3
$\begingroup$

For $x∈\mathbb{R}^n$ , let $B(x,r)$ denote the closed ball in $\mathbb{R}^n$(with Euclidean norm) of radius $r$ centered at $x$. Write $B=B(0,1)$.If $f,g:B→\mathbb{R}^n$ are continuous functions such that $f(x)≠g(x)$ for all $ x∈ B$, then which of the followings are true?

  1. $f(B)∩g(B)=\varnothing$
  2. There exist $ϵ>0$ such that $||f(x)-g(x)||> ϵ$ for all $ x∈ B$
  3. There exist $ϵ>0$ such that $ B(f(x), ϵ) ∩ B(g(x), ϵ)=\varnothing$ for all $ x∈ B$
  4. ${\rm int }(f(B)) ∩ {\rm int }(g(B))=\varnothing$ , where ${\rm int}(E)$ denotes the interior of a set $E$

How can I solve this problem? Can anyone help?

  • 2
    Please consider accept the answers they give you. [How do I accept an answer?](http://meta.math.stackexchange.com/q/3286/8271)2012-12-23

2 Answers 2

2

The assumption that $f,g$ are defined on $B(0,1)$ is entirely spurious. You can decide on the truth of these statements and get an idea of possible proofs and counterexamples by considering the case $f, g:[-1.1] \to \mathbb{R}$.

Using this simplification, draw graphs for each case and decide whether the statements are true or false.

1

Edited to reflect the new question.

(2) immediately implies (3). (Can you see why?)

(1) is not necessarily true. If $n=1$, $f(x)=x$, $g(x)=x+1$, then $g(x)\neq f(x)$, $B=[-1,1]$ and $g(B)=[0,2]$, $f(B)=[-1,1]$. So the intersection is $[0,1]$. This also shows that (4) is false, since the intersection of the interiors is $(0,1)$.

Finally, (2) is true:

Consider the function $h:B\to\Bbb R$ given by $h(x)=||f(x)-g(x)||.$ Thus $h$ is strictly positive everywhere on $B$. $B$ is compact and $h$ is continuous so $h$ attains its minimum (and its maximum). Pick $\epsilon=\min\{h(x):x\in B\}$.

  • 0
    I've merged our answers. Feel free in reverse it.2012-12-23