Closure of Preimage under Continuous Mapping is not necessarily Preimage of Closure

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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.


Then it is not necessarily the case that:

$f^{-1} \sqbrk {\map \cl H} = \map \cl {f^{-1} \sqbrk H}$


Proof

Proof by Counterexample:

Let $\R$ be the real numbers under the usual (Euclidean) topology.


Let $f: \R \to \R$ be the real function:

$\forall x \in \R: \map f x = \begin {cases} -1 & : x \le -1 \\ x & : -1 \le x \le 1 \\ 1 & : x \ge 1 \end {cases}$

It is accepted that $f$ is a continuous mapping.


Let $H \subseteq \R$ be the subset of $\R$ defined as:

\(\ds H\) \(=\) \(\ds \openint 0 1\) which is an open set in $\R$.
\(\ds \leadsto \ \ \) \(\ds \map \cl H\) \(=\) \(\ds \closedint 0 1\) Definition of Closure (Topology)
\(\ds \leadsto \ \ \) \(\ds f^{-1} \sqbrk H\) \(=\) \(\ds \openint 0 1\) Definition of $f$
\(\ds f^{-1} \sqbrk {\map \cl H}\) \(=\) \(\ds \R\) Definition of $f$
\(\ds \map \cl {f^{-1} \sqbrk H}\) \(=\) \(\ds \closedint 0 1\) Definition of Closure (Topology)

As can be seen:

$f^{-1} \sqbrk {\map \cl H} \ne \map \cl {f^{-1} \sqbrk H}$

$\blacksquare$


Sources

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