Boundary of Cartesian Product of Subsets
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Theorem
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $H \subseteq T_1$ and $K \subseteq T_2$.
Then:
- $\map \partial {H \times K} = \paren {\map \partial H \times \map \cl K} \cup \paren {\map \cl H \times \map \partial K}$
where:
- $\map \cl H$, for example, denotes the closure of $H$.
- $\map \partial H$, for example, denotes the boundary of $H$.
Proof
\(\ds \map \partial {H \times K}\) | \(=\) | \(\ds \map \cl {H \times K} \setminus \Int {H \times K}\) | Definition of Boundary (Topology) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \cl H \times \map \cl K} \setminus \Int {H \times K}\) | Closure of Cartesian Product is Product of Closures | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \cl H \times \map \cl K} \setminus \paren {\Int H \times \Int K}\) | Interior of Cartesian Product is Product of Interiors | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \cl H \times \paren {\map \cl K \setminus \Int K} } \cup \paren {\paren {\map \cl H \setminus \Int H} \times \map \cl K}\) | Set Difference of Cartesian Products | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \partial H \times \map \cl K} \cup \paren {\map \cl H \times \map \partial K}\) | Definition of Boundary (Topology) |
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 35 \ \text {(iii)}$