Base of discrete topology

The forgetful functor Γ:Top→Set\Gamma : 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.

[Disc⊣Γ⊣Codisc]:Top←Codisc→Γ←DiscSet. [Disc \dashv \Gamma \dashv Codisc] : Top \stackrel{\overset{Disc}{\leftarrow}}{\stackrel{\overset{\Gamma}{\to}}{\underset{Codisc}{\leftarrow}}} Set \,.

For S∈SetS \in Set

• 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].

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 0\pi_0 functor.

This is discussed at 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:

• The Stacks Project, Example 7.6.6

Conceptualization of the terminology via right adjoints to forgetful functors [see also at chaos] is due to

• William Lawvere, Functorial remarks on the general concept of chaos IMA preprint #87, 1984 [pdf]

and via footnote 1 [page 3] in:

• William Lawvere, Categories of spaces may not be generalized spaces, as exemplified by directed graphs, preprint, State University of New York at Buffalo, [1986] Reprints in Theory and Applications of Categories, No. 9, 2005, pp. 1–7 [tac:tr9, pdf].

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 member 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 member 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. $$X = \bigcup\limits_{B \in {\rm B}} B $$

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 $\mathcal{B}'$ is a basis, then in particular every element of $\mathcal{B}$ is a union of elements of $\mathcal{B}'$. But a singleton cannot be a union of proper subsets, so $\mathcal{B} \subset \mathcal{B}'$ and $\mathcal{B}'$ has at 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 at most $2^k$. Therefore, if a collection of $k$ sets forms a basis, we must have $2^k \geq 2^n$, so $k\geq n$.