Teaching objectives
A university lab on the second half of the continuity and convergence topic: from the ε-δ definition to uniform convergence, passing through the topological characterisation. The material is the usual content of a first topology course; what the lab contributes is that each definition is manipulated before it is written down.
The lab opens without a single Greek letter: room 1 discusses what we mean when we call a function continuous, discards the pencil answer and keeps the idea that does generalise — if a and b are close, f(a) and f(b) have to be close —, measured on two functions: an ordinary one and one with a jump.
From there, the connecting thread is a division of roles that is kept across the 21 rooms and also in the colours: ε is demanded by an adversary (always amber on screen) and δ is chosen by us (always green). Room 2 is literally a duel: the adversary lowers ε three times and the student has to respond with a δ that holds. When the formula with its quantifiers appears in room 3, the order ∀ε ∃δ is no longer an arbitrary convention to memorise.
The lab is meant to accompany the explanation, not replace it. The proofs are given in full; what changes is that the student sees them being built before reading them.
What you learn
- Respond with a δ to a required ε, in three functions of increasing difficulty: a line (δ = ε/3, the same everywhere), a parabola (δ also depends on the point) and 1/x (δ collapses when approaching the edge of the domain).
- Distinguish "continuous at each point" from "uniformly continuous", watching the graph of δ(x₀) collapse.
- Negate the definition piece by piece and know what must be exhibited to state that a function is not continuous: a witness ε₀ against which no δ works.
- Check that continuity is not a property of the formula but of the pair of metrics, and that it is not even symmetric: with f(x) = x and the circular metric, one direction is continuous and the other is not.
- Recognise the equivalence between continuity, preservation of sequence limits and preimages of open sets, and know which of the three survives without distance.
- Measure distances between functions with d₁ and d∞, and understand that "converging" depends on the chosen metric just as much as "being continuous" does.
Key mathematical ideas
- The order of the quantifiers is the content of the definition. First the adversary chooses ε, then we choose δ; that is why δ can depend on ε, and in general also on the point. Reversing the order gives uniform continuity, which is a different thing.
- Negating a definition means changing who plays first. For continuity we respond to any ε; for discontinuity we exhibit a single ε₀ against which every δ fails. The lab builds that sentence piece by piece instead of handing it over ready-made.
- The circular metric on [0;1) —
d(x,y) = min(|x−y|, 1−|x−y|)— turns the interval into a circle and gives the central counterexample: the identity from the circular metric to the Euclidean one is not continuous at 0, and it fails for every ε < 1. In the other direction it is continuous, with δ = ε. Gluing is continuous; cutting is not. - The preimage characterisation mentions no points, no ε, no δ, no distances; that is why it can be taken as the definition where there is no metric. With images it would be false: x² is continuous and maps the open set (−1;1) to [0;1). The asymmetry has a reason: f⁻¹ respects unions, intersections and complements; f does not.
- It is in the converse of the continuity-and-convergence theorem where the metric is used, and in one very specific place: when choosing δ = 1/n, that is, when a countable family of nested neighbourhoods is available. Without it the equivalence can fail and nets or filters are needed.
- The same set of functions admits metrics that give opposite verdicts: xⁿ converges with d₁ and does not converge with d∞. The useful question is not "does it converge?" but "which metric preserves what matters to me?" With d∞ both continuity and the integral are preserved; with pointwise convergence, neither is.
Notes for the classroom
- The lab covers the second half of the topic (from "Continuity" onward). It assumes balls, open sets, closed sets and boundary are already known, which make up the first half.
- Rooms 1, 2, 9, 14 and 21 work best projected. Room 1 is a class discussion: it is worth letting the bad answers come out (the pencil, the formula with no pieces) before touching anything. Room 2 can be played as a whole class: the teacher acts as the adversary and lowers ε until the class protests. Room 9 is the circular-metric counterexample, and is usually the moment when it becomes clear that continuity does not live in the formula. Room 14 builds the converse step by step and it is worth pausing it at each step. Room 21 closes with the three-thirds trick.
- Room 5 (the parabola) is the hardest: the student accepts that δ depends on ε much sooner than accepting that it depends on the point. It is worth pausing on the graph of δ(x₀) before moving on to room 6.
- Block E (rooms 18-21) can be given separately, in the function-spaces session, without having done the rest that same day.
Room-by-room contents
Room 1 · Close lands close
The opening room, without a single Greek letter. It discusses what we mean when we call a function continuous: the pencil answer is a good intuition but does not work as a definition (you cannot draw a function from ℝ² to ℝ³, nor one defined only over the rationals, nor one whose points are functions — and all three appear in this lab), and what remains is the idea that does generalise: if a and b are close, f(a) and f(b) must be close. The student fixes a at 0, moves b and reads both distances at once — the one between the arguments below in blue, the one between the images on the left in amber — on two functions: an ordinary curve, where both go to zero, and one with a jump, where the bottom one goes to zero and the left one does not budge. It closes by pointing at the loose end: “as close as desired” and “close enough” are ways of speaking, not numbers, and turning them into numbers is the next room.
Student tasks
- Compare the two distances on the curve and on the jump function, bringing b to 0 from the left.
- Decide what happens to the distance between the images as b approaches the point a.
- Decide whether, with the jump, the images can be brought within 0.5 of each other.
- Choose which of four sentences works as a definition of continuity.
Room 2 · The ε-δ game
The definition of continuity played as a duel before it is written down. The lab acts as the adversary and demands an ε (the amber horizontal corridor); the student answers with a δ (the green vertical strip), moving a slider until the piece of graph inside the strip no longer leaves the corridor. Three rounds, with a smaller ε each time. The colour split — amber for what is demanded, green for what we choose — is kept throughout all 21 rooms.
Student tasks
- Clear the three rounds by finding a valid δ for each ε.
- Answer who chooses first and why the order matters.
Room 3 · The definition, at last
What was played in the previous room, now written with symbols and drawn with the two balls: the source space with the δ ball and the target space with the ε ball, joined by the arrow f. An animation carries the whole green ball inside the amber one, and a second variant shows what it looks like when the chosen δ is too large.
Student tasks
- Identify which of four formulas is the correct definition.
- Explain what it means for the whole green ball to fit inside the amber one.
Room 4 · A line: the same δ everywhere
The first explicit computation, with the worked example in view: for f(x)=3x we have |f(x)−f(x₀)| = 3|x−x₀|, so δ = ε/3 suffices. The student then moves the point x₀ along the line and checks that the answer still holds: the slope is the same everywhere.
Student tasks
- Compute the largest working δ for two different values of ε.
- Determine what δ depends on in this case.
Room 5 · A parabola: δ also looks at where you are
A single graph of y=x², a single ε and a magnifier at fixed zoom; as x₀ moves with the slider, the system works out the largest working δ by itself and displays it next to ε, which does not move. Between x₀ = 1 and x₀ = 10 the δ drops from 0.227 to 0.025, and the green strip in the magnifier narrows to a thread. The bound from the course follows — taming |x+x₀| by first demanding |x−x₀|<1, which yields δ = min(1, ε/(1+2|x₀|)) — and the room closes with the graph of δ(x₀) up to x₀ = 10, which collapses: continuous at every point, but not uniformly continuous.
Student tasks
- Compare the largest working δ at a point near the origin and at a far one.
- Decide what happens to that δ as x₀ moves away, with the same ε.
- Apply the formula δ = min(1, ε/(1+2|x₀|)) to a concrete case.
Room 6 · 1/x and the point that is not there
The point x₀ approaches the origin and the readout of the largest valid δ collapses on screen. The bound is explained in full and in pieces: the trouble is that x itself appears in |x−x₀|/(|x||x₀|), and δ must be handed over before knowing which x the adversary will try, so first x is kept away from 0 (if |x−x₀| < x₀/2 then |x| > x₀/2 and the difference is < 2|x−x₀|/x₀², which depends only on x₀) and only then is ε brought in. Hence δ = min(x₀/2, εx₀²/2), and two numerical examples show which of the two rules: with x₀ = 3 and ε = 0.1 the second one, with ε = 1 the first. The room turns on the question of whether 1/x is discontinuous at 0: it is not, because 0 does not belong to the domain, and you cannot be discontinuous where you are not defined.
Student tasks
- Observe how the largest valid δ changes as the point approaches the origin.
- Compute the δ given by the formula for a concrete ε and x₀.
- Decide what exactly happens at x = 0.
Room 7 · Negating the definition
The negation is built piece by piece: ∀ε becomes ∃ε₀, ∃δ becomes ∀δ, and the implication keeps the hypothesis and negates the conclusion. Each correct answer writes another piece of the full statement on screen. Once it is complete, it is seen in action on the step function with ε₀ = 1/2 fixed: however small δ is, there is always a witness point whose image escapes.
Student tasks
- Build the three pieces of the negation.
- Check on the step function that no δ works.
- Determine the largest ε₀ that serves as a witness for a jump of height 1.
Room 8 · Rolling up the interval
An animation curves the segment [0;1) until it closes into a circle, and the points 0.05 and 0.95 — miles apart on the line — end up neighbours. An explorer with two movable points then shows |a−b|, 1−|a−b| and the minimum of the two at once, with the short arc highlighted. It closes with the M-40 ring road: going clockwise, the exit before number 1 is number 57.
Student tasks
- Watch the rolling up and locate which points end up as neighbours.
- Compute three circular distances, one of them with the long path as a trap.
Room 9 · The identity map that breaks
The picture from the course, now interactive: the circle on the left (the source, with the circular metric) and the interval on the right (the target, with the euclidean one). A single slider moves the green point on both at once, and the two distances to 0 are read underneath: with x = 0.98 the source reads 0.020 and the target 0.980. The slider covers exactly the zone that matters, from 0.6 to almost 1: as it is dragged, the green arc in the source shrinks while the red segment in the target grows, and the arriving point turns red as soon as it leaves the ε = 0.5 ball, which is drawn from the start. No animations: the three questions are answered on that same picture, which is not erased.
Student tasks
- Move the green point and compare both distances to 0, especially above x = 0.9.
- Decide what this identity map does to close points.
- Locate the only point where continuity fails, and for which values of ε.
Room 10 · Gluing is continuous, cutting is not
The same formula with the metrics swapped, and the previous room’s picture turned around: now the interval (Euclidean) is the source and goes on the left, and the circle (circular) is the target and goes on the right. Two movable points, a and b, live in both spaces at once: on the left their distance is the blue stretch of line, on the right the short amber arc, which never comes out longer. Two comparison bars show that the circular distance never exceeds the euclidean one, and the student may hunt for a counterexample as long as they like: there is none, which is why δ = ε always works. The conclusion is that continuity is not a property of the formula but of the pair of metrics, and it is not even symmetric.
Student tasks
- Hunt for a pair of points where the circular distance exceeds the euclidean one.
- Explain why δ = ε works in this direction.
- Mark the correct statements about what continuity depends on.
Room 11 · From the circle to the circle
Circular metric on both sides. A test bench with two circles side by side: the student picks a candidate map and watches the source point travel along with its image. There is a fixed red point at x = 0.01, right next to the gluing point, and the green one goes all the way round starting from the opposite side, slowly. Both distances to the red point are drawn and read at once: the blue arc in the source and the amber one in the target. By the end of the trip the blue one is almost nothing for all four candidates; so is the amber one for rotation, reflection and doubling, but for f(x)=x/2 it stays at 0.49 — it sends 0 to 0 and the neighbours of 0 on the other side near 0.5. The lab does not say which one breaks: the two numbers do.
Student tasks
- Try the four candidates on the test bench.
- Mark all the continuous ones.
Room 12 · A sequence and its balls
The definition of a limit read for what it says: outside each ball there are at most finitely many terms. On a real line with numbered terms, the student tightens the radius ε and sees how many are left outside and which N will do. It closes with uniqueness of the limit: with ε = d(a,b)/2 the two balls are disjoint and cannot both contain almost all the terms.
Student tasks
- Count how many terms fall outside the ball for a given ε.
- Justify why the limit of a sequence is unique.
Room 13 · If it is continuous, it preserves limits
The easy direction of the theorem, animated across the two spaces: from some N on, the terms enter the green ball and their images are trapped in the amber one. The student then rebuilds the order of the argument, which is the only thing worth being able to reproduce: fix ε, get δ from continuity, get N from convergence, conclude.
Student tasks
- Watch the tail of the sequence drag its images into the amber ball.
- Order the four steps of the proof.
Room 14 · The converse, step by step
The proof by contraposition, built one step at a time on the two-space diagram. We assume f is not continuous, the negation hands us a bad ε₀, and applying it with δ = 1, 1/2, 1/3, … the points xₙ appear inside ever smaller green balls while their images pile up outside the amber one. The sequence converges to x and its images do not converge to f(x): contradiction. This is the room most worth pausing at every step.
Student tasks
- Follow the construction step by step and watch it again.
- Rebuild the order of the steps.
- Identify where the fact that the space is metric has been used.
Room 15 · Continuity without distances
The characterisation that mentions no points, no ε, no δ and no distances. The student drags an open set U along the target axis and sees its preimage drawn on the source axis: with a continuous function it always comes out open; with the step function, placing U around the upper value makes the preimage contain its own boundary point and stop being open. It closes with the dual version for closed sets.
Student tasks
- Move U with both functions and observe the preimage.
- Place U so that the preimage stops being open.
- Explain why that proves the step function is not continuous.
Room 16 · Preimage yes, image no
Why the characterisation talks about preimages and cannot talk about images: x² is continuous and sends the open set (−1;1) to [0;1), which is not open. A point travels along the source interval and its image bounces at 0, which is precisely the endpoint that is attained. The underlying reason: f⁻¹ respects unions, intersections and complements; f does not.
Student tasks
- Determine the image of the open interval (−1;1) under x².
- Explain why the characterisation uses preimages.
Room 17 · Chaining two continuous maps
Three spaces and two arrows. An open set W in the third space is chased backwards: g⁻¹(W) in the second, f⁻¹(g⁻¹(W)) in the first, and all three are open. The topological proof takes two lines; the ε-δ version of the same result, with its intermediate η, is shown alongside so the difference in length is visible.
Student tasks
- Follow the open set being chased backwards.
- Identify what (g∘f)⁻¹(W) is equal to.
- Order the two lines of the topological proof.
Room 18 · Two ways of measuring functions
The conceptual jump of the block: the points of the space become functions. A spike of adjustable height and width is added to a constant function, and both distances are read at once: the area between the curves (d₁) and the bar at the worst point (d∞). The challenge asks for a case with large d∞ and tiny d₁, which is exactly the tall narrow spike.
Student tasks
- Make d∞ exceed 0.4 while keeping d₁ below 0.05.
- Compute d∞ for a given spike height.
Room 19 · xⁿ and its two verdicts
The curves xⁿ flatten against the axis and rise abruptly next to x = 1, which never moves. The pointwise limit is discontinuous, so it has left C[0;1]. And the verdict on convergence depends on the metric: with d₁ it equals 1/(n+1) and tends to 0, with d∞ it equals 1 for every n and does not converge.
Student tasks
- Compute the limit of 0.9ⁿ and describe the limit function.
- Compute d₁ between xⁿ and the zero function for a concrete n.
- Decide whether the sequence converges under the supremum metric.
Room 20 · The area that escapes
Spikes of height n over the interval [0;1/n]: taller and narrower each time, all of area 1. For each fixed x the sequence eventually equals 0, so the pointwise limit is the zero function and its integral is 0; yet the integral of each fₙ equals 1. The area is not lost — it escapes upwards.
Student tasks
- Compute the integral of fₙ for a concrete n.
- Determine the pointwise limit of the sequence.
- Explain why the limit cannot be exchanged with the integral.
Room 21 · The tube: uniform convergence
The closing room. Uniform convergence drawn as a tube of width 2ε around the limit: the sequence converges uniformly if it ends up entirely inside, and whatever leaves the tube is coloured red. Four sequences are compared, two that get in and two that do not. It ends with the three-thirds trick animated summand by summand and with the summary of the topic: three equivalent definitions of continuity, and only one survives without a distance.
Student tasks
- Test the four sequences with the tube and mark those that converge uniformly.
- Follow the animation of the three-thirds trick.
- Decide what uniform convergence preserves.
Rooms to project
The most striking ones to show and discuss in class.