Answered: Let f : R → R be a strictly increasing…

Let f : R → R be a strictly increasing continuous function. Show that f maps nowhere dense sets to nowhere dense sets; that is, f(E) = {f(x) : x ∈ E} is nowhere dense if E is nowhere dense.

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter1: Fundamentals

Section1.2: Mappings

Problem 27E: 27. Let , where and are nonempty. Prove that has the property that for every subset of if and...

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Question

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Let f : R → R be a strictly increasing continuous function. Show that f maps nowhere
dense sets to nowhere dense sets; that is, f(E) = {f(x) : x ∈ E} is nowhere dense if E
is nowhere dense.

Expert Solution

Step 1

Given $f : R \to R$ be a strictly increasing continuous function.

We have to show that $f$ maps sense sets to nowhere dense sets that is $f (E) = \{f (x) : x \in E\}$ is nowhere dense if $E$
is nowhere dense that is ${(\bar{f (E)})}^{\circ} = ϕ$ .

Let us suppose ${(\bar{f (E)})}^{\circ} \neq ϕ$

Let $p \in {(\bar{f (E)})}^{\circ}$

This implies $p$ is an interior point of $(\bar{f (E)})$ $\exists$ an open set $U$ in $R$ such that $p \in U \subseteq (\bar{f (E)})$ .

Hence, $p \in U \subseteq (\bar{f (E)}) \subseteq (f (\bar{E}))$

Hence, $f^{- 1} (p) \in \bar{E} and f^{- 1} (p) \in f^{- 1} (U)$ .

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