Let \((M, d_M)\), \((N, d_N)\) be metric spaces. Define \(d_{M \times N} \colon (M \times N) \times (M \times N) \to \mathbb{R}\) by \[d_{M \times N}((x_1, y_1), (x_2, y_2)) := d_M(x_1, x_2) + d_N(y_1, y_2).\]
Prove that \((M \times N, d_{M \times N})\) is a metric space.
Let \(S \subseteq M\) and \(T \subseteq N\) be compact sets in \((M, d_M)\) and \((N, d_N)\), respectively. Prove that \(S \times T\) is a compact set in \((M \times N, d_{M \times N})\).
Proof. To prove that \((M \times N, d_{M \times N})\) is a metric space we must prove that \(d_{M\times N}\) is a metric on \(M \times N\).
- Positive Definite-
Let \((x_1,y_1), (x_2,y_2) \in M \times N\). Then since \(d_M\) is a metric on \(M\), then \(d_M(x_1,x_2)\geq 0\) for all \(x_i,x_j \in M\) and \(d_N\) is a metric on \(N\) and likewise \(d_N(y_1,y_2)\geq 0\) for any \(y_i,y_j \in N.\)
Then by definition \(d_{M\times N}((x_1,y_1),(x_2,y_2))=d_M(x_1,x_2)+d_N(y_1,y_2)\geq 0 + 0 =0.\) Hence since \((x_1,y_1),(x_2,y_2)\) are arbitrary, \(d_{M\times N}((x_1,y_1),(x_2,y_2))\geq 0\) for all \((x_i,y_i),(x_j,y_j)\in M \times N\).
Suppose that \(d_{M \times N}((x_1,y_1),(x_2,y_2))=0.\) By definition \(d_{M \times N}((x_1,y_1),(x_2,y_2))=d_M(x_1,x_2)+d_N(y_1,y_2)\). Therefore \(d_M(x_1,x_2)+d_N(y_1,y_2)=0\), since \(d_M, d_N\) are metrics, then \(d_M(x_1,x_2)\geq 0, d_N(y_1,y_2)\geq 0\), which implies that \(d_M(x_1,x_2)=d_N(y_1,y_2)=0\) and also since they are metrics we have that \(x_1=x_2, y_1=y_2.\) Hence, \((x_1,y_1)=(x_2,y_2).\)
Now suppose that \((x_1,y_1)=(x_2,y_2).\) Then \(x_1=x_2, y_1=y_2\) and for the metrics \(d_M, d_N\) we would have \(d_M(x_1,x_2)=0, d_N(y_1,y_2)=0.\) Thus \(d_{M \times N}((x_1,y_1),(x_2,y_2))=d_M(x_1,x_2)+d_N(y_1,y_2)=0+0=0\).
Therefore \(d_{M \times N}((x_1,y_1),(x_2,y_2))=0\) if and only if \((x_1,y_1)=(x_2,y_2).\)
- Symmetric
Let \((x_1,y_1), (x_2,y_2) \in M \times N\). Then since \(d_M\) is a metric on \(M\), then \(d_M(x_1,x_2)=d_M(x_2,x_1)\) for all \(x_i,x_j \in M\) and \(d_N\) is a metric on \(N\) and likewise \(d_N(y_1,y_2)=d_N(y_2,y_1)\) for any \(y_i,y_j \in N.\) Therefore, \[\begin{aligned} d_{M \times N}((x_1,y_1),(x_2,y_2))&=d_M(x_1,x_2)+d_N(y_1,y_2)\\ &=d_M(x_2,x_1)+d_N(y_2,y_1)\\ &=d_{M \times N}((x_2,y_2),(x_1,y_1)). \end{aligned}\]
- Triangle Inequality
Since \(d_M, d_N\) are metrics then for all \(x_1,x_2,x_3 \in M, y_1,y_2,y_3 \in N\) we have that \(d_M(x_1,x_2)\leq d_M(x_1,x_3)+d_M(x_3,x_2)\) and that \(d_N(y_1,y_2)\leq d_N(y_1,y_3)+d_N(y_3,y_2).\) Therefore, \[\begin{aligned} d_{M \times N}((x_1,y_1),(x_2,y_2))&=d_M(x_1,x_2)+d_N(y_1,y_2)\\ d_M(x_1,x_2)+d_N(y_1,y_2) &\leq d_M(x_1,x_3)+d_M(x_3,x_2)+d_N(y_1,y_3)+d_N(y_3,y_2)\\ d_M(x_1,x_3)+d_M(x_3,x_2)+d_N(y_1,y_3)+d_N(y_3,y_2) &=d_M(x_1,x_3)+d_N(y_1,y_3)+d_M(x_3,x_2)+d_N(y_3,y_2)\\ d_M(x_1,x_3)+d_N(y_1,y_3)+d_M(x_3,x_2)+d_N(y_3,y_2)&=d_M((x_1,y_1),(x_3,y_3))+d_M((x_3,y_3),(x_2,y_2)). \end{aligned}\]
Hence \(d_{M \times N}((x_1,y_1),(x_2,y_2))\leq d_M((x_1,y_1),(x_3,y_3))+d_M((x_3,y_3),(x_2,y_2)).\)
Therefore \(d_{M \times N}\) is a metric on \(M \times N\) and \((M \times N, d_{M\times N})\) is a metric space. ◻
Proof. By part a we showed that \((M \times N, d_{M \times N})\) is a metric space. Let \(\{s_n,t_n\}\) be a sequence in \(S \times T.\) Since \(\{s_n\}\) is a sequence on a compact set \(S\) in a metric space \((M,d_M)\) then it has a convergent subsequence \({s_{n_k}}.\) Let \(\lim_{k \to \infty}{s_{n_k}}=s_0.\)
Since \(\{t_{n_k}\}\) is a sequence on a compact set \(T\) in a metric space. Thus \(\{t_{n_k}\}\) has a convergent subsequence \(\{t_{n_{k_j}}\}.\) Let \(\lim_{j\to \infty} t_{n_{k_j}}=t_0.\) Thus \(\{s_{n_{k_j}}\}\) is a subsequence of \(\{s_{n_k}\}.\) And since \(\{s_{n_k}\}\) converges to \(s_0\), then any subsequence also converges to \(s_0.\)
Let \(\epsilon >0\) be given. Then for \(\epsilon/2\) there exists \(N_1, N_2\in \mathbb{N}\) such that for all \(n_{k_j}\geq N_1, d_M(s_{n_{k_j}},s_0)<\epsilon/2\), and for all \(n_{k_j}\geq N_2, d_N(t_{n_{k_j}},t_0)<\epsilon/2\). Choose \(N=\text{Max}(\{N_1,N_2\}).\)
Then \(d_{M \times N}((s_{n_{k_j}},t_{n_{k_j}}),(s_0,t_0))=d_M(s_{n_{k_j}},s_0)+d_N(t_{n_{k_j}},t_0)<\epsilon/2 + \epsilon/2 = \epsilon.\)
Therefore \(d_{M \times N}((s_{n_{k_j}},t_{n_{k_j}}),(s_0,t_0))< \epsilon.\)
Hence \(\{(s_{n_{k_j}},t_{n_{k_j}})\) converges to \((s_0,t_0).\) Therefore \(S \times T\) is sequentially compact and \(S \times T\) is therefore compact. ◻
Solution
Part 1: \(d_{M \times N}\) is a metric.
1. \(d_{M\times N} \ge 0\), and \(d_{M\times N}((x_1,y_1),(x_2,y_2)) = 0\) iff \((x_1,y_1) = (x_2,y_2)\).
\(d_{M\times N} = d_M(x_1,x_2) + d_N(y_1,y_2) \ge 0\) with equality iff both \(d_M = 0\) and \(d_N = 0\), i.e. \(x_1 = x_2\) and \(y_1 = y_2\) (metrics separate points).
2. Symmetry: \(d_{M\times N}(p, q) = d_{M\times N}(q, p)\) for all \(p, q \in M \times N\).
\(d_M, d_N\) are symmetric, so the sum is.
3. Triangle inequality: for \(p_i = (x_i, y_i)\), \(d_{M\times N}(p_1, p_3) \le d_{M\times N}(p_1, p_2) + d_{M\times N}(p_2, p_3)\).
\(d_{M\times N}(p_1,p_3) = d_M(x_1,x_3) + d_N(y_1,y_3) \le (d_M(x_1,x_2) + d_M(x_2,x_3)) + (d_N(y_1,y_2) + d_N(y_2,y_3))\) by the triangle inequalities in \(M\) and \(N\); regroup.
4. Q.E.D.
<1>1–<1>3 are the metric axioms.
Part 2: \(S \times T\) is compact.
5. Every sequence \(((s_k, t_k)) \subseteq S \times T\) has a convergent subsequence.
5.1. \((s_k)\) has a convergent subsequence \(s_{k_j} \to s \in S\).
\(S\) is compact (metric, hence sequentially compact).
5.2. \((t_{k_j})\) has a further convergent subsequence \(t_{k_{j_l}} \to t \in T\).
\(T\) is compact.
5.3. \((s_{k_{j_l}}, t_{k_{j_l}}) \to (s, t) \in S \times T\) in \(d_{M\times N}\).
\(d_{M\times N}((s_{k_{j_l}},t_{k_{j_l}}),(s,t)) = d_M(s_{k_{j_l}}, s) + d_N(t_{k_{j_l}}, t) \to 0\).
6. Q.E.D.: \(S \times T\) is compact.
5 is sequential compactness, equivalent to compactness in metric spaces. (Alternatively: \(S \times T\) is closed and bounded in the product of complete spaces, or use the open-cover/finite-subcover argument.)