Topological Spaces

Continuing review of Kolmogorov’s book

definition: a topology of a set $X$ is a collection of subsets $\tau$ of $X$ whose elements are called open sets such that both the empty set and $X$ are open sets, the union of any subcollection of open sets is an open set, and the intersection of any finite subcollection of open sets is an open set. A topological space is a set $X$ equipped with a topology of $X$ . A closed set is the complement of an open set.

definition: a neighbourhood of a point is an open set containing the point. A contact point of a subset $E$ is a point $x$ such that every neighbourhood of $x$ contains a point of $E$ . A limit point of $E$ is a point such that every neighbourhood of the point contains an infinite number of points of $E$ . The closure of $E$ is the set $[E]$ of all contact points of $E$ .

These definitions are motivated by abstracting key properties of open sets and related concepts of metric spaces.

Topologies on a common set can be compared by set inclusion.

theorem: the intersection of any collection of topologies on a set $X$ is a topology on $X$ .

The intersection of the collection is the set of all elements common to all sets in the collection. Since the empty set and $X$ are open sets in every topology they are elements of the intersection.

Given a (finite) collection of sets in the intersection, the collection is a (finite) collection of open sets in any given topology and therefore the union (intersection) is an open set in any given topology. Hence the union (intersection) is an element of the intersection.

Hence the intersection is a topology.

corollary: for every subcollection of sets of $X$ , there is a smallest topology on $X$ containing it.

The collection of all topologies containing the collection is nonempty since $2^X$ is a topology containing any collection. So the intersection of the nonempty collection is a topology. Since the subcollection is by construction contained in every topology in the collection, it is also contained in the intersection. The intersection is the smallest topology in the collection since every other topology in the collection contains it and the intersection also belongs to the collection.

definition: A base for a topology is a collection of open sets such that the collection of unions of subcollections of the base is identical to the topology.

A base is useful in specification since a smaller collection need only be specified to identify a toplogy. A base always exists for a topology since the topology is itself a base.

theorem: If a collection $G$ of sets of a given set $X$ is such that it contains the empty set and the collection covers $X$ and the intersection of any pair of sets in the collection $G$ is the union of sets in the collection $G$ , then the collection $G$ is a base for the collection $T$ of all unions of sets in the collection.

Let $G,T$ be so identified in the statement of the theorem. If $T$ is a topology on $X$ , then $G$ is a base for $T$ . Note that

Since $G$ is a cover for $X$ , it follows that $T$ contains $X$ . It also contains the empty set since $T$ contains $G$ . Let $O$ be a subcollection of $T$ . Then every element of $O$ is a union of sets in $G$ , and so the union of sets of $O$ is a union of sets in $G$ and so is an element of $T$ . Finally, if $A = \cup O_A, B = \cup O_B$ where $O_A,O_B$ are subcollections of $G$ , then $x \in A \cap B$ iff there is an $o \in O_A, p \in O_B$ such that $x \in o \cap p$ . But since $o \cap p = \cup O_x$ for some collection $O_x$ of $G$ , it follows that $A \cap B = \cup \{\cup O_x : x \in A \cap B\}$ is the union of a collection of $G$ . Hence $T$ is a topology.

Topological spaces can be classified by properties of their base.

definition: Given a point of a topological space $(X,\tau)$ , a neighbourhood base (local base) for $x$ is a base for the neighbourhoods of $x$ .

definition: A topological space satisfies the first axiom of countability if every point of the space has a countable neighbourhood base, and satisfies the second axiom of countability if it has a countable base.

theorem: If topology $T$ on $X$ has a countable base, then every open cover has at most a countable subcover.

Let $B$ be a countable base for $T$ . Let $O$ be a subcollection of $T$ . For each $x \in \cup O$ choose $b(x) \in B$ such that there is an $o \in O$ such that $b(x) \subseteq o$ . Then $c = \{ b(x) : x \in X\}$ is a subcollection of $B$ and therefore at most countable. Let $\{c_i\}$ be an enumeration of $c$ and choose $o_i \in O$ such that $c_i \subseteq o_i$ . Then $\{o_i \}$ is an at most countable subcollection of $O$ that covers the union..

Topological Spaces

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