Existence of a ball whose images under two continuous functions are disjoint.

Question

Let $f,g: X \to \mathbb{R}^{n}$ be a continuous functions at point $a$. If $f(a)\neq g(a)$, show that exists an open ball B with center $a$ such that $\forall x,y \in B \Rightarrow f(x)\neq g(y)$.

My thoughts:

$b=|f(a)-g(a)|\neq 0.$ For $\epsilon=b/2$, there is $\delta_{1}>0$ such that $|x-a|<\delta_{1} \Rightarrow |(f(x)-g(x))-(f(a)-g(a))|<\epsilon = b/2$. Then $|f(x)-g(x) |>b/2$.

Now $|f(x)-g(y)|+|g(y)-g(a)| \geq |f(a)-g(a)|=b>0$

$c=|g(a)|$. For $\epsilon =\min\{c/2,b/2\}=d$, there is $\delta_{2}$ such that $|x-a|<\delta_{2} \Rightarrow |g(a)-g(x)|<d$.

If $x,y \in B_{\delta}(a), \delta=\min\{\delta_{1},\delta_{2}\}$, then $|f(x)-g(y)|>b-d>0$.

I believe it is correct, but I'm not sure.

Answer 1

It can also be proved by contraposition. Suppose that in every ball $B$ centered at $a$, there exists points $x, y\in B$ such that $f(x)=g(y)$. Taking balls of radius $\dfrac{1}{n}$, $n\geq 1$, we then get a sequence of points $x_n$ and $y_n$ such that $|x_n-a|, |y_n-a|<\dfrac{1}{n}$ and $f(x_n)=g(y_n)$. Obviously, $x_n\rightarrow a$ and $y_n\rightarrow a$ as $n\rightarrow\infty$. Since $f$ and $g$ are continuous, so $$\lim_{n\rightarrow \infty}f(x_n)=f(a),\;\;\; \lim_{n\rightarrow \infty}g(y_n)=g(a)$$ and therefore $f(a)=g(a)$.

Answer 2

Construct a ball $F \in \mathbb{R} ^n$ centered at $f(a) $ and another ball $G\in\mathbb{R} ^n$ centered at $g(a) $ such that $F\cap G=\emptyset$. This is possible because $f(a) \neq g(a) $. By continuity we have a ball $A\subseteq X$ centered at $a$ such that $f(A) \subseteq F$ and similarly another ball $B\subseteq X$ centered at $a$ such that $g(B) \subseteq G$. The ball $C=A\cap B$ is our desired ball such that $f(C) \cap g(C) =\emptyset$.