Equivalence of Definitions of Hereditarily Compact

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.


The following definitions of the concept of Hereditarily Compact Space are equivalent:

Definition 1

$T$ is hereditarily compact if and only if every subspace of $T$ is compact.

Definition 2

$T$ is hereditarily compact if and only if:

for each family $\left\langle{U_i}\right\rangle_{i \mathop \in I}$ of open sets of $T$, there exists a finite subset $J \subset I$ such that:
$\displaystyle \bigcup_{j \mathop \in J} U_j = \bigcup_{i \mathop \in I} U_i$


Proof

Definition 1 implies Definition 2

Let $T = \left({S, \tau}\right)$ be hereditarily compact by definition 1.

Let $\left\langle{U_i}\right\rangle_{i \mathop \in I}$ be an indexed family of open sets of $T$.

We have that:

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

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

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

Therefore $\left\langle{U_i}\right\rangle_{i \mathop \in I}$ is an open cover for $\displaystyle \bigcup_{i \mathop \in I} U_i$.

Since $\displaystyle \bigcup_{i \mathop \in I} U_i$ is compact, there exists a finite subcover of $\left\langle{U_i}\right\rangle_{i \mathop \in I}$ for $\displaystyle \bigcup_{i \mathop \in I} U_i$, say $\left\langle{U_j}\right\rangle_{j \mathop \in J}$.

Then by definition of open cover:

$\displaystyle \bigcup_{j \mathop \in J} U_j = \bigcup_{i \mathop \in I} U_i$

Thus $T = \left({S, \tau}\right)$ is hereditarily compact by definition 2.

$\Box$


Definition 2 implies Definition 1

Let $T = \left({S, \tau}\right)$ be hereditarily compact by definition 2.

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

Let $\left\langle{V_i}\right\rangle_{i \mathop \in I}$ be an open cover for $Y$:

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

Then by definition of the subspace topology:

$V_i = U_i \cap Y$

for a certain $V_i \in \tau$

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

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

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

But then:

\(\displaystyle \bigcup_{i \mathop \in I} V_i\) \(=\) \(\displaystyle \bigcup_{i \mathop \in I} \left({U_i \cap Y}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\bigcup_{i \mathop \in I} U_i}\right) \cap Y\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\bigcup_{j \mathop \in J} U_j}\right) \cap Y\)
\(\displaystyle \) \(=\) \(\displaystyle \bigcup_{j \mathop \in J} \left({U_j \cap Y}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \bigcup_{j \mathop \in J} V_j\)


Thus $\left\langle{V_j}\right\rangle_{j \mathop \in J}$ is a finite subcover of $\left\langle{V_i}\right\rangle_{i \mathop \in I}$ for $Y$.

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

Thus $T = \left({S, \tau}\right)$ is hereditarily compact by definition 1.

$\blacksquare$