Are compact subspaces of compact subspaces are compact subsubspaces?

Solution 1:

1 & 2 are true, and 3 is false.

I refer to Topology by James Munkres. Let $A, B$ and $C$ have respective topologies $\mathscr T_A, \mathscr T_B$ and $\mathscr T_C$, where $\mathscr T_A$ is the subspace topology of $A$ in $B$, and $\mathscr T_B$ is the subspace topology of $B$ in $C$. Note that by Exer 16.1 (*), $\mathscr T_A'$, the subspace topology of $A$ in $C$, is equal to $\mathscr T_A$.

Pf: Let $A$ be covered by sets open in C:

$$A \subseteq \bigcup_{D \in \mathscr D} D, \tag{1}$$

for some $\mathscr D \subseteq \mathscr T_C$. By Lemma 26.1 (*), $A$ is a compact subspace of $C$, if we can find find some finite subcollection of $\mathscr D' \subseteq \mathscr D$ that covers $A$ i.e.

$$A \subseteq \bigcup_{D \in \mathscr D'} D.$$

Observe that by taking intersection of both sides of $(1)$ with $B$, we get

$$A = A \cap B \subseteq [\bigcup_{D \in \mathscr D} D] \cap B = \bigcup_{D \in \mathscr D} [D \cap B],$$

where, for every $D \in \mathscr D$, $D \cap B$ is open in $B$, by definition of $\mathscr T_B$. Thus, $A$ is covered by sets open in $B$. Hence, since $A$ is a compact subspace of $B$, we have a finite subcover $\mathscr D'' \subseteq \mathscr D$ of $A$:

$$A \subseteq \bigcup_{D \in \mathscr D''} [D \cap B].$$

Observe that

$$\bigcup_{D \in \mathscr D''} [D \cap B] \subseteq \bigcup_{D \in \mathscr D''} D$$

Therefore, we can choose $\mathscr D' = \mathscr D''$. QED

Pf: Let $A$ be covered by sets open in B:

$$A \subseteq \bigcup_{D \in \mathscr D} D$$

for some $\mathscr D \subseteq \mathscr T_B$. By Lemma 26.1 (*), $A$ is a compact subspace of $B$, if we can find find some finite subcollection of $\mathscr D' \subseteq \mathscr D$ that covers $A$ i.e.

$$A \subseteq \bigcup_{D \in \mathscr D} D.$$

By definition of $\mathscr T_B$, every $D \in \mathscr D$ has a partner $E_D \in \mathscr T_C$ s.t. $D = B \cap E_D$. Thus, we can collect all the $E_D$'s in some $\mathscr E \subseteq \mathscr T_C$, namely $\mathscr E: = \{E_D | D = B \cap E_D \ \text{for some} \ D \in \mathscr D \}$ s.t.

$$\bigcup_{D \in \mathscr D} D = \bigcup_{E_D \in \mathscr E} [B \cap E_D].$$

Observe that $$A \subseteq \bigcup_{D \in \mathscr D} D = \bigcup_{E_D \in \mathscr E} [B \cap E_D] \subseteq \bigcup_{E_D \in \mathscr E} E_D.$$

Thus, $A$ is covered by sets open in $C$. Hence, since $A$ is a compact subspace of $C$, we have a finite subcover $\mathscr E' \subseteq \mathscr E$ of $A$:

$$A \subseteq \bigcup_{E_D \in \mathscr E'} E_D.$$

Observe that by taking intersection of both sides with $B$, we get

$$A = A \cap B \subseteq \bigcup_{E_D \in \mathscr E'} [B \cap E_D] = [\bigcup_{E_D \in \mathscr E'} E_D] \cap B.$$

Therefore, we can choose $\mathscr D'$ to be made up of the $D$'s in $\mathscr D$ that have a partner $E_D \in \mathscr E'$. QED