Show that exist open ball B, such that $f(B)\cap g(B)=\emptyset$ $M,N$ metric space. Let $f,g:M\to N$ continuous in a point $a\in M$. If $f(a)\neq g(a)$, then exist a open ball B of center a, such that $f(B)\cap g(B)=\emptyset$. In particular, if $x\in B$, then $f(x)\neq g(x)$.
My approach, if $f,g:M\to N$ are continuous in a point $a\in M$, with $f(a)\neq g(a)$, then given $\epsilon_f>0$ for the open ball $B(f(a),\epsilon_f)$, we can find a $\delta_f$, such that $f(B(a,\delta_f))\subset B(f(a),\epsilon_f)$. Analogous with g, then $g(B(a,\delta_g))\subset B(g(a),\epsilon_g)$. Now defined, $\epsilon=min\{\epsilon_f,\epsilon_g\}$, then $B(f(a),\epsilon)\cap B(g(a),\epsilon)=\emptyset$, any idea pls!. regards!
 A: Hint: A metric space is a Hausdorff topological space.  The preimage of an open set under a continuous mapping is open.  The intersection of two open sets is open.  For any nonempty open set $U$ and $a\in U$, there exists an open ball $B$ centered at $a$ such that $B\subseteq U$.
A: Metrizable spaces are Hausdorff spaces. So given two distinct points $f(a), g(a)\in N$, we can find disjoint neighborhoods $U$ and $V$ of $f(a)$ and $g(a)$, respectively. Then by continuity of $f$ and $g$, $f^{-1}(U)$ and $g^{-1}(V)$ are open in $M$, so $f^{-1}(U)\cap g^{-1}(V)$ is a neighborhood of $a$ such that 
\begin{align}
&\ f(f^{-1}(U)\cap g^{-1}(V)) \cap g(f^{-1}(U)\cap g^{-1}(V))\\
\subset &\ f(f^{-1}(U))\cap f(g^{-1}(V))\cap g(f^{-1}(U))\cap g(g^{-1}(V))\\
\subset &\ U\cap f(g^{-1}(V))\cap g(f^{-1}(U))\cap V\\
\subset &\ U\cap V = \varnothing.
\end{align}
As for an "$\epsilon-\delta$" proof: Let $\varepsilon = d_N(f(a), g(a))$. Choose $\delta_f>0$ such that $d_M(a,b)<\delta_f$ implies $d_N(f(a), f(b))<\frac\varepsilon2$, and $\delta_g>0$ such that $d_M(a,b)$ implies $d_N(g(a), g(b))<\frac\varepsilon2$. Put $\delta=\min\{\delta_f, \delta_g\}$, then
$$f(B(a, \delta))\cap g(B(a,\delta)) \subset B\left(f(a), \frac\varepsilon2\right)\cap B\left(g(a), \frac\varepsilon2\right)=\varnothing. $$
A: Your set up is very good, but something you should notice is that you have power in choosing how small $\epsilon_f$ and $\epsilon_g$ can be.  In your setup, take $\epsilon_f < d(f(a),g(a))$ to be arbitrary, but instead of letting $\epsilon_g$ be arbitrary, let 
$$
\epsilon_g\;\; < \;\; d(f(a), g(a)) - \epsilon_f
$$
where $d:N \times N \to [0,\infty)$ is the metric on $N$.  Now we can simply let $x \in B(g(a), \epsilon_g)$ and show that $x$ is not in the ball of radius $\epsilon_f$ around $f(a)$.  Observe that
$$
d(f(a), g(a)) \;\; \leq \;\; d(f(a), x) + d(x, g(a)) \;\; < \;\; d(f(a),x) + d(f(a), g(a)) - \epsilon_f.
$$
The above inequality simplifies to $d(f(a),x) > \epsilon_f$.  
