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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

$\mid y-x \mid < \delta$ $\implies$ $\mid f(y) - f(x) \mid < \epsilon$

terrible at math
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2 Answers2

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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!

Ben Millwood
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    I am working with the metric definition so thank you! That does make sense and I can easily choose such a ball of radius $\delta$.

    Then I have that $\mid f(x) - f(y) \mid$ is just zero, correct?

    That's an awesome proof, thanks :)

     – terrible at math Mar 08 '14 at 20:20
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    Correct!${}{}{}{}{}{}$ – Ben Millwood Mar 08 '14 at 20:24
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    interesting question -- does this hold for ANY discrete set or are there exceptions?

    Consider this one: $D := { \frac{1}{n} \mid n \in \mathbb{N}} \cup {0} \subset \mathbb{C}$

    This is a discrete set but it seems like there may be a function that is not continuous on it due to the $0$ ?

     – terrible at math Mar 08 '14 at 20:28
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    That's not a discrete set. There's no ball around $0$ that doesn't capture at least one other point. – Ben Millwood Mar 08 '14 at 20:34
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    actually yeah, you're right. Then I should ask, if a set is not discrete does there exist at least one discontinuous function from that set? – terrible at math Mar 08 '14 at 20:35
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    Yep. If a set is not discrete you can find a sequence in it which converges to some point not in the sequence – e.g. $0$ in your example has a sequence of nonzero points converging to it. Then just make a function that takes one value on the sequence, and a different value on the limit. – Ben Millwood Mar 09 '14 at 14:41
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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.

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Jeff Snider
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