Equivalence of Definitions of Hereditarily Compact

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Definition 1 implies Definition 2
Let $T = \struct {S, \tau}$ be hereditarily compact by definition 1.

Let $\family {U_i}_{i \mathop \in I}$ be an indexed family of open sets of $T$.

We have that:


 * $\ds \bigcup_{i \mathop \in I} U_i \subset S$

By hypothesis, $\ds \bigcup_{i \mathop \in I} U_i$ is compact when considered as a subspace of $T$.

Furthermore, $\ds U_i = U_i \cap \bigcup_{i \mathop \in I} U_i$ is open in $\ds \bigcup_{i \mathop \in I} U_i$, by definition of the subspace topology.

Therefore $\family {U_i}_{i \mathop \in I}$ is an open cover for $\ds \bigcup_{i \mathop \in I} U_i$.

Since $\ds \bigcup_{i \mathop \in I} U_i$ is compact, there exists a finite subcover of $\family {U_i}_{i \mathop \in I}$ for $\ds \bigcup_{i \mathop \in I} U_i$, say $\family {U_j}_{j \mathop \in J}$.

Then by definition of open cover:
 * $\ds \bigcup_{j \mathop \in J} U_j = \bigcup_{i \mathop \in I} U_i$

Thus $T = \struct {S, \tau}$ is hereditarily compact by definition 2.

Definition 2 implies Definition 1
Let $T = \struct {S, \tau}$ be hereditarily compact by definition 2.

Let $Y \subset S$ be a subspace of $T$.

Let $\family {V_i}_{i \mathop \in I}$ be an open cover for $Y$:


 * $\ds \bigcup_{i \mathop \in I} V_i = Y$

Then by definition of the subspace topology:


 * $V_i = U_i \cap Y$

for a certain $V_i \in \tau$

But then $\family {U_i}_{i \mathop \in I}$ is an indexed family of open sets of $T$.

By hypothesis, there exists a finite subset $J \subset I$ such that:


 * $\ds \bigcup_{j \mathop \in J} U_i = \bigcup_{i \mathop \in I} U_i$

But then:

Thus $\family {V_j}_{j \mathop \in J}$ is a finite subcover of $\family {V_i}_{i \mathop \in I}$ for $Y$.

Thus $Y$ is a compact subspace of $T$.

Thus $T = \struct {S, \tau}$ is hereditarily compact by definition 1.