Compactness
The single most useful property a set can have. Its definition — every open cover has a finite subcover — looks strange until you see what it does: it turns infinitely many local statements into one global statement. Nearly every existence theorem this year runs through it.
- State the open cover definition and use it directly.
- Prove the Heine–Borel theorem in $\mathbb{R}^n$.
- Move between covering, sequential, and closed-and-bounded.
- Show compact sets are closed and bounded, and find where the converse fails.
Local information is easy to get and hard to use. Compactness is the hypothesis that lets you take a property holding near every point and conclude it holds uniformly on the whole set. Uniform continuity, attainment of maxima, and partitions of unity are all this move.
Optimisation over a compact feasible region is guaranteed to have a solution, which is why constraints matter as much as objectives. Numerical methods converge on compact domains and can wander off on non-compact ones. Existence of equilibria in economics is a fixed point theorem on a compact convex set.
Heine and Borel's theorem was assembled between 1872 and 1895, and the modern open-cover definition took until Alexandroff and Urysohn in 1929. That slow arrival is a hint: it is not the obvious definition, it is the one that turned out to work. In infinite dimensions closed and bounded stops implying compact, which reshapes all of functional analysis.
The definition
An open cover of $K\subseteq\mathbb{R}^n$ is a family $\{U_\alpha\}$ of open sets with $K\subseteq\bigcup_\alpha U_\alpha$. The set $K$ is compact if every open cover of $K$ has a finite subcover: finitely many $U_{\alpha_1},\dots,U_{\alpha_m}$ already cover $K$.
Cover $\mathbb{R}$ by $(-n,n)$ for $n\in\mathbb{N}$. A finite subfamily has a largest $n$ and misses everything beyond. Here the obstruction is unboundedness.
Heine–Borel
This direction holds in any metric space.
In $\mathbb{R}^n$, a set is compact if and only if it is closed and bounded.
In an infinite-dimensional normed space the closed unit ball is closed and bounded but never compact. The sequence of standard basis vectors satisfies $\|e_i-e_j\|=\sqrt2$ for $i\ne j$, so it has no convergent subsequence. Heine–Borel is a theorem about $\mathbb{R}^n$, not a tautology.
Sequential compactness
$K$ is sequentially compact if every sequence in $K$ has a subsequence converging to a point of $K$.
For $K\subseteq\mathbb{R}^n$, compact and sequentially compact are equivalent. Both are equivalent to closed and bounded.
If $K$ is compact and $F\subseteq K$ is closed, then $F$ is compact.
Given an open cover of $F$, add $\mathbb{R}^n\setminus F$ to obtain an open cover of $K$. Extract a finite subcover and drop the added set; what remains still covers $F$.
$[0,1]$ is compact whether viewed inside $\mathbb{R}$ or inside $\mathbb{R}^2$ as a segment. Open and closed are relative to an ambient space; compactness is not, because a cover by relatively open sets can always be traded for one by ambient open sets.
• Proving compactness by exhibiting one cover with a finite subcover.
The definition quantifies over all covers.
• Quoting "closed and bounded implies compact" outside $\mathbb{R}^n$. It
is exactly what fails in infinite dimensions.
• Forgetting that continuous images are compact but preimages need not be.
$\sin$ maps the non-compact $\mathbb{R}$ into the compact $[-1,1]$.
• Assuming an infinite intersection of compact sets could be empty. A
nested family of nonempty compact sets always has a common point.
Practice
Original problems written for this guide in the style of Spivak and Munkres exercises. Write a full solution before opening the answer.
Is $\{(x,y):x^2+y^2\le1,\ y>0\}$ compact?
Answer
Exhibit an open cover of $[0,1)$ with no finite subcover.
Answer
Prove every finite set is compact.
Solution
Are finite unions and arbitrary intersections of compact sets compact?
Solution
If $K_1\supseteq K_2\supseteq\cdots$ are nonempty compact sets, show $\bigcap_kK_k\ne\emptyset$.
Solution
For compact $K$ and a point $p\notin K$, show there is a nearest point of $K$ to $p$.
Solution
Show the closed unit ball of the space of bounded sequences is not compact.
Solution
If $f:[0,1]\to\mathbb{R}$ is continuous, is its graph compact in $\mathbb{R}^2$?
Solution
Expanded topic checklist
Course coverage. This unit is part of the complete MAT257Y1 topic map. It should cover the following items before you move on:
- Open cover definition. Use it directly, in both directions.
- Heine–Borel. Prove it by bisection and know that it is specific to $\mathbb{R}^n$.
- Sequential compactness. Prove the equivalence via Bolzano–Weierstrass.
- Hereditary properties. Closed subsets of compact sets, nested intersections.
Use the current UofT MAT257Y1 calendar topics as the coverage spine, and pair lecture notes with Spivak's Calculus on Manifolds or Munkres' Analysis on Manifolds. The exact assigned textbook should still be checked against the instructor syllabus.
Compact gives closed and bounded always; the converse is Heine–Borel and needs $\mathbb{R}^n$. Sequential compactness is the equivalent form that is easiest to apply. Unit 3 spends the whole page cashing these in.
Deep practice extensions
Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.
State compactness in terms of covers.
Guide
Which half of Heine–Borel holds in any metric space?
Guide
Name a closed bounded set that is not compact.
Guide
What makes the bisection proof terminate?