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Separated Union Topological Space 📂Topology

Separated Union Topological Space

Definition

Let {Xα}α∈A\left\{ X_\alpha \right\}_{\alpha \in A} be an arbitrary topological space index family. Let us say u⊂⨆α∈AXαu \subset \bigsqcup \limits_{\alpha \in A} X_\alpha. Then, for all α∈A\alpha \in A, if u∩Xαu \cap X_\alpha is an open set in Xα X_\alpha, then uu is said to be an open set ∗^{\ast} in ⨆α∈AXα\bigsqcup \limits_{\alpha \in A} X_\alpha.

  • The so-called open∗^{\ast} here is not exactly open in the sense of topology. However, by collecting subsets uu that satisfy such conditions, one can verify that it actually becomes a topology of ⨆α∈AXα\bigsqcup \limits_{\alpha \in A}X_\alpha. Thus, it is called open.

Theorem

The Disjoint Union as a Topology

Let T\mathcal{T} be a collection of open∗^{\ast} subsets in ⨆α∈AXα\bigsqcup \limits_{\alpha \in A}X_\alpha. And let Tα\mathcal{T}_\alpha be the topology of XαX_\alpha.

  • (0)(0): Then T\mathcal{T} is the topology of ⨆α∈AXα\bigsqcup \limits_{\alpha \in A}X_\alpha and this is called the disjoint union topology.

Properties of the Disjoint Union Topology

Let ⨆α∈AXα\bigsqcup \limits_{\alpha \in A} X_\alpha be a disjoint union topological space. Then

  • (a)(a): Let YY be any topological space. Then, it is equivalent that f : ⨆α∈AXα→Yf\ :\ \bigsqcup \limits_{\alpha \in A} X_\alpha \rightarrow Y is continuous and both ∀ β∈A\forall\ \beta\in A, f∘ι : Xβ→Yf\circ \iota\ :\ X_\beta \rightarrow Y are continuous.
  • (b)(b): The disjoint union topology is the unique topology on ⨆α∈AXα\bigsqcup \limits_{\alpha \in A} X_\alpha that satisfies (a)(a).
  • (c)(c): A subset F⊂⨆α∈AXαF \subset \bigsqcup \limits_{\alpha \in A} X_\alpha being a closed set is equivalent to F∩XαF\cap X_\alpha being a closed set in XαX_\alpha for all α∈A\alpha \in A.

Proof

(0)(0)

Strategy: The proof directly confirms whether it satisfies the three conditions to be a topology.

(1)(1)

For ∅,⨆α∈AXα\varnothing, \bigsqcup \limits_{\alpha \in A}X_\alpha and all α∈A\alpha \in A, we have ∅∩Xα=∅∈Tα(⨆α∈AXα)∩Xα=Xα∈Tα \varnothing \cap X_\alpha=\varnothing \in \mathcal{T}_\alpha \\ \left( \bigsqcup \limits_{\alpha \in A}X_\alpha \right) \cap X_\alpha=X_\alpha \in \mathcal{T}_\alpha therefore ∅,⨆α∈AXα∈T \varnothing, \bigsqcup \limits_{\alpha \in A}X_\alpha \in \mathcal{T}

(2)(2)

Let us denote ui∈T,∀ i∈Nu_{i} \in \mathcal{T},\quad \forall\ i\in \mathbb{N}. Then, by definition, we have ui∩Xα∈Tα,∀ i∈N u_{i}\cap X_\alpha \in \mathcal{T}_\alpha,\quad \forall\ i\in \mathbb{N} However, (⋃i=1∞ui)∩Xα=⋃i=1∞(ui∩Xα) \left( \bigcup \limits_{i=1}^\infty u_{i}\right)\cap X_\alpha=\bigcup \limits_{i=1}^\infty \left( u_{i} \cap X_\alpha \right) and since the countable union of open sets is an open set, (⋃i=1∞ui)∩Xα∈Tα \left( \bigcup \limits_{i=1}^\infty u_{i} \right) \cap X_\alpha \in \mathcal{T}_\alpha thus ⋃i=1∞ ui∈T \bigcup \limits_{i=1}^\infty\ u_{i} \in \mathcal {T}

(3)(3)

Let us denote u1,u2∈Tu_{1}, u_2 \in \mathcal{T}. Then, by definition, we have ui∩Xα∈Tα,i=1,2 u_{i}\cap X_\alpha \in \mathcal{T}_\alpha,\quad i=1,2 However, (u1∩u2)∩Xα=u1∩u2∩Xα∩Xα=(u1∩Xα)∩(u2∩Xα) (u_{1} \cap u_2)\cap X_\alpha=u_{1}\cap u_2\cap X_\alpha \cap X_\alpha=\left( u_{1}\cap X_\alpha \right) \cap \left(u_2\cap X_\alpha \right) and since the intersection of open sets is also an open set, (u1∩u2)∩Xα∈Tα \left( u_{1} \cap u_2 \right) \cap X_\alpha \in \mathcal{T}_\alpha thus u1∩u2∈T u_{1}\cap u_2 \in \mathcal {T}

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