Topological spaces · Topology

Intuition

With the axioms in hand, a topology can be written down directly: name a set, name a collection of its subsets, and check three conditions. Spaces built this way need not come from any distance at all, and the small ones can be listed in full — which makes them the cheapest place to test a claim.

A committee decides which groups of people count as close, subject to three procedural rules. Nobody measures anything; the minutes are the whole of the structure.

sets diagram: a-space-with-three-open-sets

The Sierpiński space: two points, and the open sets $\varnothing$ , $\{a\}$ and $X$ . The axioms hold — the union and the intersection of any of these is again one of them — and no metric produces it, since $b$ lies in no open set except the whole space. A topology this small can be checked by listing everything.

Writing a topology down

A topological space is a pair $(X,\tau)$ : a set, and a collection of its subsets satisfying the three axioms. On a finite set the collection can be listed, and the axioms checked case by case. Two families are worth meeting at once: the Sierpiński space above, and the cofinite topology, in which a set is open when it is empty or leaves out only finitely many points.

\tau_{\text{cof}}=\{\,U\subseteq X: U=\varnothing\ \text{or}\ X\setminus U\ \text{is finite}\,\}

Small spaces, and what they show

On $X=\{a,b\}$ there are exactly four topologies: the indiscrete one, the discrete one, and the two Sierpiński spaces $\{\varnothing,\{a\},X\}$ and $\{\varnothing,\{b\},X\}$ .

The cofinite collection is a topology

\tau_{\text{cof}}\ \text{satisfies the three axioms}

Complements turn the conditions into statements about finite sets. The complement of the whole space is empty and so finite, and the empty set is admitted outright. The complement of a union is the intersection of the complements, which is contained in any one of them and so is finite as soon as one of the sets is non-empty. The complement of a finite intersection is a finite union of finite sets, which is finite. Each axiom becomes a fact about finite sets that needs no topology at all.

Proof steps

X\setminus X=\varnothing\ \text{is finite},\quad \varnothing\in\tau_{\text{cof}}

The two sets the first axiom demands are there: one by the finiteness test, the other by the definition itself.

X\setminus\textstyle\bigcup_i U_i=\bigcap_i(X\setminus U_i)

To test a union, look at its complement, which De Morgan turns into an intersection of the complements.

\bigcap_i(X\setminus U_i)\subseteq X\setminus U_{i_0}\ \text{finite}

That intersection sits inside the complement of any one non-empty member, which is finite, so it is finite too.

X\setminus(U_1\cap\cdots\cap U_n)=(X\setminus U_1)\cup\cdots\cup(X\setminus U_n)

For a finite intersection the complement is a finite union of finite sets.

\text{finite union of finite sets is finite}

So that complement is finite as well, and all three axioms hold.

Applications

Practice

Three Checks and No More

To decide whether a listed collection is a topology, check the two forced sets, then unions, then intersections of finitely many.

\varnothing,X\in\tau;\quad \textstyle\bigcup U_i\in\tau;\quad U_1\cap\cdots\cap U_n\in\tau

Try it

On $X=\{a,b,c\}$ , which collection is a topology?