 # 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!

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