Kinh Nghiệm Hướng dẫn Base of discrete topology Mới Nhất
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The forgetful functor Γ:Top→SetGamma : Top to Set from Top to Set that sends any topological space to its underlying set has a left adjoint Disc:Set→TopDisc : Set to Top and a right adjoint Codisc:Set→TopCodisc : Set to Top.
Nội dung chính
- The left adjoint of the discrete space functor
Disc(S)Disc(S) is the topological space on SS in which every subset is an open set,
this is called the discrete topology on SS, it is the finest topology on SS; Disc(S)Disc(S) is called a discrete space;
Codisc(S)Codisc(S) is the topological space on SS whose only open sets are the empty set and SS itself, which is called the indiscrete topology on SS (rarely also antidiscrete topology or codiscrete topology or trivial topology or chaotic topology (SGA4-1, 1.1.4)), it is the coarsest topology on SS; Codisc(S)Codisc(S) is called a indiscrete space (rarely also antidiscrete space, even more rarely codiscrete space).
- The Stacks Project, Example 7.6.6
- William Lawvere, Functorial remarks on the general concept of chaos IMA preprint #87, 1984 (pdf)
- William Lawvere, Categories of spaces may not be generalized spaces, as exemplified by directed graphs, preprint, State University of Tp New York Buffalo, (1986) Reprints in Theory and Applications of Categories, No. 9, 2005, pp. 1–7 (tac:tr9, pdf).
(Disc⊣Γ⊣Codisc):Top←Codisc→Γ←DiscSet. (Disc dashv Gamma dashv Codisc) : Top stackreloversetDiscleftarrowstackreloversetGammatoundersetCodiscleftarrow Set ,.
For S∈SetS in Set
For an axiomatization of this situation see codiscrete object.
Properties
The left adjoint of the discrete space functor
The functor DiscDisc does not preserve infinite products because the infinite product topological space of discrete spaces may be nondiscrete. Thus, DiscDisc does not have a left adjoint functor.
However, if we restrict the codomain of DiscDisc to locally connected spaces, then the left adjoint functor of DiscDisc does exist and it computes the set of connected components of a given locally connected space, i.e., is the π 0pi_0 functor.
This is discussed locally connected spaces – cohesion over sets and cosheaf of connected components.
References
For Grothendieck topologies, the terminology “chaotic” is due to
reviewed, e.g., in:
Conceptualization of the terminology via right adjoints to forgetful functors (see also chaos) is due to
and via footnote 1 (page 3) in:
Let $$left( X,tau right)$$ be a topological space, then the sub collection $$rm B $$ of $$tau $$ is said to be a base or bases or open base for $$tau $$ if each thành viên of $$tau $$ can be expressed as a union of members of $$rm B$$.
In other words let $$left( X,tau right)$$ be a topological space, then the sub collection $$rm B$$ of $$tau $$ is said to be a base if for a point $$x$$ belonging to an open set $$U$$ there exists $$B in rm B$$ such that $$x in B subseteq U$$.
Example:
Let $$X = left a,b,c,d,e right$$ and let $$tau = left phi ,left a,b right,left c,d right,left a,b,c,d right,X right$$ be a topology defined on $$X$$. $$rm B = left left a,b right,left c,d right,X right$$ is a sub collection of $$tau $$, which meets the requirement for a base, because each thành viên of $$tau $$ is a union of members of $$rm B$$.
Remarks:
• It should be noted that there may be more than one base for a given topology defined on that set.
• Since the union of an empty sub collection of members of $$rm B$$ is an empty set, so an empty set $$phi in tau $$.
For Discrete Topology
Let $$X = left 1,2,3 right$$ and let $$tau = left phi ,left 1 right,left 2 right,left 3 right,left 1,2 right,left 2,3 right,left 1,3 right,X right$$ be a topology defined on $$X$$. $$rm B = left left 1 right,left 2 right,left 3 right right$$ is a base for $$tau $$. Check whether $$rm B$$ is a base or not,and take all possible unions of $$rm B$$ there must become $$tau $$.
Possible unions $$ = left phi ,left 1 right,left 2 right,left 3 right,left 1,2 right,left 2,3 right,left 1,3 right,X right$$
In this case, the discrete topological space, the collection of all singletons subsets of $$X$$, forms a base for a discrete topological space.
For Indiscrete Topology
Let $$X = left 1,2,3 right$$ and let $$tau = left phi ,X right$$ be a topology defined on $$X$$. $$rm B = left X right$$ is a base for $$tau $$.
Theorem
Let $$left( X,tau right)$$ be a topological space, then a sub collection $$rm B$$of $$tau $$ is a base for $$tau $$ if and only if:
1. $$X = bigcuplimits_B in rm B B $$
2. If $$B_1$$ and $$B_2$$ belongs to $$rm B$$, then $$B_1 cap B_2$$ can be written as a union of members of $$rm B$$; i.e. for $$x in B_1 cap B_2$$ then there exist $$B$$ of $$rm B$$ such that $$x in B subseteq B_1 cap B_2$$.
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Let $S$ be a set.
Let $tau$ be the discrete topology on $S$.
Let $BB$ be the set of all singleton subsets of $S$:
$BB := set set x: x in S$.
Then $BB$ is a basis for $T$.
Proof
Let $T = struct S, tau$ be the discrete space on $S$.
Let $U in tau$.
Then:
$ds U = bigcup_x mathop in U set x$
Hence:
$forall x in U: exists set x in BB: set x subseteq U$
Thus $U$ is the union of elements of $BB$.
Hence by definition $BB$ is a basis for $T$.
$blacksquare$
If $mathcalB’$ is a basis, then in particular every element of $mathcalB$ is a union of elements of $mathcalB’$. But a singleton cannot be a union of proper subsets, so $mathcalB subset mathcalB’$ and $mathcalB’$ has least $n$ elements.
As an alternative proof, we could observe that the number of possible unions that we can form from a collection of $k$ subsets is most $2^k$. Therefore, if a collection of $k$ sets forms a basis, we must have $2^k geq 2^n$, so $kgeq n$.
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