Open Sets, Closed Sets and Limit Points
Rebuild openness, closure and limit points from balls alone, and restate continuity as preimages of open sets.
What a learner can do afterwards
- Prove that arbitrary unions and finite intersections of open sets are open
- Compute the closure and the interior of a given subset
- Prove that a function is continuous exactly when preimages of open sets are open
1 · Read
Work in a metric space X. Write B_X(a,r)={x in X:d(a,x)<r}, where r>0. A is open if every a in A has a ball contained in A. Its interior consists of these points and is its largest open subset. On R use d(x,y)=|x-y|. For u<v, (u,v) excludes both endpoints and [u,v] includes them. The interior of [u,v] is (u,v): endpoints have no ball inside, but every point between them does.
A point in a union of open sets belongs to some member. That member supplies a ball contained in the union. For a nonempty finite intersection, choose a ball at the point in each member; the minimum of their positive radii is positive and works in all members. Thus arbitrary unions and finite intersections are open. The empty union is empty; the empty intersection is X. Both are open. Infinite intersections need not be open.
In R, let U_n=(-2/n,2/n) for positive integers n. Every U_n is open and contains 0. For x!=0, choose n>2/|x|; then x is outside U_n. Their intersection is {0}. Every ball B(0,r) contains the nonzero point r/2, so no positive-radius ball at 0 fits inside {0}. This infinite intersection is not open.
A limit point of A is a point x such that every positive-radius ball at x meets A at a point other than x. A is closed if X without A is open, equivalently if A contains all its limit points. The closure is A plus all its limit points, the smallest closed set containing A. Arbitrary intersections of closed sets are closed: their complements are unions of open complements, hence open.
For u<v in R, closure((u,v))=[u,v]. Every ball at either endpoint meets the interval; outside points have balls missing it. The half-open interval [u,v) has interior (u,v) and closure [u,v], so it is neither open nor closed. If u<v<w, the set (u,v) union (v,w) has closure [u,w]. Every ball at the missing point v meets the set. The interior of that closure is (u,w), filling only the middle gap.
A finite subset of R has no limit points. At a member, choose a ball excluding all other members. At an outside point, choose a ball missing the whole finite set. In the discrete metric, distinct points have distance 1 and identical points have distance 0. Here B_X(x,1/2)={x}, so every subset is open. Every complement is open too, making every subset closed.
In R, let A={3/n:n>=1}. Every ball B(0,r) meets A: choose n>3/r. Thus 0 is a missing limit point. If x is outside A and x!=0, the tail lies below |x|/2 and stays away from x. Choose a ball avoiding that tail and the finitely many earlier terms. Hence closure(A)=A union {0}. The same argument works for {c/n:n>=1} with any c>0.
If x lies in B(a,R), let r=R-d(a,x)>0. For d(x,y)<r, the triangle inequality gives d(a,y)<=d(a,x)+d(x,y)<R. Thus B(x,r) lies inside B(a,R), proving that metric balls are open. For finitely many balls containing x, use the smallest margin. In R this gives the largest possible radius: a larger ball crosses a nearest excluded endpoint.
For f:X->Y, f^-1(V)={x in X:f(x) is in V}. This is a set of inputs, not an inverse function. At a, continuity means: for every epsilon>0, some delta>0 makes d_X(a,x)<delta imply d_Y(f(a),f(x))<epsilon for all x in X. A continuous map satisfies this at every a.
For continuous f and open V, take a in f^-1(V). Choose epsilon>0 with B_Y(f(a),epsilon) inside V. Continuity gives a source ball mapped into that target ball, so f^-1(V) is open. Conversely, if all open preimages are open, f^-1(B_Y(f(a),epsilon)) contains B_X(a,delta) for some delta>0. This gives the epsilon-delta implication. Hence f is continuous exactly when every open target set has an open preimage.
A constant f:R->R has preimage R when its value lies in V, and empty otherwise; both are open. Its image of any nonempty open set is a singleton, not open in R. For g:R->R with g(x)=2 for x<=3 and 5 for x>3, the open target (1,3) has preimage (-infinity,3]. No ball at 3 fits inside, since each contains inputs greater than 3. This nonopen preimage proves g is not continuous.
For f(x)=kx on R with k!=0, |f(x)-f(a)|=|k||x-a|. Therefore f^-1(B(f(a),epsilon))=B(a,epsilon/|k|). The largest source radius whose ball maps into this target ball is delta=epsilon/|k|. A larger radius includes an input whose image is on the excluded target boundary.
Balls decide openness, limit points decide closedness, and preimages decide continuity.
Lesson teaching, questions and worked solutions written for this dataset. Course and problem records were consulted for scope and question design; no external prose was adapted.
Supplementary readings, questions, worked solutions and study cards written for this dataset.
More to read
Choosing a ball for a proof
To prove a set open, start with an arbitrary point of that set. Your task is to find a positive radius whose entire ball stays inside. The radius may depend on the point.
For a union, membership gives one open member containing the point. A ball inside that member also lies inside the union. Other members need not contain the point, and no shared radius is required.
For a finite intersection, the point belongs to every member. Choose a positive radius for each member and take their minimum. This minimum remains positive because there are only finitely many radii. Its ball lies in all the members.
The same step can fail for an infinite family. In R, the open intervals (-2/n,2/n) intersect to {0}. Every nonzero point is eventually excluded as n grows. Although 0 remains, every ball around it contains a nonzero point. The intersection is therefore not open.
Complements give a related result for closed sets. The complement of an intersection is the union of the complements. If all the original sets are closed, their complements are open. That union is open, so the original intersection is closed, even for an infinite family.
Membership, interior and limit points
Three questions about a point are different: does it belong to A, does it have a ball inside A, and does every ball around it meet A at another point?
The second question defines an interior point. The third defines a limit point. A limit point need not belong to A. Conversely, a point of A need not be a limit point. A finite real set illustrates this: each member can be isolated from the others by a small ball, so the set has no limit points. Its closure still contains all its original members.
Closure is A together with its limit points. Interior contains the points with a whole positive-radius ball inside A. In the usual real metric, [u,v) has interior (u,v) and closure [u,v], when u<v. The included endpoint u is not interior; the omitted endpoint v belongs to the closure.
The ambient metric matters. In a discrete metric, a radius-1/2 ball contains only its center. Every subset is therefore open, and its complement is open too. Thus every subset is also closed. A singleton in this metric behaves differently from a singleton in R with its usual metric.
Recovering continuity from preimages
A preimage collects inputs. For f:X->Y and V contained in Y, f^-1(V) contains exactly the x for which f(x) belongs to V. No inverse function is needed.
Suppose f is continuous and V is open. At a point a in f^-1(V), choose a target ball around f(a) contained in V. Continuity supplies a source ball around a mapped into that target ball. Hence the source ball lies in f^-1(V). This proves that the preimage is open.
For the converse, assume every open target set has an open preimage. Fix a and epsilon>0. The target ball B_Y(f(a),epsilon) is open. Its preimage is open and contains a, so it contains B_X(a,delta) for some delta>0. Every input within delta of a then maps within epsilon of f(a). This is the continuity condition.
The radius can sometimes be computed exactly. For f(x)=kx on R with k nonzero, output distances are |k| times input distances. The preimage of the target epsilon-ball at f(a) is the source ball of radius epsilon/|k| at a. Increasing that radius admits an input whose output reaches the excluded target boundary.
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