description/proof of subset of Euclidean metric topological space is compact iff subset is closed and bounded (Heine-Borel theorem)
Topics
About: topological space
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of Euclidean metric space.
- The reader knows a definition of compact topological space.
- The reader knows a definition of closed subset of topological space.
- The reader knows a definition of bounded subset of metric space.
- The reader admits the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger.
- The reader admits the proposition that any compact subset of any Hausdorff topological space is closed.
- The reader admits the proposition that the product of any finite number of compact topological spaces is compact.
- The reader admits the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.
- The reader admits the proposition that the \(d\)-dimensional Euclidean topological space is homeomorphic to the product of any combination of some lower-dimensional Euclidean spaces whose (the product's) dimension equals \(d\).
- The reader admits the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
- The reader admits the proposition that any closed subset of any compact topological space is compact.
- The reader admits the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
Target Context
- The reader will have a description and a proof of the proposition that any subset of any Euclidean metric topological space is compact if and only if the subset is closed and bounded (the Heine-Borel theorem).
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean metric topological space }\)
\(S\): \(\subseteq \mathbb{R}^d\)
//
Statements:
\(S \in \{\text{ the compact subsets of } \mathbb{R}^d\}\)
\(\iff\)
\(S \in \{\text{ the closed subsets of } \mathbb{R}^d\} \cap \{\text{ the bounded subsets of } \mathbb{R}^d\}\)
//
2: Proof
Whole Strategy: Step 1: see that any closed interval, \([r_1, r_2] \subseteq \mathbb{R}\), is compact; Step 2: suppose that \(S\) is compact; Step 3: see that \(S\) is closed; Step 4: see that \(S\) is bounded; Step 4: suppose that \(S\) is closed and bounded; Step 5: see that \(S \subseteq [-n, n]^d\), that \(S\) is compact on \([-n, n]^d\), and that \(S\) is compact on \(\mathbb{R}^d\).
Step 1:
Let us see that any closed interval, \([r_1, r_2] \subseteq \mathbb{R}\), is a compact subset.
Let \(\widetilde{U} := \{U_j \subseteq \mathbb{R} \vert j \in J\}\) be any open cover of \([r_1, r_2]\).
Let us define \(Q := \{r \in [r_1, r_2] \vert [r_1, r] \text{ has a finite subcover of } \widetilde{U}\}\): \(\widetilde{U}\) is an open cover of \([r_1, r]\).
\(Q\) is not empty, because \(r_1 \in Q\) (there is a \(U_j\) such that \(r_1 \in U_j\), and \(\{U_j\}\) is a finite subcover), and \(Q\) is upper bounded, because \(r \le r_2\).
So, the supremum, \(r' := Sup (Q)\), exists, as a property of \(\mathbb{R}\), as is well known.
\(r_1 \le r' \le r_2\), because if \(r_2 \lt r'\), there would be an \(r'' \in Q\) such that \(r_2 \lt r''\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger, a contradiction.
Let us see that \(r' \in Q\).
Let us suppose that \(r' \notin Q\).
\(r_1 \lt r'\), because \(r' \neq r_1\), because \(r_1 \in Q\).
There would be a \(U_j \in \widetilde{U}\) such that \(r' \in U_j\), because \(r' \in [r_1, r_2]\), there would be a \(B_{r', \epsilon} \subseteq U_j\), so, there would be an \(r''\) such that \(r_1 \lt r'' \lt r'\) and \(r'' \in B_{r', \epsilon} \subseteq U_j\), but there would be an \(r''' \in Q\) such that \(r'' \lt r'''\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger, which would mean that \([r_1, r''']\) had a finite subcover, then, the subcover plus \(U_j\) would be a finite subcover for \([r_1, r']\), because for each \(r \in [r_1, r''']\), \(r\) was contained in the subcover for \([r_1, r''']\) while for each \(r \in (r''', r']\), \(r \in B_{r', \epsilon} \subseteq U_j\), which would mean that \(r' \in Q\), a contradiction against \(r' \notin Q\).
Let us suppose that \(r' \lt r_2\).
There would be a \(U_j \in \widetilde{U}\) such that \(r' \in U_j\), and there would be a \(B_{r', \epsilon} \subseteq U_j\) and there would be an \(r'' \in B_{r', \epsilon}\) such that \(r' \lt r'' \lt r_2\).
Then, while \([r_1, r']\) had a finite subcover of \(\widetilde{U}\), \([r_1, r'']\) would have the finite subcover plus \(U_j\) as a finite subcover, because for each \(r \in [r_1, r']\), \(r\) was contained in the subcover for \([r_1, r']\) while for each \(r \in (r', r'']\), \(r\) would be contained in \(B_{r', \epsilon} \subseteq U_j\), which would mean that \(r'' \in Q\), a contradiction against that \(r'\) was the supremum.
So, \(r' = r_2\).
So, \([r_1, r_2]\) has a finite subcover.
So, \([r_1, r_2]\) is a compact subset.
Step 2:
Let us suppose that \(S\) is compact.
Step 3:
\(\mathbb{R}^d\) is Hausdorff, as is well known.
\(S \subseteq \mathbb{R}^d\) is closed, by the proposition that any compact subset of any Hausdorff topological space is closed.
Step 4:
\(\{B_{0, n} \subseteq \mathbb{R} \vert n \in \mathbb{N} \setminus \{0\}\}\) is an open cover of \(S\), because for each \(s \in S\), \(s \in B_{0, n}\) for an \(n\).
As \(S\) is compact, there is a finite subcover, \(\{B_{0, n} \vert n \in N\}\), where \(N \subseteq \mathbb{N} \setminus \{0\}\) is a finite subset.
There is \(n' := Max (N)\), and \(S \subseteq B_{0, n'}\).
So, \(S\) is bounded.
Step 4:
Let us suppose that \(S\) is closed and bounded.
Step 5:
There is an \(r \in \mathbb{R}\) such that \(S \subseteq [-r, r]^d\).
\([-r, r]\) is compact, by Step 1.
\([-r, r]^d\) as the product topological space is compact, by the proposition that the product of any finite number of compact topological spaces is compact.
\([-r, r]^d \subseteq \mathbb{R}^d\) as the topological subspace is \([-r, r]^d\) as the product topological space, by the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces and the proposition that the \(d\)-dimensional Euclidean topological space is homeomorphic to the product of any combination of some lower-dimensional Euclidean spaces whose (the product's) dimension equals \(d\).
So, \([-r, r]^d \subseteq \mathbb{R}^d\) as the topological subspace is compact.
\(S\) closed on \(\mathbb{R}^d\) is closed on \([-r, r]^d\), by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset: \(S = S \cap [-r, r]^d\).
So, \(S\) is compact on \([-r, r]^d\), by the proposition that any closed subset of any compact topological space is compact.
Then, \(S\) is compact on \(\mathbb{R}^d\), by the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
