---
title: "Open Sets, Closed Sets and Limit Points"
description: "Rebuild openness, closure and limit points from balls alone, and restate continuity as preimages of open sets."
canonical: https://lightmysky.com/learn/mathematics/open-sets-closed-sets-and-limit-points-mt_9bBdT38dbB
source: https://lightmysky.com/learn/mathematics/open-sets-closed-sets-and-limit-points-mt_9bBdT38dbB.md
retrieved: 2026-11-12
---

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# Open Sets, Closed Sets and Limit Points

Rebuild openness, closure and limit points from balls alone, and restate continuity as preimages of open sets.

Subject: Mathematics · Area: Topology · Ages 20 to 21
Page: https://lightmysky.com/learn/mathematics/open-sets-closed-sets-and-limit-points-mt_9bBdT38dbB

## Ready when they can

- 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

## Lesson: Open sets, closed sets, and continuity

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.

**Example.** 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.

**Example.** 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.

*(drawing: Closure keeps every original point and adds the missing limit points.)*

**Example.** 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.

**Example.** 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.

**Recap.** 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.*

## Practice

17 questions on this page, each with its working shown.

## Needs first

- [Metric Spaces: Distance as an Axiom](https://lightmysky.com/learn/mathematics/metric-spaces-distance-as-an-axiom-mt_A3TMb8wL6k)

## Opens up

- [Completeness and the Contraction Mapping Theorem](https://lightmysky.com/learn/mathematics/completeness-and-the-contraction-mapping-theorem-mt_DYsLzg-IaK)