# Compactness in a Nutshell

Compactness is an important topological property – A property that when seen in a certain perspective can make an infinite string become a ring, that you will give it to your girlfriend while proposing her for marriage. Therefore, compactness really intrigues me.

Let’s make our way to mathematics, without further adieu.

A space A is called ** Compact** under an ambient space X if any open cover of A has a finite subcover.

**Theorem 1**: A closed set of a compact set is compact.

**Theorem 2**: The continuous image of a compact space is compact.

**Theorem 3**: A compact subset of a Hausdorff Space ( eg: Metric Space ) is closed.

**Theorem 4**: A bijective continuous map from a compact space X to a Hausdorff space Y is a homeomorphism.

**Theorem 5 ( Heine Borel Theorem )**: X \(\in R^{n}\) is compact <=> X is closed and bounded.

**Theorem 6**: X is compact <=> X is complete and totally bounded.

**Theorem 7**: A continuous real-valued function defined on a compact space is bounded and attains its bounded. (Theorem 2 + Theorem 5)

A space X is called ** Sequentially Compact** (

**SC**) if every sequence in X has a convergent subsequence.

A space X is called ** Totally Bounded (TB)** if, for every \(\epsilon > 0 \), there are finite open balls of radius \(\epsilon\), that cover X.

A Space X satisfies ** Bolzano Weierstrass Property (BWP)** if every infinite subset of X has a limit point in X.

A Space X is called **Separable** if X has a countable dense subset.

A Space X is called ** Complete** if every Cauchy sequence in X has a limit point in X.

**Super Theorem 8:**

**Theorem 9**: X is Sequentially Compact *Metric Space* <=> X is a Compact *Metric Space*.

**Theorem 10**: For a Sequentially Compact Space, Lesbegue Covering Lemma holds.

#### Topological Algebraic Properties of Compact Sets

**Theorem 1**: A is compact, B is compact, then \( A \cup B \) is compact.

**Theorem 2**: A is compact, B is closed then \( A \cap B \) is compact.

**Note**: In general, A is compact, B is compact, then \( A \cap B \) is not compact. But for metric spaces, this is true due to **Theorem 3.**

**Counterexample 1:** A is compact, B is compact, then \( A \cap B \) may not compact.

X = **N**∪{−∞,∞}. A=**N**∪{∞}; B=**N**∪{−∞}. A∩B=**N** has the discrete topology and is not compact.

**Counterexample 2**: In X, A is compact, then \( \overline{X} \) may not be compact.

In Hausdorff Space, obviously, it is true. Hence, we have to take a non-Hausdorff Space.

X = Z, where the open sets are precisely the sets containing 0. Then \( \overline{0} \) = Z, which is not compact (why?).

**Theorem 3**: X x Y is compact <=> X and Y are compact.

More to come!

Stay Tuned!