A sequence tests neighborhoods in a countable order. To reach every closure point in an arbitrary space, use one index for each neighborhood of the point and move toward smaller neighborhoods. This produces a net.
The ordering is not a time scale and need not be countable. Its purpose is to combine finitely many eventual requirements: after two neighborhood tests, there must be a later index satisfying both.
Predict
Predict the correct order on neighborhoods. Smaller neighborhoods must count as later, so the order is reverse inclusion. Their intersection is a common refinement and supplies a later index.
Predict whether a directed set must have a largest element. It need not. The natural numbers are directed without a greatest member, and neighborhood systems often behave similarly.
Experiment
Compare two index sets.
Directed set
Order
Common upper bound
Natural numbers
Usual order
Maximum of two indices
Finite subsets of a set
Inclusion
Union
Neighborhoods of x
Reverse inclusion
Intersection
Pairs of neighborhoods
Coordinatewise refinement
Pair of intersections
A net is simply a function from a directed set into the space. Its values need not be distinct, and its indices need not be points of the space.
Understand
A directed set is a nonempty preordered set in which every two indices have a common upper bound. A net x_d converges to x if every neighborhood U of x contains all values beyond some index d₀: whenever d≥d₀, x_d∈U. Directedness lets one combine finitely many such eventual statements.
If x∈closure A, index by the open neighborhoods U of x, ordered by reverse inclusion. Choose a_U∈A∩U. These intersections are nonempty by closure. Given a neighborhood V, every later neighborhood U⊆V gives a_U∈V. Thus the net converges to x. Conversely, any net in A converging to x makes each neighborhood meet A. This proves the exact net characterization of closure.
Continuity is equivalent to preserving all convergent nets. Continuity pulls target neighborhoods back to source neighborhoods. If it fails at x, choose a fixed target-open V around f(x) such that every source neighborhood U contains x_U with f(x_U) outside V. The neighborhood-indexed net converges to x but its image does not converge to f(x).
Hausdorffness is equivalent to uniqueness of net limits. The forward direction uses disjoint neighborhoods and a common later index. Conversely, if distinct x,y cannot be separated, index pairs of their neighborhoods by refinement and choose a point in each intersection. The resulting net converges to both, contradicting uniqueness. Unlike the sequence-only version, this converse is valid in every space.
Apply
Construct the neighborhood net approaching zero from R without zero in the cocountable topology. Its index collection is too rich to be replaced by an ordinary sequence that converges to zero, as the preceding lesson proved.
Repair “A net is just a sequence with a different name.” A sequence is a special net, but a general directed set need not have a countable cofinal subset. That extra indexing flexibility is exactly what makes the closure and continuity characterizations universal.
Continue
Compactness has useful net formulations, but open covers and filters will be our primary proof tools. Optional depth: formulate a subnet carefully using an order-preserving cofinal indexing map, and explain why merely selecting arbitrary net values does not guarantee preservation of a limit. The distinction matters in general compactness arguments.
Apply · explain · revise
Practice and reasoning
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How many finite subsets does a three-element set have, when all are used as indices ordered by inclusion?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
A directed index set must provide:
Prove every closure point x of A is a limit of a net in A.
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Prove f is continuous exactly when it preserves all convergent nets.
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Prove a space is Hausdorff exactly when every net has at most one limit.
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In cocountable R, let A=R without zero. Construct a net in A converging to zero and explain why no sequence in A can do so.
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Reflect and continue
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