Section 4.2 Total Boundedness
Definition 4.7.
A subset \(S\) of a metric space \((X,d)\) is called totally bounded if for every \(r \gt 0\text{,}\) there exist finitely many points \(x_1, \dotsc, x_N \in S\) such that
\begin{equation*}
S \subseteq \bigcup_{n=1}^N B_r(x_n).
\end{equation*}
If \(S=X\) is totally bounded, we say that the metric space \((X,d)\) is totally bounded.
Lemma 4.8. Total boundedness and metric subspaces.
A subset \(S\) of a metric space \((X,d)\) is totally bounded if and only if the associated metric subspace \((S,d')\) is totally bounded.
Proof.
Lemma 4.9.
Every totally bounded subset of a metric space is bounded.
Proof.
Let \(S\) be a totally bounded subset of a metric space \((X,d)\text{.}\) Then there exist finitely many points \(x_1, \dotsc, x_N \in S\) such that
\begin{equation*}
S \subseteq \bigcup_{n=1}^N B_1(x_n).
\end{equation*}
Set \(a = \max_{m,n \in \{1, \dotsc, N\}} d(x_m, x_n)\text{.}\) Then for any pair of points \(x,y \in S\text{,}\) there exist \(m,n \in \{1, \dotsc, N\}\) with \(d(x, x_m) \lt 1\) and \(d(y, x_n) \lt 1\text{.}\) So
\begin{equation*}
d(x, y) \le d(x, x_m) + d(x_m, x_n) + d(x_n, y) \le 2 + a.
\end{equation*}
Thus \(\diam S \le 2 + a\text{,}\) which means that \(S\) is bounded.
Theorem 4.10.
If a subset of a metric space is sequentially compact, then it is totally bounded.
Proof.
Let \(K\) be a sequentially compact subset of a metric space \((X,d)\text{.}\) Fix \(r \gt 0\text{.}\) We construct finitely many points \(x_1, \dotsc, x_N\) in \(K\) as follows.
If \(K = \varnothing\text{,}\) then set \(N = 0\) (i.e., we consider an empty set of points). Otherwise, choose a point \(x_1 \in K\) arbitrarily.
If \(K \subseteq B_r(x_1)\text{,}\) then we stop here. Otherwise, choose a point \(x_2 \in K \setminus B_r(x_1)\text{.}\)
If \(K \subseteq B_r(x_1) \cup B_r(x_2)\text{,}\) then we stop here. Otherwise, choose a point \(x_3 \in K \setminus (B_r(x_1) \cup B_r(x_2))\text{.}\)
Continue this process until it terminates. It will terminate after finitely many steps, for if it didn’t, it would produce a sequence \((x_n)_{n \in \N}\) in \(K\) such that \(d(x_m, x_n) \ge r\) whenever \(m \ne n\text{.}\) Such a sequence cannot have a convergent subsequence, and is therefore impossible in a sequentially compact set.
Therefore, we eventually find that
\begin{equation*}
K \subseteq \bigcup_{n=1}^N B_r(x_n)
\end{equation*}
for some \(N \in \N \cup \{0\}\text{.}\)
Lemma 4.11.
Let \((X,d)\) be a totally bounded metric space. Then every subset \(Y \subset X\) is totally bounded.
Proof.
Let \(r \gt 0\) and choose finitely many points \(x_1, \dotsc, x_N \in X\) such that
\begin{equation*}
X \subseteq \bigcup_{n=1}^N B_{r/2}(x_n).
\end{equation*}
We now separate the points with \(B_{r/2}(x_n) \cap Y \ne \varnothing\) from the points with \(B_{r/2}(x_n) \cap Y = \varnothing\text{.}\) We can relabel \(\{x_1, \dotsc, x_N\}\) such that there exists \(M \in \{0, \dotsc, N\}\) with
\begin{equation*}
B_{r/2}(x_n) \cap Y \ne \varnothing \quad \text{for }n=1, \dotsc, M
\end{equation*}
and
\begin{equation*}
B_{r/2}(x_n) \cap Y = \varnothing \quad \text{for }n = M + 1, \dotsc, N.
\end{equation*}
For each \(n=1, \dotsc, M\text{,}\) choose a point \(y_n \in B_{r/2}(x_n)\text{.}\) Recall that for every \(y \in Y\text{,}\) there exists \(n \in \{1, \dotsc M\}\) such that \(d(y, x_n) \lt \frac{r}{2}\text{.}\) Hence \(d(y, y_n) \le d(y, x_n) + d(x_n, y_n) \lt r\text{.}\) It follows that
\begin{equation*}
Y \subseteq\bigcup_{n=1}^M B_r(y_n).
\end{equation*}
Thus \(Y\) is totally bounded.
Proposition 4.12.
For \(n \in \N\text{,}\) every bounded subset of \(\R^n\) is totally bounded.
Proof.
We first show that the set \([-M, M]^n\) is totally bounded for any \(M \in \N\text{.}\) To this end, fix \(r \gt 0\text{.}\) Choose \(k \in \N\) such that \(2^{-k} \lt \frac{r}{2\sqrt{n}}\) and let
\begin{equation*}
A = 2^{-k}\Z^n \cap [-M, M]^n.
\end{equation*}
Then it is clear that \([-M, M]^n \subseteq \bigcup_{a \in A} B_r(a)\text{.}\) As \(A\) is finite, this means that \([-M, M]^n\) is totally bounded.
Finally, given an arbitrary bounded set \(S \subset \R^n\text{,}\) choose \(M \in \N\) such that \(S \subseteq [-M, M]^n\) and apply Lemma 4.11.
Exercises Exercises
1. Total boundedness for subsets.
Let \((X,d)\) be a metric space and \(Y \subseteq X\text{.}\) Show that \(Y\) is totally bounded as a subset of \(X\) if and only if the metric subspace \((Y,d')\) is totally bounded.
Hint.
Part a of Exercise 1.2.2 is useful here.
Solution.
Suppose that the metric subspace \((Y,d')\) is totally bounded and let \(r \gt 0\text{.}\) Then we can find finitely many points \(y_1,\ldots,y_N \in Y\) such that
\begin{gather}
Y \subseteq \bigcup_{n=1}^N B_r'(y_n)
\subseteq \bigcup_{n=1}^N B_r(y_n)\text{,}\tag{✶}
\end{gather}
where here \(B_r'(y_n)\) denotes the open ball in \((Y,d')\) and in the last step we have used Part a of Exercise 1.2.2. Thus \(Y\) is totally bounded as a subset of \(X\text{.}\)
Conversely, suppose that \(Y\) is totally bounded as a subset of \(X\text{.}\) Then for every \(r \gt 0\) we can find \(y_1,\ldots,y_N \in Y\) such that
\begin{gather}
Y \subseteq \bigcup_{n=1}^N B_r(y_n),\tag{✶✶}
\end{gather}
holds, where here \(B_r(y)\) denotes the open ball in \((X,d)\) with radius \(r\) and centre \(y\text{.}\) Intersecting both sides with \(Y\) and using Part a of Exercise 1.2.2 again we find
\begin{equation*}
Y
\subseteq \bigcup_{n=1}^N B_r(y_n) \cap Y
= \bigcup_{n=1}^N B_r'(y_n)\text{.}
\end{equation*}
Thus the metric subspace \((Y,d')\) is totally bounded.
