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.

$\blacksquare$


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