Testing every open set of the target is impossible for the line, where there are as many open sets as there are subsets of the rationals. Two shortcuts make continuity checkable: test only a basis — or even a subbasis — and test closed sets instead if they are more convenient.

A locksmith does not test every key ever cut. Testing the few blanks the others are copied from is enough, because everything else is built from them.

Checking $f(x)=x^2$ on a basis: the preimage of the basic interval $(a,b)$ with $0<a<b$ is $(-\sqrt b,-\sqrt a)\cup(\sqrt a,\sqrt b)$, a union of intervals and so open. Doing this for every interval is a finite piece of work; doing it for every open set is not.

Let $\mathcal{B}$ be a basis and $\mathcal{S}$ a subbasis for the topology on $Y$. Then $f$ is continuous exactly when the preimage of every basic set is open, exactly when the preimage of every subbasic set is open, and exactly when the preimage of every closed set is closed. Each is proved by the same two identities about preimages.

• On a map into $\mathbb{R}$ the usual test is the rays: $f$ is continuous exactly when $\{x:f(x)<b\}$ and $\{x:f(x)>a\}$ are open for all $a,b$.
• The closed-set test is the convenient one for a map defined by an equation: $\{x:f(x)=c\}$ is closed whenever $f$ is continuous into a space where points are closed.
• The three tests are shortcuts and not new definitions. Anything proved with one of them is a statement about the original condition.
• Preimages behave perfectly: they respect unions, intersections and complements. Images do none of this, which is why every test in this course is written with $f^{-1}$.
• A map out of a subspace is usually checked by writing it as a restriction, which the next lesson turns into a rule of its own.

Every open set of the target is a union of basic sets, and the preimage of a union is the union of the preimages. So the preimage of any open set is a union of open sets, which is open. The same argument runs for a subbasis with one extra step, since a basic set is then a finite intersection and preimages respect intersections as well; a finite intersection of open sets is open.

Write an arbitrary open set of the target as a union of basic sets.

The preimage of a union is the union of the preimages, with no condition on the map.

Each piece is open by hypothesis, and the union axiom finishes it.

For a subbasis, a basic set is a finite intersection and preimages respect intersections.

A finite intersection of open sets is open, so the basis test applies and the map is continuous.

Since every open set is a union of basic sets and preimages respect unions, testing the basis is testing everything.

• Testing preimages of basic sets is enough; testing subbasic sets is enough as well.
• A map is continuous exactly when preimages of closed sets are closed.
• Preimages respect unions, intersections and complements; images respect none of them.
• For a map into $\mathbb{R}$ the usual test is the two families of rays.

Great work! You now know how to:

• check continuity on a basis or a subbasis;
• use the closed-set test;
• compute a preimage of an interval;
• say why preimages and not images carry the definition.