Every function from a discrete subset is continuous Let $D \subset \mathbb{C}$ be a discrete subset and let $f : D \mapsto \mathbb{C}$ be a function. Show that $f$ is continuous.
What's the best way to do this? I was thinking a proof by contradiction since a direct proof seems a little tricky...
Definitions i am using:
$D \subset \mathbb{C}$ is a discrete subset if $\forall z \in D$ there exists a ball of radius $r>0$ such that $D \cap B_r(z)$ = $\{z\}$.
$f: A \mapsto B$ is continuous if $\forall x,y \in A$ , given $\epsilon > 0$ there exists $\delta>0$ such that
$\mid y-x \mid < \delta$ $\implies$ $\mid f(y) - f(x) \mid < \epsilon$
 A: There are several possible approaches here. In light of the edits to your question, you probably want the third.
Topological: we wish to establish if every preimage in $f$ of an open set in $\mathbb C$ is open. But one definition of "discrete set" is "one whose subspace topology is discrete", where the discrete topology is the one in which every set is open. So clearly every preimage is open.
Sequential: a function is continuous iff it preserves limits of convergent sequences. But a convergent sequence in a discrete set must be eventually constant, and such sequences are always preserved by any function.
Metric: we wish to show that for all $\epsilon > 0$ and $x \in D$ there exists $\delta$ so that if $y \in D$ is close to $x$, i.e. has $|x - y| < \delta$, then $f(y)$ is close to $f(x)$, i.e. $|f(x) - f(y)| < \epsilon$. But $D$ discrete means that for each $x$ there is some $\delta$ so that the only point $y$ within $\delta$ of $x$ is $x$ itself, and then certainly $|f(x) - f(y)|$ is small!
A: Let $E:=f[D]$ be the range of $f$, so that $E\subset\mathbb{C}$.
Under $f$, the preimage of any open set $U\subset E$ is a set in $D$.  That is, $$f^{-1}[U]\subset D.$$  All sets in $D$ are open, thus $f^{-1}[U]$ is open.  Hence $f$ is continuous.
As a reference, this answer shows that all sets in a discrete space are open.
