set inclusion

ender · **Joined:** Thu, 22 Jun 2006 01:08:00 UTC **Posts:** 123

hi, can you give me an idea of this?

let $(\Omega, d)$ and $(\Omega \prime , d\prime )$ be two metric spaces, let $f:\Omega \mapsto \Omega \prime$ continuous, for $A\subseteq \Omega$

$f(\overline A)\subseteq \overline{f(A)}$

give an example where the inclusion is strict.

I´m not looking for someone to give me the answer, just an idea, thanks

outermeasure · **Posted:** Thu, 13 Oct 2011 17:10:02 UTC

ender wrote:

hi, can you give me an idea of this?

let $(\Omega, d)$ and $(\Omega \prime , d\prime )$ be two metric spaces, let $f:\Omega \mapsto \Omega \prime$ continuous, for $A\subseteq \Omega$

$f(\overline A)\subseteq \overline{f(A)}$

give an example where the inclusion is strict.

I´m not looking for someone to give me the answer, just an idea, thanks

You can find them inside $\mathbb{R}$ .

qleak · **Posted:** Thu, 13 Oct 2011 17:58:56 UTC

Let f:R->R be a function where R is the reals.

Okay, the closure of f(A) is basically points approached by y-values of the graph.
Let $y \in \overline{f(A)}$ . Suppose $y \in f(\overline{A})$ , then f(x)=y for some x in $\overline{A}$ .

So if you don't want y in $f(\overline{A})$ , you can't have f(x)=y for any x in A and you can't have f(x)=y for any x real with $x_n \to x$ for some sequence $x_n \in A$ .

Thus, a continuous function with a horizontal asymptote it never crosses should be fine, there are tons of functions like this.

S.O.S. Mathematics CyberBoard

set inclusion

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