Real Function is Continuous at Isolated Point
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Theorem
Let $A \subseteq \R$ be any subset of the real numbers.
Let $f: A \to \R$ be a real function.
Let $x \in A$ be an isolated point of $A$.
Then $f$ is continuous at $x$.
Proof
By definition of isolation point:
- $\exists \delta \in \R_{> 0}: \openint {x - \delta} {x + \delta} \cap A = \set x$
Pick any $\epsilon > 0$.
We have that for any $z \in \openint {x - \delta} {x + \delta} \cap A = \set x$:
- $\size {\map f z - \map f x} = \size {\map f x - \map f x} = 0 < \epsilon$.
This satisfies the condition of continuity at $x$,
and the limit in this case is trivially equal to $\map f x$.
$\blacksquare$