3.3 Compactness
Definition 3.21. A set \(\, S\subseteq \mathbb {C}\,\) is said to be bounded if there is an \(\,M > 0 \ni \left |z\right | < M\, \forall z \in S\).
Definition 3.22. Let \(\, S\subseteq \mathbb {C}\). An open covering of \(S\) is a collection of open sets \(\, G \ni \bigcup \limits _{g \in G}g\supseteq S\).
Definition 3.23. A set \(\, S\subseteq \mathbb {C}\,\) is said to be compact if every open covering of \(S\) has a finite sub-cover.
\(B(0; 1) \quad G = \big \{B(0; r): \, r \in [ 0,1) \subseteq \mathbb {R}\big \}\)
\(\bigcup \limits _{r\in [0,1)}B(0; r) \supset B(0; 1)\)
No subcollection of \(G\) contains \(\, B(0; 1)\)
E.g \(\, \big \{ n + i0: \, n \in \mathbb {N}\big \} \subset \mathbb {C}\)
\(G = \big \{ B(0; n):\, n\in \mathbb {N}\big \}\)
\(B(0; 2)\quad B(0; n)\quad B(0; n- 1)\)
Theorem 3.25 (Heine–Borel). A set \(\, S \subseteq \mathbb {C}\,\) is compact if and only if it is closed and bounded in \(\mathbb {C}\).
Proof.
Compact \(\implies \) closed and bounded
Suppose \(S\) is compact. The open discs \(D_n=\{z:\left |z\right |<n\}\) cover \(\mathbb {C}\), hence cover \(S\); a finite subcover is contained in the largest of them, so \(S\) is bounded.
For closedness, let \(w\notin S\). For each \(z\in S\) put \(r_z=\frac {1}{2}\left |z-w\right |>0\); the discs \(B(z;r_z)\) cover \(S\), so finitely many \(B(z_k;r_{z_k})\) do. Then the disc about \(w\) of radius \(\min _k r_{z_k}\) misses every one of them, hence misses \(S\). So the complement of \(S\) is open and \(S\) is closed.
Closed and bounded \(\implies \) compact
Since \(S\) is bounded it lies in some closed square \(Q\). Suppose an open cover of \(S\) had no finite subcover. Bisect \(Q\) into four closed subsquares; at least one of them meets \(S\) in a part still admitting no finite subcover, or else the four finite subcovers would combine into one. Choose such a subsquare and repeat, obtaining nested squares \(Q\supset Q_1\supset Q_2\supset \cdots \) whose diameters halve at each step.
Their intersection is a single point \(z_0\), and \(z_0\in S\) because \(S\) is closed and each \(Q_k\) meets \(S\). Some member \(U\) of the cover contains \(z_0\), and since the diameters tend to \(0\), \(Q_k\subseteq U\) for \(k\) large. But then \(Q_k\cap S\) is covered by the single set \(U\) — a finite subcover — which contradicts the choice of \(Q_k\). Hence a finite subcover exists. □
Remark 3.26. Both halves are needed, and each can fail alone: the open disc \(\left |z\right |<1\) is bounded but not closed, and the real axis is closed but not bounded. Neither is compact.
- 1.
- Unit circle in \(\mathbb {C}\) is compact
Theorem 3.28 (Cantor’s Theorem). Let \(\, K_j'\)s be compact subsets of \(\mathbb {C},\, j \in \mathbb {N}\,\) with \(\, K_1\supset K_2\supset K_3\supset \cdots \,\) Then \(\bigcap \limits _{j= 1}^{\infty }K_j\,\) is nonempty.
Proof. Suppose that \(\, \bigcap \limits ^{\infty }_{j = 1} K_j\,\) is empty. Then \(\, G = \big \{K^c_j : \, j\in \mathbb {N}\big \},\quad \mathbb {C} = \bigcup \limits ^{\infty }_{j= 1}K^c_j\,\) so that \(\,K_1\subset \bigcup \limits ^{\infty }_{j = 1} K^c_j\)
Since \(K_1\) is compact, there are numbers
\[\, j_1, \, j_2,\ldots , j_n,\quad j_1< j_2< \cdots < j_n \quad \ni \quad K_1 \subset \bigcup \limits ^n_{m = 1}K^c_{j_m} \subset K^c_{j_{n+1}}\]
\[\Big (\therefore \quad K_{j_{n+1}} \subset K_{j_n} \subset \cdots \subset K_{j_1}\Big )\]
\(\implies \, K_1 \cap K_{j_{n+1}} = \emptyset \). Thus a contradiction to the given hypothesis. So \(\bigcup \limits ^{\infty }_{j = 1}K_j\,\) has to be nonempty. □
Proof. Let \(F\subseteq S\) with \(F\) closed and \(S\) compact, and let \(\mathcal {U}\) be an open cover of \(F\).
The complement \(\mathbb {C}\setminus F\) is open, so adjoining it to \(\mathcal {U}\) gives an open cover of the whole of \(S\): any point of \(S\) either lies in \(F\), and is covered by \(\mathcal {U}\), or lies outside \(F\) and is covered by the new set.
By compactness of \(S\) this enlarged cover has a finite subcover. Discarding \(\mathbb {C}\setminus F\) from it if present — it contains no point of \(F\) — leaves finitely many members of \(\mathcal {U}\) still covering \(F\). Hence \(F\) is compact.
The trick of adding the complement to the cover and removing it afterwards is worth remembering; it is the standard way of passing compactness to a closed subset. □
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