Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 i have corrected it – daichi Dec 20 '12 at 2:32
Please consider accept the answers they give you. How do I accept an answer? – leo Dec 23 '12 at 21:28

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.

share|cite|improve this answer

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

share|cite|improve this answer
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). – Gyu Eun Lee 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. – Gyu Eun Lee Dec 20 '12 at 3:44
I've merged our answers. Feel free in reverse it. – leo Dec 23 '12 at 21:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.