Metric Spaces: Distance as an Axiom
Keep only the properties distance must have, and check how much of the analysis of the real line survives on that alone.
What a learner can do afterwards
- State the metric axioms and verify them for an unfamiliar example such as a function space
- Redefine convergence and continuity using only the metric
- Give two metrics on one set that disagree about which sequences converge
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A metric on a set X assigns a finite real distance d(x,y) to every pair. For all x,y,z in X, it satisfies four axioms: d(x,y)>=0; d(x,y)=0 exactly when x=y; d(x,y)=d(y,x); and d(x,z)<=d(x,y)+d(y,z). These are nonnegativity, identity of indiscernibles, symmetry and the triangle inequality. Verification must cover every allowed point. One counterexample disproves a proposed metric.
A formula resembling ordinary distance may fail an axiom. On the reals, d(x,y)=|x-y|^2 is nonnegative, symmetric and zero exactly when x=y, but its triangle inequality can fail. The formula d(x,y)=||x|-|y|| has another problem: opposite nonzero numbers have distance zero. This violates identity of indiscernibles.
For three distinct points with positive symmetric distances u,v,r, check all three triangle inequalities: r<=u+v, u<=r+v and v<=r+u. Together these give |u-v|<=r<=u+v, with r>0 also required. If two distances are 4 and 6, the third may be any value from 2 through 10. Set every diagonal distance to zero.
The usual real metric is d(x,y)=|x-y|. The discrete metric is 0 for equal points and 1 for distinct points. It obeys the triangle inequality: when x differs from z, at least one leg through y has distance 1. An open ball B(a,r)={x:d(x,a)<r} uses r>0 and a strict inequality. A discrete open ball with radius at most 1 contains only its center.
A sequence x_n converges to x if, for every epsilon>0, some positive integer N makes d(x_n,x)<epsilon for all n>=N. Equivalently, its distances to x tend to zero. Indices start at n=1. Find a tail inside an open ball by solving this strict inequality. For distance 1/n and radius 0.02, solve n>50. For distance 1/n^2, solve n^2>1/epsilon.
In the usual real metric, 1/n converges to 0. Discrete convergence to x requires eventual equality with x: apply epsilon=1/2. Eventual equality also suffices. Thus 1/n has no discrete limit. Different metrics need not disagree: e(x,y)=2d(x,y) is a metric because multiplying by 2 preserves each axiom. Its convergent sequences and limits agree with those of d, since 2d(x_n,x) tends to zero exactly when d(x_n,x) does.
For f:(X,d_X)->(Y,d_Y), continuity at a means that every epsilon>0 has a delta>0 such that d_X(x,a)<delta implies d_Y(f(x),f(a))<epsilon for every x in X. For f(x)=kx with usual real metrics and k!=0, the output gap is |k||x-a|. Choose delta=epsilon/|k|. Any larger delta admits an input gap epsilon/|k|, which gives output gap epsilon and fails the strict bound.
In metric spaces, f is continuous at a exactly when every sequence x_n tending to a has f(x_n) tending to f(a). Continuity controls late terms using its delta. If continuity fails, one epsilon>0 has a violating input inside every positive radius. Choose x_n with input distance below 1/n and output distance at least epsilon. Then x_n tends to a, but its images do not tend to f(a).
Here e denotes the fixed positive epsilon from failure of continuity.
Every function from a discrete metric space to any metric space is continuous. At any input a, fix epsilon>0 and choose delta=1. The condition d_X(x,a)<1 forces x=a. The output distance is then d_Y(f(a),f(a))=0<epsilon. This argument uses the discrete input metric and places no restriction on the function's values.
Let C([a,b]) be the real continuous functions on a closed bounded interval, with a<=b. Define d(f,g)=max_{t in [a,b]} |f(t)-g(t)|. The extreme value theorem gives a finite attained maximum. This distance is nonnegative and symmetric. A zero maximum means every gap is zero, so f=g. On other domains, continuity alone need not give a finite attained maximum.
To prove the uniform metric's triangle inequality, first use |f(t)-h(t)|<=|f(t)-g(t)|+|g(t)-h(t)| at each t. Bound the two right-hand gaps by their separate maxima d(f,g) and d(g,h). This bound is independent of t. Taking the maximum on the left gives d(f,h)<=d(f,g)+d(g,h), completing the metric verification.
To calculate a uniform distance, find the absolute gap function and maximize it on the stated interval. For f(t)=2t and g(t)=t on [0,3], the gap is t. Its maximum is 3 at t=3. More generally, a gap ct with c>=0 on [0,b], where b>=0, has maximum cb.
The four metric axioms let us define convergence and continuity. The chosen metric determines which sequences converge.
This lesson, its questions and its flashcards were written for this dataset.
These readings, practice questions and flashcards were written for this dataset.
More to read
A distance rule must pass every axiom
A proposed metric must assign a finite real number to every pair of points in its set. Check nonnegativity, zero distance exactly for equal points, symmetry, and the triangle inequality. Each condition must hold for every allowed choice of points.
The triangle inequality compares a direct distance with two distances through another point. It requires d(x,z)<=d(x,y)+d(y,z). A formula can pass the other three conditions and still fail this one. For squared real distance, the points 0, 2 and 4 give direct distance 16 and a two-distance sum of 8.
On a three-point set, specify zero diagonal distances and positive symmetric distances between distinct points. Then check the three triangle inequalities. If two distances are u and v, the third distance r must lie between |u-v| and u+v, inclusive, while remaining positive. Equality in a triangle inequality is allowed.
Use distances to define a limit
The open ball B(a,r) contains exactly the points x for which d(x,a)<r. Its boundary points are excluded. A sequence converges to a when every such ball contains all sufficiently late terms. The starting index of that tail may depend on the radius.
Continuity uses distances in both the input and output spaces. For every positive output tolerance epsilon, a positive input radius delta must control every input within that radius. For a real linear function f(x)=kx with k nonzero and usual distances, delta=epsilon/|k| works.
Continuity also has a sequence test in metric spaces: every sequence tending to a must have images tending to f(a). To see why failure is detected, fix the positive output tolerance that defeats continuity. For each n, choose a violating input within distance 1/n of a. These inputs converge to a, but their output distances from f(a) never fall below that tolerance.
Two metrics can treat one sequence differently
In the discrete metric, distinct points have distance 1 and equal points have distance 0. A convergent sequence must eventually equal its limit. Indeed, convergence with tolerance 1/2 leaves no other possible distance than 0. Conversely, eventual equality makes every later distance zero, which proves convergence.
The real sequence 1/n has distinct terms, so it cannot converge discretely. With the usual real metric, it converges to 0. The points and their order have not changed; the distances have.
Changing distances does not always change convergence. Given any metric d, define e(x,y)=2d(x,y). Multiplication by 2 preserves all metric axioms. For each proposed limit x, the e-distances tend to zero exactly when the d-distances do. These two metrics therefore have the same convergent sequences with the same limits.
Measure the largest gap between functions
For real continuous functions on a nonempty closed bounded interval [a,b], define their uniform distance as max |f(t)-g(t)| over the interval. The absolute gap is continuous, so the extreme value theorem provides a finite attained maximum. Continuity on an unrestricted domain does not provide that guarantee.
Absolute value gives nonnegativity and symmetry. The maximum gap is zero exactly when the functions agree at every input. For the triangle inequality, bound |f(t)-h(t)| by |f(t)-g(t)|+|g(t)-h(t)|. Each term on the right is bounded by its own maximum. The sum of those maxima bounds every left-hand gap and therefore also bounds its maximum.
For a numerical distance, calculate the gap function before maximizing it. On [0,b] with b>=0, a gap ct with c>=0 has maximum cb at the right endpoint. Maximizing either function alone would answer a different question.
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