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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?

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Is it supposed to read $f,g:\Bbb R^n\to\Bbb R^n$? Also, $f(x)$ and $g(x)$ are not sets, as far as I can tell here. What should $1.$ read? –  Pedro Tamaroff Dec 20 '12 at 2:27
    
sorry for my mistakes.now i have corrected it –  daichi Dec 20 '12 at 2:32
2  
Please consider accept the answers they give you. How do I accept an answer? –  leo Dec 23 '12 at 21:28
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2 Answers

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.

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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 $\epsilon=\min\{h(x):x\in B\}$.

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then all the 4 options are correct. am i right? –  daichi Dec 20 '12 at 3:26
    
Wait. This was a true/false type of question? If so you should have mentioned this...I haven't thought about (1) that deeply, so I don't know if (1) is true yet. I felt like I should at least let you do that part (and not confuse you with an answer I would have to come up with on the spot). –  kigen Dec 20 '12 at 3:37
    
very sorry i have to mention that.yes its a true/false type question –  daichi Dec 20 '12 at 3:42
    
Please edit your question in that case. –  kigen Dec 20 '12 at 3:44
    
I've merged our answers. Feel free in reverse it. –  leo Dec 23 '12 at 21:26
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