Compactness & Connectedness Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice compactness and connectedness: open covers and finite subcovers, the Heine-Borel test in \(\mathbb{R}^n\), compact sets as closed and bounded in Euclidean space, sequential compactness in metric spaces, closed subsets and finite unions of compact sets, continuous images, extreme values, uniform continuity on compact metric spaces, separations, intervals as connected subsets of \(\mathbb{R}\), connected unions with nonempty intersection, and the intermediate value theorem. If you want a refresher, open the lesson for small examples and quick checks.
Answer the question set and review your mistakes at the end.
How this compactness and connectedness practice works
1. Take the practice set: answer questions about compact sets, connected sets, continuous images, intervals, and common counterexamples.
2. Open the lesson: review definitions, recognition tests, worked examples, and single-answer checks.
3. Retry: return to the question set and use the compactness or connectedness test that matches each problem.
What you will learn in the compactness and connectedness lesson
Compactness tests
Open-cover definition: every open cover has a finite subcover
Heine-Borel: in \(\mathbb{R}^n\), compact means closed and bounded
Examples such as \([0,1]\), \((0,1)\), \([0,\infty)\), and \(\{0\}\cup\{1/n:n\ge1\}\)
Sequences and set operations
In metric spaces, compactness gives convergent subsequences
Closed subsets of compact spaces are compact; finite unions of compact sets are compact
Missing limit points and arbitrary unions are common compactness traps
Connectedness tests
A separation splits a set into two nonempty separated open pieces
Intervals are connected in \(\mathbb{R}\); separated gaps break connectedness
If connected sets share a point, their union remains connected
Continuous-image theorems
Continuous images of compact sets are compact
Continuous images of connected sets are connected
Continuous real functions on compact metric spaces are bounded, attain extrema, and are uniformly continuous
Purpose: Build a fast, accurate way to decide whether a set is compact, connected, both, or neither. You will move between definitions, Euclidean shortcuts, sequences, continuous images, and standard counterexamples.
Success criteria
State compactness using open covers and finite subcovers.
Use Heine-Borel in \(\mathbb{R}^n\): compact means closed and bounded.
Use sequential compactness in metric spaces.
Know which operations preserve compactness.
State connectedness using separations.
Recognize intervals as connected subsets of \(\mathbb{R}\).
Use continuous images of compact and connected sets.
Combine compactness and connectedness to get closed intervals or points in \(\mathbb{R}\).
Key vocabulary
Compact: every open cover has a finite subcover.
Heine-Borel: in Euclidean space, compact is the same as closed and bounded.
Limit point: a point approached by other points of the set.
Connected: cannot be split into a separation.
Separation: two nonempty disjoint pieces that are open in the subspace and cover the set.
Continuous image: the set \(f(A)=\{f(a):a\in A\}\).
Quick pre-check
Pre-check 1: Which condition characterizes compact subsets of \(\mathbb{R}^n\)?
Hint: In Euclidean space, use the Heine-Borel theorem.
Pre-check 2: What is the continuous image of a connected set?
Hint: Continuity preserves connectedness.
Compactness means finite control of open covers
Learning goal: Use the open-cover definition and the Euclidean closed-and-bounded shortcut without mixing them up.
Key idea
A set \(K\) is compact if every open cover of \(K\) contains finitely many open sets that still cover \(K\). In \(\mathbb{R}^n\), Heine-Borel converts this definition into the practical test: \(K\) is compact exactly when it is closed and bounded.
Recognition checklist
In \(\mathbb{R}^n\), first ask: is the set closed?
Then ask: is it bounded?
For intervals, endpoints matter: \([0,1]\) is compact but \((0,1)\) is not.
For rays such as \([0,\infty)\) and for \(\mathbb{R}\) or \(\mathbb{Z}\), closed is not enough because the set is unbounded.
Finite subsets such as \(\{1,2,3\}\) are compact in \(\mathbb{R}\).
Worked example
Example: Decide whether \([0,1]\), \((0,1)\), and \([0,\infty)\) are compact in \(\mathbb{R}\).
\([0,1]\) is closed and bounded, so it is compact. The interval \((0,1)\) is bounded but not closed, so it is not compact. The ray \([0,\infty)\) is closed but not bounded, so it is not compact.
Try it
Try it 1: If a metric space is compact, every open cover has:
Hint: Compactness extracts a finite part from the cover.
Try it 2: Is \((0,1)\) compact in \(\mathbb{R}\)?
Hint: A bounded set can still fail to be compact if it misses boundary points.
Compact metric sets catch subsequences and limit points
Learning goal: Use sequential compactness, closed subsets, finite unions, and missing-limit-point examples.
Key idea
In metric spaces, compactness is equivalent to sequential compactness: every sequence in the set has a subsequence converging to a point still in the set. This makes missing limit points easy to detect.
Closure rules
Every compact subset of a metric space is closed and bounded.
A closed subset of a compact space is compact.
A finite union of compact sets is compact.
A finite intersection of compact sets in a metric space is compact.
An arbitrary union of compact sets need not be compact.
Worked example
Example: Compare \(A=\{1/n:n\ge1\}\) and \(B=\{0\}\cup\{1/n:n\ge1\}\) in \(\mathbb{R}\).
The set \(A\) is bounded but not closed because \(1/n\to0\) and \(0\notin A\), so \(A\) is not compact. The set \(B\) includes its only missing limit point and is bounded, so \(B\) is compact.
Try it
Try it 1: Is \(\{1/n:n\ge1\}\) compact in \(\mathbb{R}\)?
Hint: Look at the sequence \(1/n\) and its limit.
Try it 2: A finite union of compact sets is:
Hint: A finite number of finite subcovers can be combined into one finite subcover.
Continuous functions preserve compactness
Learning goal: Use compactness to control values and uniform continuity of continuous functions.
Key idea
If \(f:X\to Y\) is continuous and \(K\subset X\) is compact, then \(f(K)\) is compact in \(Y\). For real-valued functions, compactness of \(f(K)\subset\mathbb{R}\) means the function is bounded and attains both a maximum and a minimum. Heine-Cantor adds another compactness payoff: a continuous function on a compact metric space is uniformly continuous.
Worked example
Example: Why must a continuous function \(f:[0,1]\to\mathbb{R}\) attain a maximum?
The interval \([0,1]\) is compact, and \(f([0,1])\) is compact in \(\mathbb{R}\). A compact subset of \(\mathbb{R}\) is closed and bounded, so it contains its supremum. That supremum is the maximum value of \(f\).
Try it
Try it 1: A continuous real-valued function on a compact set must:
Hint: The image is a compact subset of \(\mathbb{R}\).
Try it 2: What is the continuous image of a compact set?
Hint: Compactness is preserved by continuous maps.
Try it 3: A continuous function on a compact metric space is:
Hint: Heine-Cantor says compactness makes the local continuity control uniform over the whole space.
Connectedness means no clean split into two pieces
Learning goal: Recognize connected sets in \(\mathbb{R}\) and detect gaps that create separations.
Key idea
A set is connected if it has no separation. In \(\mathbb{R}\), the connected sets are exactly intervals and single points. A missing point inside a real interval often creates two separated pieces.
Common examples
\([0,1]\), \((0,1)\), \([0,\infty)\), \(\mathbb{R}\), and singletons are connected.
\(\{0,1\}\), \(\mathbb{Z}\), \(\mathbb{R}\setminus\{0\}=(-\infty,0)\cup(0,\infty)\), and \([0,1]\cup[2,3]\) are disconnected.
\((0,1)\cup\{1\}\cup(1,2)=(0,2)\), so adding the missing point reconnects the two intervals.
Worked example
Example: Why is \([0,1]\cup[2,3]\) disconnected?
There is a gap between \(1\) and \(2\). The parts \([0,1]\) and \([2,3]\) are nonempty and separated inside the union, so the union has a separation.
Try it
Try it 1: Is \([0,1]\cup[2,3]\) connected?
Hint: Look for a gap between the two pieces.
Try it 2: Is every interval in \(\mathbb{R}\) connected?
Hint: Intervals are the model connected subsets of the real line.
Continuous images and overlapping unions keep connectedness
Learning goal: Use the two most common connectedness preservation tools.
Key idea
Continuity preserves connectedness: if \(C\) is connected, then \(f(C)\) is connected. Also, if \(A\) and \(B\) are connected and \(A\cap B≠\emptyset\), then \(A\cup B\) is connected. In \(\mathbb{R}\), a connected image is an interval.
Worked example
Example: Suppose \(A=[0,1]\) and \(B=[1,2]\). Why is \(A\cup B\) connected?
Both sets are intervals, hence connected, and they share the point \(1\). Their union is \([0,2]\), an interval, so it is connected.
Try it
Try it 1: A continuous function from a connected set to \(\mathbb{R}\) has image:
Hint: Connected subsets of \(\mathbb{R}\) are intervals or points.
Try it 2: If \(A\) and \(B\) are connected and \(A\cap B≠\emptyset\), then \(A\cup B\) is:
Hint: A shared point prevents the two connected pieces from being separated.
In the real line, compact and connected means closed interval or point
Learning goal: Combine compactness and connectedness to read the shape of continuous real images.
Key idea
A compact connected subset of \(\mathbb{R}\) is a closed interval \([a,b]\) or a single point. Thus if \(K\) is compact and connected and \(f:K\to\mathbb{R}\) is continuous, then \(f(K)\) is compact and connected in \(\mathbb{R}\), so it is a closed interval or a point.
Worked example
Example: What can the image \(f([0,1])\) look like for continuous \(f:[0,1]\to\mathbb{R}\)?
The domain \([0,1]\) is compact and connected. The image is therefore compact and connected in \(\mathbb{R}\), so it is a compact interval \([m,M]\) or a point if \(m=M\).
Try it
Try it 1: The continuous image of a compact connected set in \(\mathbb{R}\) is:
Hint: Preserve both properties, then use the real-line classification.
Try it 2: If \(f:[0,1]\to\mathbb{R}\) is continuous and \(f(0)\lt0\lt f(1)\), then \(f\) has:
Hint: The image of a connected interval is an interval, so it contains all values between the endpoint values.
Avoid the common compactness and connectedness mix-ups
Learning goal: Finish with counterexamples that separate similar-looking statements.
Common traps
Compact does not imply connected: \(\{0,1\}\) is compact but disconnected.
Connected does not imply compact: \((0,1)\) is connected but not compact in \(\mathbb{R}\).
Closed and bounded is a Euclidean shortcut: outside \(\mathbb{R}^n\), it is not a universal compactness test.
Infinite unions can fail compactness: \(\bigcup_{n\ge1}[0,n]=[0,\infty)\).
Intersections of connected sets can fail outside real intervals, but intervals in \(\mathbb{R}\) intersect in an interval or the empty set.
Hausdorff matters: compact subsets of Hausdorff spaces are closed.
Worked example
Example: Give a compact subset of \(\mathbb{R}\) that is not connected.
The set \(\{0,1\}\) is finite, hence compact in \(\mathbb{R}\), but it has a separation into the two singleton pieces. Therefore compactness alone does not force connectedness.
Try it
Try it 1: Is every compact subset of \(\mathbb{R}\) connected?
Hint: A finite set can be compact without being an interval.
Try it 2: A compact subset of a Hausdorff space is:
Hint: Hausdorff separation lets compact sets contain all their limit behavior.
Final recap
Compact means every open cover has a finite subcover.
In \(\mathbb{R}^n\), compact means closed and bounded.
In metric spaces, compactness is equivalent to sequential compactness.
Closed subsets of compact spaces and finite unions of compact sets are compact.
Continuous images of compact sets are compact, and continuous functions on compact metric spaces are uniformly continuous.
Connected means no separation; intervals are connected in \(\mathbb{R}\).
Continuous images of connected sets are connected.
A compact connected subset of \(\mathbb{R}\) is a closed interval or a point.
Next step: Close this lesson and try the quiz again. For compactness questions, ask which compactness test applies. For connectedness questions, look for intervals, gaps, continuous images, or overlapping connected unions.
Practice set
Compactheid en samenhang practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
0/10answered
Question 1Not answered
In \(\mathbb{R}\), is \([0,1]\) compact?
Correct answer:
Explanation: In \(\mathbb{R}\), closed and bounded sets are compact.
Question 2Not answered
Is in \(\mathbb{R}\) de verzameling \([0,1]\) compact?
Correct answer: B. Ja
Explanation: In \(\mathbb{R}\) zijn gesloten en begrensde verzamelingen compact.
Question 3Not answered
Is in \(\mathbb{R}\) de verzameling \((0,1)\) compact?
Correct answer: D. Nee, omdat ze niet gesloten is
Explanation: Ze is begrensd maar niet gesloten, dus ze is niet compact in \(\mathbb{R}\).
Question 4Not answered
Welke voorwaarde karakteriseert compacte deelverzamelingen van \(\mathbb{R}^n\)?
Correct answer: A. Gesloten en begrensd
Explanation: Volgens Heine-Borel zijn compacte deelverzamelingen van \(\mathbb{R}^n\) precies de gesloten en begrensde verzamelingen.
Question 5Not answered
Wat is het continue beeld van een compacte verzameling?
Correct answer: B. Compact
Explanation: Continuïteit behoudt compactheid.
Question 6Not answered
Is elk interval in \(\mathbb{R}\) samenhangend?
Correct answer: C. Ja
Explanation: Intervallen zijn precies de samenhangende deelverzamelingen van \(\mathbb{R}\) met meer dan één punt.
Question 7Not answered
Is \([0,1]\cup[2,3]\) samenhangend?
Correct answer: B. Nee
Explanation: Er zit een gat tussen de twee intervallen, dus de unie is niet samenhangend.
Question 8Not answered
Wat is het continue beeld van een samenhangende verzameling?
Correct answer: D. Samenhangend
Explanation: Continuïteit behoudt samenhang.
Question 9Not answered
In een metrische ruimte is elke compacte verzameling:
Correct answer: C. Gesloten
Explanation: Compacte deelverzamelingen van metrische ruimten zijn gesloten.
Question 10Not answered
Welke verzameling is samenhangend in \(\mathbb{R}\)?
Correct answer: D. \([0,1]\)
Explanation: Het interval \([0,1]\) kan niet worden gesplitst in twee niet-lege open delen.
