Closed Image of Closure of Set under Continuous Mapping equals Closure of Image
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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 $H \subseteq S_1$ be a subset of $S_1$.
Let $\map \cl H$ denote the closure of $H$.
Let $f: T_1 \to T_2$ be a continuous mapping.
Let $f \sqbrk {\map \cl H}$ be closed in $T_2$.
Then:
- $f \sqbrk {\map \cl H} = \map \cl {f \sqbrk H}$
Proof
By Continuity Defined by Closure:
- $f \sqbrk {\map \cl H} \subseteq \map \cl {f \sqbrk H}$
$\Box$
\(\ds H\) | \(\subseteq\) | \(\ds \map \cl H\) | Set is Subset of its Topological Closure | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds f \sqbrk H\) | \(\subseteq\) | \(\ds f \sqbrk {\map \cl H}\) | Image of Subset is Subset of Image | |||||||||||
We are given that $f \sqbrk {\map \cl H}$ is closed in $T_2$, hence: | |||||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \cl {f \sqbrk H}\) | \(\subseteq\) | \(\ds f \sqbrk {\map \cl H}\) | Closure of Subset of Closed Set of Topological Space is Subset |
$\Box$
The proof follows by definition of set equality.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 26$