Let $X$ be a topological space and $A \subset X$. Adams and Franzosa's Introduction to Topology states the definition
$A$ is said to be compact in $X$ or a compact subset of $X$ if $A$ is compact in the subspace topology inherited from $X$.
I want to make sure I understand what this means. Earlier in the text, a compact topological space is defined as follows:
A topological space $X$ is compact if every open cover of $X$ has a finite subcover.
So, how I'm interpreting the definition of $A$ being compact is as follows.
Let $U_i$, $i \in I$ (for some index set $i$) be open sets in $X$, and
$$\mathcal{O} = \left\{A \cap U_i \mid i \in I \right\}\text{.}$$
These sets are open under the subspace topology applied to $A$ inherited by $X$. For notational convenience, let $A_i = A \cap U_i$, $i \in I$. Suppose
$$A \subset \bigcup_{A_i \in \mathcal{O}}A_i\text{.}$$
Then if there exists a finite subcollection $\mathcal{P} \subset \mathcal{O}$ such that
$$A \subset \bigcup_{A_i \in \mathcal{P}}A_i$$
we say that $A$ is a compact subset of $X$.
Is this correct?
Best Answer
Now we claim that if $X$ is a topological space, a subset $A$ of $X$ is compact if and only if for every collection of open sets (in $X$) $\mathcal U$ such that $A \subseteq \bigcup \mathcal U$ there exists a finite subset $\mathcal U'$ of $\mathcal U$ such that $A \subseteq \bigcup \mathcal U'$.
Proof : First suppose that $A$ is compact, and let $\mathcal U$ be a collection of open sets (in $X$) such that $A \subseteq \bigcup \mathcal U$. Note that the collection $\{A \cap U : U \in \mathcal U\}$ is a collection of open sets (in $A$) whose union is $$\textstyle \bigcup \{A \cap U : U \in \mathcal U\} = A \cap \bigcup \mathcal U = A \quad \textrm{since } A \subseteq \bigcup \mathcal U.$$ Thus, since $A$ is compact, by definition there is a finite subset of $\{A \cap U : U \in \mathcal U\}$ whose union is also $A$, but a finite subset of $\{A \cap U : U \in \mathcal U\}$ is a set of the form $\{A \cap U : U \in \mathcal U'\}$ for some finite subset $\mathcal U'$ of $\mathcal U$. In other words, there is a finite subset $\mathcal U'$ of $\mathcal U$ such that $$\textstyle A = \bigcup \{A \cap U : U \in \mathcal U'\} = A \cap \bigcup \mathcal U',$$ which of course implies that $A \subseteq \bigcup \mathcal U'$. Can you show now the reverse implication?
I hope all of this clarify the things!