Open Balls and Continuity When we define continuity using open balls, we define $$\forall \epsilon >0, \exists\delta>0:f(B_\delta(a))\subset B_\epsilon(f(a))$$
Let us consider everything unspecified to be in $\mathbb{R^n}$ i.e. $f:\mathbb{R^n}\rightarrow \mathbb{R^n}$and usual distance functions if these are necessary to be specified.
What is so special about open balls here? It seems closed balls work fine too.
 A: Let $(X,d),(Y,\rho)$ be metric spaces and $f:X\to Y$. For the rest of the post we will say $f$ is Cont1 at $a\in X$ if
$$\forall \epsilon>0\exists \delta>0:f(B_{\delta}(a))\subseteq B_{\epsilon}(f(a))$$
and Cont2 if
$$\forall \epsilon>0\exists \delta>0:f(\overline{B}_{\delta}(a))\subseteq \overline{B}_{\epsilon}(f(a))$$
We will prove that $f$ is Cont1 $\iff$ $f$ is Cont2. 
Cont1$\implies$ Cont2:
Let $\epsilon>0$. Then,
$$\exists \delta>0:f(B_{\delta}(a))\subseteq B_{\epsilon}(f(a))$$
Let $0<\delta^{\prime}<\delta$. Then,
$$f(\overline{B}_{\delta^{\prime}}(a))\subseteq f(B_{\delta}(a))\subseteq B_{\epsilon}(f(a))\subseteq \overline{B}_{\epsilon}(f(a))$$
which is Cont2
Cont2$\implies$ Cont1:
Let $\epsilon>0$ and $0<\epsilon^{\prime}<\epsilon$. Then,
$$\exists \delta>0:f(\overline{B}_{\delta}(a))\subseteq \overline{B}_{\epsilon^{\prime}}(f(a))$$
Then,
$$f(B_{\delta}(a))\subseteq f(\overline{B}_{\delta}(a))\subseteq \overline{B}_{\epsilon^{\prime}}(f(a))\subseteq B_{\epsilon}(f(a))$$
which is Cont1
Therefore whether you define the balls in the definition of continuity to be open or closed makes no difference
A: As long as your require the radius of the ball to be strictly positive note that closed balls contain open balls, so the requirement for "open balls" is satisfied.
Generally, however, the topological definition for a continuous map is "the preimage of an open set is open". It is enough to require this for basic open sets, which in metric spaces corresponds to open balls. Equally, however, one can require this for closed sets as well, namely the preimage of a closed set is closed. Do note that in this case the notion is not transferred immediately to closed balls, there are closed sets which are not the union/intersection of closed balls (for example the $x$-axis in the plane).
