Proposition. Let \(X\) be a topological space, the following are equivalent:

\(X\) is compact
If \(\{V_i:i∈I\}\) is a family of closed subsets of \(X\) s.t. \(\bigcap_{i∈I} V_i=∅\) then \(∃\) finite \(J⊆I\) s.t. \(\bigcap_{j∈J} V_j=∅\)

Proof. (1)\(⇒\)(2) Let \(V_i, i∈I\) be s.t. \(\bigcap_{i∈I} V_i=∅\).

\(\bigcup_{i∈I} (X∖V_i)=X∖\bigcap_{i∈I} V_i=X\)

So \(\{X∖V_i:i∈I\}\) is open cover of \(X\).

\(X\) compact\(⇒ X=(X∖V_{i_1})∪⋯∪(X∖V_{i_n})=X∖\bigcap_{j=1}^n V_{i_j}⇒ \bigcap_{j=1}^n V_{i_j}=∅\) i.e. (2) holds.

(2)\(⇒\)(1) Let \(\{U_i:i∈I\}\) be open cover of \(X\).

\(\bigcup_{i∈I} U_i=X⇒ \bigcap_{i∈I} X∖U_i=∅, X∖U_i\) closed.

By (2) \(∃U_{i_1},…, U_{i_n}\) s.t. \(\bigcap_{j=1}^n X∖U_{i_j}=∅\)

\(X=\bigcup_{j=1}^n U_{i_j}\) finite cover\(⇒ X\) compact. ◻

Definition. A subset \(A\) of a topological space is compact if it is a compact topological space when endowed with the subspace topology.

Definition. \(\{U_i:i∈I\}\) is a cover of \(A\) if \(A⊆\bigcup_{i∈I} U_i\)

Equivalent definition of compact subset:

\(A\) is compact\(\Leftrightarrow\)For every family \(\{U_i:i∈I\}\) which is a cover of \(A\) with \(U_i\) open in \(X\), \(∃U_{i_1},…, U_{i_n}, i_j∈I\) s.t. \(A⊆\bigcup_{j=1}^n U_{i_j}\)

Example.

Any finite topological space is compact. A discrete space is compact if and only if it is finite.
The indiscrete space is compact
If a metric space \(X\) is compact then \(X\) is bounded
Proof. \(X⊆\bigcup_{n∈ℕ} B (x_0, n)⇒ X⊆B (x_0, N)\) for some \(N⇒ X\) is bounded. ◻
Let \(X\) be any space with cofinite topology. Then any \(A⊆X\) is compact.
Proof. Say \(A⊆\bigcup_{i∈I} U_i\) where \(U_i\) are open
If \(i_0∈I\), \(A∖U_{i_0}=\{a_1,…, a_k\}, a_j∈U_{i_j}, i_j∈I\). So \(A⊆U_{i_0}∪U_{i_1}∪⋯∪U_{i_k}\). ◻
(Heine-Borel) \([a, b]∈\mathbb{R}\) is compact
Proof. see Metric Space course. ◻
Remark. \((a, b)\) is not compact: \((a, b)=\bigcup_{n∈ℕ} \left( a + \frac{1}{n}, b - \frac{1}{n} \right)\), ∄finite subcover

Proposition. Any closed subset \(A\) of a compact space \(X\) is compact.

Proof. Say \(A⊆\bigcup_{i∈I} U_i, U_i\) open, then \(X⊆(X∖A)∪\bigcup_{i∈I} U_i, X\) compact\(⇒ X⊆(X∖A)∪(U_{i_1}∪⋯∪U_{i_n})\) for some \(i_j∈I⇒ A⊆U_{i_1}∪⋯∪U_{i_n}⇒ A\) is compact. ◻

Remark. The converse is not true in general for topological spaces, so compact subsets might not be closed, e.g. \(X\) with indiscrete topology. Then \(\{x\}⊆X\) is compact but \(\{x\}\) is not closed.

Proposition. Let \(X\) be Hausdorff and \(K⊆X\) be compact. If \(x∈X∖K\), there are open disjoint sets \(U, V\) such that \(K⊆U, x∈V\).

Proof. For each \(y∈K\) pick \(U_y, V_y\) open disjoint such that \(y∈U_y, x∈V_y, K⊆\bigcup_{y∈K} U_y⇒ K⊆U_{y_1}∪⋯∪U_{y_n}\). Let \(V=\bigcap_{i=1}^n V_{y_i}\).
\(∀i:V_y∩U_{y_i}=∅, V⊆V_{y_i}⇒V \cap U_{y_i}=∅⇒V∩\underbrace{(U_{y_1}∪⋯∪U_{y_n})}_U=∅\). Also \(K⊆U\). ◻

Corollary. Any compact subset \(K\) of a Hausdorff space \(X\) is closed in \(X\).

Proof. We show \(X∖K\) is open

By Prop. \(∀x∈X∖K∃U_x \ni x\) open such that \(U_x \cap K=∅\).

So \(X∖K\) is union of \(U_x\)(open)

So \(X∖K\) is open\(⇒ K\) is closed ◻

Proposition. If \(f:X→Y\) is continuous map of topological spaces and \(A⊆X\) is compact, then \(f (A)\) is compact.

Proof. Say \(f (A)⊆\bigcup_{i∈I} U_i, U_i\) open, then \(A⊆\bigcup_{i∈I} f^{- 1} (U_i)\), \(f^{- 1} (U_i)\) open since \(f\) continuous.

Since \(A\) compact, \(A⊆f^{- 1} (U_{i_1})∪⋯∪f^{- 1} (U_{i_k})\) for some \(i_j∈I⇒ f (A)⊆U_{i_1}∪⋯∪U_{i_n}⇒ f (A)\) compact. ◻

Remark. If \(K⊆A⊆X\), \(X\) topological space, \(K\) compact in \(A\) (with subspace topology) then \(K\) is compact in \(X\).

Proof. \(i:A→X, i (a)=a\) is continuous. \(K⊆A\) is compact\(⇒ i (K)=K\) is compact in \(X\). ◻

Remark. Not true for closed e.g. \(A=(0, 2)⊆\mathbb{R}, K=(0, 1], K\) is closed in \(A\) but not in \(\mathbb{R}\). compactness is an inherent notion

The converse implication holds both for ‘closed’ and for ‘compact’, i.e. if \(K⊆A⊆X\) and \(K\) is closed (compact) in \(X\), then \(K\) is closed (compact) in \(A\).

Proposition. Let \(X\) be a topological space.

If \(K_1,…, K_n\) are compact subsets of \(X\) then \(K_1∪⋯∪K_n\) is compact.
If \(\{K_i, i∈I\}\) are compact and \(X\) is Hausdorff then \(\bigcap_{i∈I} K_i\) is compact.

Proof.

Say \(\{U_i:i∈I\}\) is a cover of \(K_1∪⋯∪K_n\). \(∃J_1,…, J_n⊆I\) finite such that \(K_r⊆\bigcup_{j∈J_r} U_j⇒ K_1∪⋯∪K_n⊆\bigcup_{r=1}^n\bigcup_{j∈J_r} U_j\)
Since \(X\) is Hausdorff, compact\(⇒\)closed, so \(K_i\) closed\(∀i⇒ \bigcap_{i∈I} K_i\) is closed.
Pick \(K_{i_0}, i_0∈I, \bigcap_{i∈I} K_i⊆K_{i_0}\) (closed)
\(K_{i_0}\) compact\(⇒ \bigcap_{i∈I} K_i\) is compact in \(K_{i_0}⇒ \bigcap_{i∈I} K_i\) is compact in \(X\).

◻

Remark. (1) not true for infinite union e.g. \((0, 1)=\bigcup_{n∈ℕ} \left[ \frac{1}{n}, 1 - \frac{1}{n} \right]\)

(2) not true if \(X\) is not Hausdorff (Ex.5 sheet3)

Theorem. The product \(X×Y\) is compact iff \(X, Y\) are compact.

Proof. (\(⇒\)) easy \(P_X (X×Y)=X⇒ X\) compact. Same for \(Y\).

(\(⇐\)) \(X×Y⊆\bigcup_{i∈I} W_i\). Since the rectangular open sets compose a basis for the product topology, replace \(W_i\) by sets \(U×V\), \(U\) open in \(X\), \(V\) open in \(Y\).\[X×Y⊆\bigcup_{j∈J} U_j×V_j\]Therefore it is enough if we prove that \(\{U_j×V_j:j∈J\}\) has a finite subcover.

For every \(y∈Y\), the compact set \(X×\{y\}\) is covered by \(\{U_j×V_j:j∈J\}\). Therefore \(∃\) finite subset \(F_y⊆J\) such that\[X×\{y\}⊆\bigcup_{j∈F_y} U_j×V_j\]The set \(V_y=\bigcap_{j∈F_y} V_j\) is an open set containing \(y\). The family \(\{V_y:y∈Y\}\) is an open cover of \(Y\), which is compact.

It follows that there exist \(y_1,…, y_m\) such that \(Y=V_{y_1}∪⋯∪V_{y_m}\).

We state that \(X×Y=\bigcup_{k=1}^m \bigcup_{j∈F_{y_k}} U_j×V_j\). Indeed consider an arbitrary element \((x, y)\) in the product. Then there exists \(k∈\{1,…, m\}\) such that \(y∈V_{y_k}\). In particular \(y∈V_j\) for all \(j∈F_{y_k}\). On the other hand \(X×\{y_k\}⊆\bigcup_{j∈F_{y_k}} U_j×V_j\), whence there exists \(j_0∈F_{y_k}\) such that \(x∈U_{j_0}\). It follows that \((x, y)∈U_{j_0}×V_{j_0}\). ◻