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[parent] $Y$ is compact if and only if every open cover of $Y$ has a finite subcover (Theorem)

Theorem.
Let $ X$ be a topological space and $ Y$ a subset of $ X$. Then the following statements are equivalent.

  1. $ Y$ is compact as a subset of $ X$.
  2. Every open cover of $ Y$ (with open sets in $ X$) has a finite subcover.

Proof. Suppose $ Y$ is compact, and $ \{U_i\}_{i\in I}$ is an arbitrary open cover of $ Y$, where $ U_i$ are open sets in $ X$. Then $ \{U_i\cap Y\}_{i\in I}$ is a collection of open sets in $ Y$ with union $ Y$. Since $ Y$ is compact, there is a finite subset $ J\subset I$ such that $ Y=\cup_{i\in J} (U_i\cap Y)$. Now $ Y=(\cup_{i\in J} U_i)\cap Y \subset \cup_{i\in J} U_i$, so $ \{U_i\}_{i\in J}$ is finite open cover of $ Y$.

Conversely, suppose every open cover of $ Y$ has a finite subcover, and $ \{U_i\}_{i\in I}$ is an arbitrary collection of open sets (in $ Y$) with union $ Y$. By the definition of the subspace topology, each $ U_i$ is of the form $ U_i = V_i\cap Y$ for some open set $ V_i$ in $ X$. Now $ U_i \subset V_i$, so $ \{V_i\}_{i\in I}$ is a cover of $ Y$ by open sets in $ X$. By assumption, it has a finite subcover $ \{V_i\}_{i\in J}$. It follows that $ \{U_i\}_{i\in J}$ covers $ Y$, and $ Y$ is compact. $ \Box$

The above proof follows the proof given in [1].

Bibliography

1
B.Ikenaga, Notes on Topology, August 16, 2000, available online http://www.millersv.edu/˜bikenaga/topology/topnote.html.



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Cross-references: cover, subspace topology, union, collection, subcover, finite, open sets, open cover, compact, equivalent, subset, topological space
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This is version 3 of $Y$ is compact if and only if every open cover of $Y$ has a finite subcover, born on 2003-04-12, modified 2003-05-17.
Object id is 4179, canonical name is YIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcover.
Accessed 3328 times total.

Classification:
AMS MSC: 54D30 (General topology :: Fairly general properties :: Compactness)
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