Compactness

Intuition: An open cover is a blanket of open sets laid over $A$ -- every point of $A$ lies in at least one set of the family. The blanket may use uncountably many pieces. The question compactness asks is whether, no matter how wastefully the blanket is chosen, we can always pull out finitely many of its pieces and still cover every point of $A$. This is a strong "finiteness" requirement on the set $A$ itself; it has nothing to do with the size of the index set of the cover.

Intuition: Compactness is a universal-quantifier statement: every open cover must admit a finite subcover. To show a set is compact, you start with an arbitrary cover and must whittle it down to finitely many sets. To show a set is not compact, you need to exhibit just one open cover with no finite subcover. This asymmetry is typical: universal statements are proved by carefully controlling an arbitrary input; existential refutations need only one counterexample.

Intuition: Finite sets are the trivial case: you just pick one open set per point. This is the template for all compactness arguments -- reduce an arbitrary cover to a finite one by making finitely many choices. More interesting compact sets mimic this by forcing a finiteness condition (total boundedness, sequential behavior) that lets us choose finitely many sets even when the set itself is infinite.

Intuition: Closed subsets inherit compactness because being closed is a "sealed edges" condition. When you cover $F$ by open sets, you can add the one extra open set $F^c$ to cover all of $K$, use $K$'s compactness to extract finitely many sets, and then discard $F^c$ -- it wasn't covering any point of $F$ anyway. The trick is that the complement of a closed set is open and thus available for use in an open cover.

Intuition: The Hausdorff property lets us "separate" $y$ from each $x \in K$ individually. Compactness then allows us to combine these infinitely many separations into a single finite one. Without Hausdorff, compact sets need not be closed: in the trivial topology $\tau = \{\emptyset, X\}$ every subset is compact but only $\emptyset$ and $X$ are closed. In metric spaces, which are always Hausdorff, compactness does imply closedness, which is a key ingredient of Heine-Borel.

Intuition: In a metric space, compactness is at least as strong as being closed and bounded. The Heine-Borel theorem shows that in $\mathbb{R}^n$ the converse also holds. In general metric spaces (e.g. infinite-dimensional ones) it does not: the closed unit ball of $\ell^2$ is closed and bounded but not compact, because sequences can escape via high-frequency oscillations that admit no convergent subsequence.

Intuition: The proof is a "least upper bound sweeps right" argument, using the completeness of $\mathbb{R}$ crucially (step 3 is where sup $S$ is used). The set $S$ tracks how far to the right we can cover from $a$ with a finite subcover; we show we can always push the boundary a bit further, so it must reach $b$. This is the only compactness proof in this chapter that genuinely uses the order structure of $\mathbb{R}$; everything else in $\mathbb{R}^n$ reduces to products of intervals.

Intuition: The "tube lemma" packaged inside this proof is the key idea: if $\{x\} \times K_2$ is covered by a set $U$, then a whole "tube" $V \times K_2$ around $x$ is also covered (for some open neighborhood $V$ of $x$), provided $K_2$ is compact. You prove this by using compactness of $K_2$ to get a finite subcover and intersecting the corresponding $V$s. Once you have tubes, you cover $K_1$ by tubes and reduce to finitely many using compactness of $K_1$. The argument is a template for Tychonoff's theorem in infinite products.

Three classical theorems from MATH/MTHE 281, all phrased here via Heine-Borel for subsets of $\mathbb{R}$ (and remaining true for $\mathbb{R}^n$ with the Euclidean topology):

1. Heine-Borel: $K \subseteq \mathbb{R}$ is compact $\iff$ $K$ is closed and bounded.
2. Sequential compactness (infinite-subset form): $K \subseteq \mathbb{R}$ is compact $\iff$ every infinite subset of $K$ contains a nonconstant convergent sequence with limit in $K$.
3. Closed subsets of compact sets: if $K \subseteq \mathbb{R}$ is compact and $F \subseteq K$ is closed, then $F$ is compact.

Intuition: Heine-Borel is the bridge between the clean abstract definition of compactness and the concrete recognition test for subsets of $\mathbb{R}^n$. In $\mathbb{R}^n$, compactness is exactly closed and bounded, a condition you can usually check by inspection. This equivalence fails in infinite-dimensional spaces, which is why functional analysis needs finer notions like weak compactness. The route through products of intervals is characteristic: compactness questions in $\mathbb{R}^n$ reduce to the one-dimensional case by induction on dimension.

Intuition: Sequential compactness is the dynamic cousin of compactness: it says sequences cannot escape. Every infinite list of points in $K$ has an accumulation point still inside $K$. This is often the most useful form in analysis, since we reason about convergence constantly. In general topological spaces, compactness and sequential compactness are not equivalent -- there are compact spaces whose sequences do not all admit convergent subsequences, and sequentially compact spaces that are not compact. In metric spaces, the two notions coincide (proved below).

Intuition: The forward direction is a covering argument: if no subsequence converges, each point has a "quarantine ball" missing most of the sequence, and finitely many quarantine balls cannot hold the whole sequence. The reverse direction needs to show open-cover compactness from a hypothesis that controls only sequences, which requires building a dense structure of "small" approximations -- that is where total boundedness enters.

Intuition: Total boundedness is a quantitative refinement of boundedness. "Bounded" says $A$ fits in one big ball; "totally bounded" says for every resolution $\varepsilon$, $A$ can be blanketed by finitely many tiny balls of radius $\varepsilon$. The finite $\varepsilon$-net is a kind of low-resolution approximation of $A$. In $\mathbb{R}^n$ the two notions coincide: a bounded set is totally bounded because a big cube can be partitioned into finitely many small cubes. In infinite dimensions they differ sharply -- the unit ball of $\ell^2$ is bounded but not totally bounded, because the orthonormal basis vectors are pairwise at distance $\sqrt 2$ and can never be covered by finitely many small balls.

Intuition: Sequential compactness prevents two bad behaviors at once. It forbids escape to infinity (forces boundedness, indeed total boundedness), and it forbids leaking through gaps (forces completeness). A space that is sequentially compact is thus "small in all directions and sealed tight."

Intuition: This is the key existence theorem for convergent subsequences. The finite $(1/k)$-nets give us pigeonhole rooms of shrinking radius; a sequence with infinitely many terms is forced to concentrate in one such room at each scale, and the Cantor diagonal argument then picks a single subsequence that concentrates at every scale. Completeness then delivers the limit. The argument is the quantitative content of Bolzano-Weierstrass.

Intuition: This is the "metric space" version of Heine-Borel. In $\mathbb{R}^n$ completeness is automatic (every closed subset is complete) and total boundedness reduces to boundedness, recovering the classical "closed and bounded" slogan. The general statement is indispensable in infinite-dimensional analysis: to show a set in a Banach space is compact, you usually check that it is closed and show it is totally bounded directly (often via an equicontinuity argument, as in Arzelà-Ascoli).

Intuition: Any open cover of a sequentially compact set has a "universal scale" $\delta$: anything smaller than $\delta$ is small enough to be hidden inside a single set of the cover. This fails on non-compact sets -- the cover $\{(1/(n+1), 1/n) : n \ge 1\} \cup \{(1/2, 2)\}$ of $(0, 1]$ has no Lebesgue number, because near $0$ the sets of the cover become arbitrarily thin. The lemma is the tool that lets us upgrade "locally the cover is nice" to "uniformly the cover is nice."

Intuition: Total boundedness gives us finitely many small balls covering $K$, and the Lebesgue number lemma says each small ball fits inside some single $U_\alpha$. Combining, finitely many $U_\alpha$'s cover $K$. Sequential compactness gives both ingredients -- the finite-net structure and the uniform scale at which open covers become "well-behaved." Together with the earlier direction, this establishes: in a metric space, the three conditions compact, sequentially compact, and complete + totally bounded are all equivalent.

Intuition: Compactness is preserved by continuous maps, the same way connectedness and path-connectedness are. This makes compactness a topological invariant and an extremely robust property -- once you have one compact set, applying any continuous map gives another. This is the origin of many applications: we produce compact sets by applying continuous maps to known compact sets such as closed intervals.

Intuition: This is perhaps the single most useful consequence of compactness. Continuous real-valued functions on compact sets attain their extremes -- the infimum and supremum are not just approached, they are achieved at actual points of the domain. The theorem fails without compactness: $f(x) = x$ on $(0, 1)$ has $\inf = 0$ and $\sup = 1$, neither attained; $f(x) = \arctan x$ on $\mathbb{R}$ has $\sup = \pi/2$, never attained. Compactness is the right hypothesis: it prevents both the domain from "escaping" (so suprema exist) and the supremum from "slipping out" (so they are attained).

Intuition: Pointwise continuity means the $\delta$ you need may depend on the point -- near a steep part of the graph you need a smaller $\delta$. Uniform continuity means one $\delta$ works everywhere. On a compact domain, finitely many local $\delta$s suffice (by compactness), so you can take their minimum and get a uniform $\delta$. This is another instance of compactness upgrading local control to global control. An alternative proof uses the Lebesgue number lemma on the cover by $\varepsilon/2$-preimage balls directly.

Intuition: The full statement allows any indexing set, including uncountable ones. It uses the axiom of choice in an essential way (in fact it is equivalent to the axiom of choice over ZF). The theorem is the foundation for the abstract approach to compactness in functional analysis (e.g. Banach-Alaoglu, which says the closed unit ball of a dual Banach space is weak* compact, is proved via Tychonoff on a product of closed intervals).

Intuition: The finite case is essentially the same tube argument used for Heine-Borel, and it needs no choice. The infinite case genuinely requires deeper machinery (Alexander subbase theorem, ultrafilters, or Zorn's lemma). For this course the finite case is enough -- it is already enough to deduce that $[-R, R]^n$ is compact and hence to conclude Heine-Borel.