Topological spaces, bases, subspace/product/quotient

In one sentence

A topology is a choice of which sets are open, everything else is defined from that choice, and the one construction that behaves unexpectedly is the infinite product — where the useful topology is the coarser one.

Why the exam asks it

Because generalising from metric spaces is where the intuition breaks: the definitions look familiar and the theorems have different hypotheses. And because the product topology on an infinite product is genuinely counterintuitive — the natural-looking definition is the wrong one.

The idea

A topology on $X$ is a collection $\tau$ of subsets containing $\emptyset$ and $X$, closed under arbitrary unions and finite intersections. Members are open; complements are closed.

A basis is a collection $\mathcal{B}$ whose unions give all open sets; equivalently, for each open $U$ and $x\in U$ there is $B \in \mathcal{B}$ with $x \in B \subseteq U$. Specifying a basis is how a topology is usually presented — open balls give the metric topology, half-open intervals give the Sorgenfrey line.

A topology $\tau_1$ is coarser than $\tau_2$ if $\tau_1 \subseteq \tau_2$. The indiscrete topology is the coarsest and the discrete the finest, and most constructions here are asking for the coarsest or finest topology with some property.

The three constructions

Subspace. $A \subseteq X$ gets $\{U \cap A : U \in \tau\}$. Note that open in $A$ is not open in $X$: $[0,\tfrac12)$ is open in $[0,1]$ and not in $\mathbb{R}$.

Product. For finitely many factors, open boxes form a basis and there is no subtlety. For infinitely many, two candidates exist:

• Box topology: all products $\prod U_i$ with each $U_i$ open.
• Product topology: those with $U_i = X_i$ for all but finitely many $i$.

The product topology is coarser, and it is the one meant by "product" unless stated otherwise. The reason is what it makes continuous:

$f : Y \to \prod X_i$ is continuous iff every coordinate $\pi_i \circ f$ is continuous.

That characterisation is the point of the definition, and it fails for the box topology. $f(t) = (t,t,t,\dots) : \mathbb{R}\to\mathbb{R}^\omega$ has every coordinate continuous; in the box topology $\prod(-\tfrac1n,\tfrac1n)$ is open and $f^{-1}$ of it is $\{0\}$, which is not. So $f$ is continuous into the product and not into the box.

Almost every good property — Tychonoff, connectedness, metrisability of countable products — holds for the product topology and fails for the box.

Quotient. For a surjection $q : X \to Y$, declare $V \subseteq Y$ open iff $q^{-1}(V)$ is open. This is the finest topology making $q$ continuous, and it is how the circle, the torus, the Möbius band and projective space are built. Quotients are where separation goes wrong: a quotient of a Hausdorff space need not be Hausdorff, and the line with two origins is the standard example.

Closure, interior, boundary

Defined exactly as in metric spaces, but only via open sets — sequences are not enough in general, which is why nets exist. In a first countable space (metric spaces included) sequences do suffice, and that is precisely the hypothesis under which metric intuition transfers.

Where intuition breaks

"The product topology has all products of open sets as basic opens." That is the box topology. The product topology restricts all but finitely many coordinates, and being coarser is a feature.

"Open in a subspace means open in the whole space." $[0,\tfrac12)$ in $[0,1]$. Openness is always relative to a topology, and the subspace has its own.

"A quotient of a Hausdorff space is Hausdorff." The line with two origins — glue two copies of $\mathbb{R}$ along $\mathbb{R}\setminus\{0\}$ — is a quotient of a Hausdorff space and is not Hausdorff. Quotients preserve very little.

"Arbitrary intersections of open sets are open." Finite only, exactly as in metric spaces. $\bigcap(-\tfrac1n,\tfrac1n) = \{0\}$.

"Any collection of subsets is a basis." It must cover $X$, and the intersection of two basis elements must contain a basis element around each of its points.

"Sequences determine the closure." Only in first countable spaces. In general a point can be in the closure of $A$ with no sequence from $A$ converging to it.

The exam's angle

1. Ask which topology is on the set. Half the options in this topic are the same set with a different topology, and the answer changes.
2. For products, use the coordinate criterion. Continuity into a product is continuity of each coordinate — that one line answers most product questions.
3. If an option is about an infinite product, check whether it needs the box topology to be false. Usually it does, and usually the intended topology is the product.
4. For quotients, expect separation to fail and check Hausdorffness explicitly.
5. Recall that finite intersections is the axiom whenever an option ranges over infinitely many open sets.

The night before

• Topology: arbitrary unions, finite intersections, contains $\emptyset$ and $X$.
• Basis generates; coarser $=$ fewer open sets.
• Subspace: intersect with $A$. Open in $A$ $\ne$ open in $X$.
• Product topology: all but finitely many coordinates unrestricted. Box: all restricted. Product is coarser and is the default.
• $f$ into a product is continuous $\iff$ each coordinate is. Fails for the box — $t\mapsto(t,t,\dots)$.
• Quotient: finest topology making $q$ continuous. Need not be Hausdorff — the line with two origins.
• Sequences suffice only in first countable spaces.