# 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 ${U\in\mathcal T(\mathcal B)}$. $\Box$