I think that a). is valid. The definition of continuity, at a point $\displaystyle c \in M$, adapted to this case is:Quote:

Is at least one of the following two implications valid for every map $\displaystyle f:M \rightarrow N$ of a metric space $\displaystyle M$ to a metric space $\displaystyle N$?

a). $\displaystyle f$ is continuous $\displaystyle \Rightarrow \ f(U)$ is open for every $\displaystyle U \subset M$.

b). $\displaystyle f(U)$ is open for every open $\displaystyle U \subset M \Rightarrow \ f$ is continuous.

$\displaystyle (\forall \varepsilon >0)(\exists \delta >0)[d_M(c,x)< \delta \ \Rightarrow \ d_N(f(c),f(x))< \varepsilon]$

$\displaystyle d_N(f(c),f(x))< \varepsilon$ is just $\displaystyle B_{\varepsilon}(f(c),d_N)$ valid $\displaystyle \forall c,x \in M$.

Hence it does imply that $\displaystyle f(U)$ is open for every $\displaystyle U \subset M$.

So, i've answered the question with a yes! However, for the sake of curiosity I will try and figure out b).

Since a). is true, i'll try and find a counterexample.

Consider the French Railway Metric. This is open because a ball can be constructed around every point.

However, $\displaystyle f$ is not continuous.

Suppose that $\displaystyle x=(1,0)$ and $\displaystyle x_n=(1+\frac{1}{n},0)$.

As $\displaystyle d(x,x_n) \rightarrow 0$, $\displaystyle d_F(x,x_n) \rightarrow 2$ so it cannot be continuous.

Is this the correct way to think about the problem?