# Proof Validation Function From Integers to Rationals is Continuous

I am teaching myself real analysis, so any help is greatly appreciated.

Let the function be defined as $F : Z \rightarrow Q$ where $Z$ is the set of integers and $Q$ is the set of rational numbers, such that

$F(m) = \frac{1}{m}$

Show that $F$ is continuous except for $m=0$.

Proof By definition of continuous we have that a function defined on a set $D$ is continuous at a point $p \in D$ if given any number $\epsilon > 0$, there is a neighborhood $U$ about $p$ such that $|F(q)-F(p)| < \epsilon$ for every point $q \in U \cap D$. The function $F$ is said to be continuous on $D$ if it is continuous at each point in $D$

$Z$ is disconnected with $Z={\{1}\} \cup \{{2}\} \cup ...$

Which means that any neighbor hood $B(p,r)$ centered at point $p \in Z$ and with radius $r < 1$ will contain only the point $p$ itself. This implies given such a sufficient neighborhood $B$ and $p,q \in B$ we have

$|F(q)-F(p)| = |F(p)-F(p)| = 0 < \epsilon$

This is true for all $\epsilon > 0$, and therefore it must be true that $F$ is continuous.

Q.E.D.

I would like to verify the correctness of this proof. That is because there is another function $F: Z^2 \rightarrow Q$ that I am trying to prove to be uniformly continuous, but I am analyzing my way up to it.

• Your proof is right. More generally, any function from $\Bbb Z$ (or any totally disconnected space) to any codomain is continuous, indeed uniformly continuous, wherever it's defined. Jun 1, 2014 at 22:20
• Thank you very much for your feedback. That is very exciting, and I will try to prove that. It's always beneficial to know what to expect when proving something. Jun 2, 2014 at 1:14
• @GregMartin Would you mind posting your comment as a question, so that it can be removed from the unanswered que? Dec 14, 2015 at 16:42

Your proof is right. More generally, any function from $\Bbb Z$ (or any totally disconnected space) to any codomain is continuous, indeed uniformly continuous, wherever it's defined.