Topology, Part I

Time for some basics of point-set topology. Let {X} be a nonempty set. A collection of subsets {\mathcal T} of {X} is called a topology on {X}, and the pair {(X,\mathcal T)} a topological space if

  • {\varnothing, X\in\mathcal T},
  • {U,V\in\mathcal T\Rightarrow U\cap V\in\mathcal T},
  • {U_\alpha, \alpha\in I\in\mathcal T\Rightarrow \bigcup_{\alpha\in I}U_\alpha\in\mathcal T}.

The members of {\mathcal T} are called open sets. The collection {\{\varnothing, X\}} is called the trivial topology, and the power set of {X},

\displaystyle 2^X = \{S: S\subset X \}

is the discrete topology. If {\mathcal T} and {\mathcal T'} are topologies on {X} with {\mathcal T\subset\mathcal T'}, then {\mathcal T} is said to be coarser than {\mathcal T'}, and {\mathcal T'} is said to be finer than {\mathcal T}.

Theorem 1 The intersection of topologies is a topology.

Proof: Let {\mathcal T_\beta} be topologies on {X}, for {\beta\in J}. Put {\mathcal T=\bigcap_{\beta\in J}\mathcal T_\beta}. Since {\varnothing,X\in\mathcal T_\beta} for each {\beta}, {\varnothing,X\in\mathcal T}. If {U,V\in\mathcal T}, then {U,V\in\mathcal T_\beta} for each {\beta}, so {U\cap V\in\mathcal T_\beta} for each {\beta}, and hence {U\cap V\in\mathcal T}. If {U_\alpha\in\mathcal T}, for {\alpha\in I} then {U_\alpha\in\mathcal T_\beta}, so {\bigcap_{\alpha\in I}\in \mathcal T_\beta} for each {\beta} and hence {\bigcap_{\alpha\in I}U_\alpha\in\mathcal T}. It follows that {\mathcal T} is a topology. \Box

A simple corollary of this statement is that every set {S\subset X} has a unique generated topology, namely the intersection of all topologies on {X} containing {S}:

\displaystyle \mathcal T(S) = \bigcap_{\mathcal T\supset S}\mathcal T.

We say that a collection {\mathcal B} of subsets of {X} is a basis for a topology on {X} if

  • For each {x\in X}, there exists {B\in\mathcal B} such that {x\in B}.
  • If {x\in X} and {B_1,B_2\in\mathcal B} with {x\in B_1\cap B_2}, then there exists {B\in\mathcal B} such that {x\in B\subset B_1\cap B_2}.

The topology {\mathcal T} generated by {\mathcal B} is defined as

\displaystyle \mathcal T(\mathcal B) = \{U\subset X : \forall x\in U\;\exists B\in\mathcal B \text{ such that } x\in B\subset U \}.

Let’s verify that this actually defines a topology!

Theorem 2 The topology generated by a basis is indeed a topology.

Proof: The statement {\varnothing\in\mathcal T(\mathcal B)} holds vacuously. For any {x\in X}, there exists {B\in\mathcal B} such that {x\in B\subset X}, so {X\in\mathcal T(\mathcal B)}. If {U,V\in\mathcal T(\mathcal B)} then if {x\in U\cap V}, there exist {B_U,B_V\in\mathcal B} such that {x\in B_U\subset U} and {x\in B_V\subset V}. Hence there exists {B\in\mathcal B} such that {B\subset B_U\cap B_V\subset U\cap V}, so that {U\cap V\in\mathcal T(\mathcal B)}. If {U_\alpha\in\mathcal T(\mathcal B)}, {\alpha\in I}, let {U=\bigcup_{\alpha\in I}U_\alpha}. If {x\in U}, then {x\in U_\alpha} for some {\alpha\in I}, so there exists {B\in\mathcal B} with {x\in B\subset U_\alpha\subset U}. It follows that {U\in\mathcal T(\mathcal B)}. \Box

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