the reversed order), and inclusion forms a partial order on the set of all basis for a topological space $(X,mathcal T)$. In fact, if you are given any set $S$, and $Tsubset P(S)$ a collection of subsets of $S$, then $T$ is naturally equipped with the partial order given by set inclusion.
answered Mar 30 at 10:48
lEmlEm
3,4321921
answered Mar 30 at 10:48
lEmlEm
3,4321921
answered Mar 30 at 10:48
lEmlEm
3,4321921
answered Mar 30 at 10:48
lEmlEm
3,4321921
3,4321921
add a comment |
add a comment |
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Yes, sure you can define a partial order $mathcalB le mathcalB'$ iff $mathcalB' subseteq mathcalB$ on the set of all bases of the topology $(X, mathcalT)$. What's your purpose with it, Zorn?
$endgroup$
– Henno Brandsma
Mar 30 at 13:41