Definition 2.2.1. Let \((M,d)\) be a metric space and let \(A\subseteq M.\) A collection \(\{U_i\}_{i\in I}\) of subsets of \(M\) is called a cover of \(A\) if and only if
\[A\subseteq\bigcup_{i\in I}U_i.\]A cover of \(A\) is called an open cover of \(A\) if and only if all of its elements are open subsets of \(M.\)
Definition 2.2.2. Let \((M,d)\) be a metric space and let \(A\subseteq E\subseteq M.\) The set \(A\) is compact relative to \(E\) if and only if, for every collection \(\{U_i\}_{i\in I}\) of subsets open relative to \(E\) satisfying \(A\subseteq\bigcup_{i\in I}U_i,\) there exist \(i_1,\dots,i_n\in I\) such that
\[A\subseteq\bigcup_{k=1}^{n}U_{i_k}.\]Theorem 2.2.3. Let \((M,d)\) be a metric space and let \(A\subseteq E\subseteq M.\) Then \(A\) is compact relative to \(M\) if and only if \(A\) is compact relative to \(E.\)
Proof. Suppose that \(A\) is compact relative to \(M.\) Let \(\lbrace U_i\rbrace_{i\in I}\) be a collection of sets, open relative to \(E,\) such that \(A\subseteq \bigcup_{i\in I}U_i.\) By Theorem 2.1.23, there is open set \(V_i\) such that \(U_i = V_i \cap E,\) for each \(i\in I\); and since \(A\) is compact relative to \(M,\) we have
for some finitely many indices \(i_1,\dots,i_n\in I.\) Since \(A\subseteq E,\) \((\ast)\) implies
\[A\subseteq \bigcup_{k=1}^{n}U_{i_k}.\tag{\(\ast\ast\)}\]Therefore \(A\) is compact relative to \(E.\)
Conversely, suppose that \(A\) is compact relative to \(E.\) Let \(\lbrace V_i\rbrace_{i\in I}\) be a collection of open sets which covers \(A,\) and let \(U_i=V_i\cap E\) for each \(i\in I.\) Then \(U_i\) is open relative to \(E\) for each \(i\in I,\) and \(\lbrace U_i\rbrace_{i\in I}\) covers \(A.\) Since \(A\) is compact relative to \(E,\) \((\ast\ast)\) holds for some finitely many indices \(i_1,\dots,i_n\in I.\) Since \(U_i\subseteq V_i\) for each \(i\in I,\) \((\ast\ast)\) implies \((\ast).\) Therefore \(A\) is compact relative to \(M.\)\(\square\)
Definition 2.2.4. Let \((M,d)\) be a metric space and let \(A\subseteq M.\) The set \(A\) is compact if it is compact relative to \(M.\)
Theorem 2.2.5. Let \((M,d)\) be a metric space and let \(A\subseteq M.\) If \(A\) is compact then \(A\) is closed.
Proof.
References
- Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.