Continuity of a Function with a Finite Domain I was wondering if a function with a finite domain could be continuous, particularly, if the domain is only a single element. For example, is $f(x)= x$ continuous on $D=\{ x \mid x=1\}$?
Sorry if this is a really basic question. Thank you for your time. 
 A: By definition of continuity at $a$, $\forall \epsilon > 0, \exists \delta > 0$ such that $|x-a| < \delta$ implies $|f(x) - f(a)| < \epsilon$.
For a function of finite domain, enumerate the points and $L$ to be the minimum distance between every pair of points. For any $\epsilon$ take $\delta = \min(\epsilon, L)$ and you find that the definition holds.
So yes, every function with finite domain is continuous, and furthermore continuous everywhere in its domain.
A: In $\mathbb{R}$ with standard topology yes, as already proven by @Camille.
Although your question has real-analysis tag, let me mention that, this does not have to be true in different topologies. In general, function $f$ is continuous if the preimage of any open set is open. Thus, for any set $A$ and topological spaces 
\begin{align}X &= (A, \{\varnothing,A\}) &\text{(trivial topology)} \\
Y &= (A, 2^A) & \text{(discrete topology)}
\end{align}
any non-constant function $f : X \to Y$ is discontinuous.
I hope this helps $\ddot\smile$
