Definition of compactness of metric spaces

Question

In my lecturer's notes it says the following:

Let $A \subseteq M$ and let $B = \{U_i : i \in I\}$ be an open cover of $A$. When determining the compactness or not of $A$, we might question whether it matters whether the $U_i$ are open in $A$ or open in $M$. In fact, it does not matter whether the $U_i$ are open in $A$ and $A=\cup_{i \in I} U_i$ or whether the the $U_i$ are open in $M$ and $A \subseteq\cup_{i \in I} U_i$.

Why is this?

Answer 1

This is because compactness is independent of the ambient space. More formally stated, if $A \subset E \subset M$, then $A$ is compact in $E$ iff $A$ is compact in $M$. In your case, $E = A$ but the result still holds. This is because given an open (in $A$) cover of $A$, one can easily show from it the existence of an open (in $M$) cover of $A$, and vice versa.

Answer 2

Assume every $M$-open cover of $A$ has a finite subcover. Take $A$-open cover $B_A$ of $A$. Then $B_A$ is a collection of sets of the form $V_i = U_i \cap A$, where $U_i$ are open in $M$. Now the collection $B_M = \{U_i\}$ is an $M$-open cover of $A$ and thus has a finite subcover $B_M'$. Now we can define a finite $A$-open subcover of $A$ by setting $$ B_A' = \{U_i \cap A: U_i \in B_M'\}\,. $$

Conversely, assume every $A$-open cover of $A$ has a finite subcover. Now let $B_M = \{ U_i\}$ be an arbitrary $M$-open cover of $A$. Define $$ B_A = \{ U_i \cap A: U_i \in B_M\}\,. $$ Now $B_A$ is an $A$-open cover of $A$ and thus has a finite subcover $B_A'$. This then defines a finite subcover of $B_M$ by $$ B_M' = \{ U_i : U_i \cap A \in B_A'\}\,. $$

Answer 3

Assume that $A\subseteq M$ is compact, and assume that $U_i$ is an open cover of $A$, where the $U_i$ are open in $A$. By definition of the subspace topology, there exist sets $V_i$ that are open in $M$, such that $U_i = V_i\cap A$.

Since

$$A = \bigcup_{i} U_i = (\bigcup_{i} V_i)\cap A\subseteq \bigcup_{i} V_i,$$ we see that the $V_i$ is an open cover of $A$ of sets that are open in $M$. Since $A$ is compact in $M$, there exists a finite subcover $V_1,\ldots,V_n$, such that

$$A \subseteq V_1\cup\cdots\cup V_n.$$

Since $U_i = V_i\cap A$, we can take the intersection with $A$ on both sides to get $$A = U_1\cup\cdots\cup U_n.$$ The cover of sets $U_i$ that were open in $A$ thus has a finite subcover.

This proves that if $A$ is compact, then any cover of $A$ of sets that are open in $A$ have a finite subcover.

For the other direction, assume that any cover of $A$ of sets that are open in $A$ have a finite subcover. We want to prove that $A$ is compact, so let $\{V_i\}$ be an open cover of $A$ of sets that are open in $M$. Then $$A\subseteq \bigcup_i V_i.$$ Taking the intersection with $A$ on both sides, we get $$A = \bigcup_i (A\cap V_i),$$ so by definition of being open in $A$, the sets $A\cap V_i$ constitute an open cover of $A$ (of sets that are open relative to $A$). By our hypothesis, this has a finite subcover, so $$A = (A\cap V_1)\cup\cdots\cup (A\cap V_n)\subseteq V_1\cup\cdots\cup V_n.$$ This proves that the cover $\{V_i\}$ has a finite subcover, whence $A$ is compact.