Subbases and initial topologies

Try

Suppose a set X is observed through two real-valued measurements. If each measurement is to be continuous, the inverse image of every real-open interval must be open in X. Those forced sets may not yet form a basis, because intersecting two observations imposes simultaneous conditions.

A subbasis records the initial conditions before intersections are added. The topology they generate includes finite simultaneous tests and arbitrary alternatives, expressed by intersections and unions respectively.

Predict

Predict why finite intersections, rather than only unions, are needed. On a three-point set, the observations that select the pairs a,b and b,c jointly select only b. That singleton must become open.

Predict which topology makes the measurement maps continuous with the fewest added opens. It should be the coarsest topology containing their forced inverse images, not the finest. The discrete topology would always make them continuous but would discard the minimality question.

Experiment

Generate the small example explicitly.

Operation	Result
First subbasic set	Pair a,b
Second subbasic set	Pair b,c
Their intersection	Singleton b
Empty intersection	Whole space
Empty union	Empty set

These five sets already form a topology. More complicated subbases can create many distinct finite intersections and many further unions.

Understand

Given any family S of subsets of X, take all finite intersections of members of S, including the empty intersection X. This family covers X and is closed under finite intersections, so it is a basis. Arbitrary unions of these finite intersections form the topology generated by S. This convention works even if the original S does not cover X.

For maps f_i:X→Y_i into topological spaces, take the subbasic sets f_i⁻¹(U) with U open in Y_i. The generated topology is called the initial topology for the maps. Each f_i is continuous because all its required inverse images are included. Conversely, any topology making all the maps continuous must contain every such inverse image, hence all their finite intersections and unions. This proves the claimed coarsest property.

There is a mapping criterion worth proving now. A map h:Z→X into this initial topology is continuous exactly when all composites f_i∘h are continuous. Necessity is composition. For sufficiency, the inverse image of a subbasic set is (f_i∘h)⁻¹(U), open by assumption. Since inverse images commute with unions and finite intersections, the same is true for every generated open set.

This criterion is the reason product topologies are defined by their coordinate projections. It lets one establish continuity of an entire map by checking its observed coordinates.

Apply

Let the sole measurement be f(x)=x² from the set of real numbers to the usual real line. Every induced open set is symmetric under x↦−x. The points one and minus one therefore have identical neighborhoods in this induced topology.

Diagnose “The induced topology is usual because the formula is a usual continuous function.” Continuity in the usual topology only proves that the induced topology is no finer than the usual one. The nonsymmetric interval (1,2) is usual-open but absent from the induced topology, proving strictness.

Continue

The next lesson is another initial construction: a subset inherits the smallest topology making its inclusion continuous. Optional depth: express the Euclidean plane topology using inverse images of real-open intervals under the two coordinate projections, then identify the finite intersections as open rectangles.