1114: Closure of Continuous Map Preimage of Subset Is Contained in but Not Necessarily Equal to Preimage of Closure of Subset

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description/proof of that closure of continuous map preimage of subset is contained in but not necessarily equal to preimage of closure of subset

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of closure of subset.
• The reader knows a definition of continuous map.
• The reader knows a definition of Euclidean topology.
• The reader admits the proposition the preimage of any closed subset under any continuous map is a closed subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain is contained in but not necessarily equal to the preimage of the closure of the subset.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the continuous maps }\}$
$S$: $\subseteq T_2$
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Statements:
$\overline{f^{-1}(S)}\subseteq f^{-1}(\bar{S})$
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2: Proof

Whole Strategy: Step 1: see that $f^{-1}(S)\subseteq f^{-1}(\bar{S})$ and $f^{-1}(\bar{S})\subseteq T_1$ is closed, and think of the definition of closure to see that $\overline{f^{-1}(S)}\subseteq f^{-1}(\bar{S})$; Step 2: see an example that $\overline{f^{-1}(S)}\subset f^{-1}(\bar{S})$.

Step 1:

$f^{-1}(S)\subseteq f^{-1}(\bar{S})$, obviously.

$f^{-1}(\bar{S})$ is closed, by the proposition the preimage of any closed subset under any continuous map is a closed subset.

As $\overline{f^{-1}(S)}$ is the smallest closed subset that contains $f^{-1}(S)$ by the definition of closure of subset, $\overline{f^{-1}(S)}\subseteq f^{-1}(\bar{S})$, because the right hand side is one of such closed subsets.

Step 2:

As an example that the equality does not hold, let $T_1$ be $\mathbb{R}$ with the discrete topology, $T_2$ be $\mathbb{R}$ with the Euclidean topology, $f$ be the identity map, $S$ be $(0,1)$. $f$ is continuous, because the preimage of any open subset is open. $\overline{f^{-1}(S)}=(0,1)$, because $[1,\mathrm{\infty })$ and $(-\mathrm{\infty },0]$ are open on $T_1$, while $f^{-1}(\bar{S})=[0,1]$, so, $\overline{f^{-1}(S)}\subset f^{-1}(\bar{S})$.

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References

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