Continuity of a function at an isolated point

Suppose $c$ is an isolated point in the domain $D$ of a function $f$. In the delta neighbourhood of $c$, does the function $f$ have the value $f(c)$?

Solution 1:

You can also see this is true using the topological definition of continuity at a point: a function is continuous at a point $f(x)$ if for any neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U)$ is contained in $V$. For an isolated point, you can take the neighborhood consisting of just the point $c$, so its image $f(c)$ will obviously be contained in $V$, as $V$ is a neighborhood of $f(c)$.

Solution 2:

I see now that the comments above provide essentially the answer, with whatever definition of continuity you have. The following ties everything together.

Let's use as the definition of continuity $\lim_{x \rightarrow c}\,f(x) = f(c)$. Expand: For all $\varepsilon > 0$ there exists $\delta > 0$ such that whenever $0 < |x-c| < \delta$ it is true that $|f(x)-f(c)| < \varepsilon$. When $\delta$ is small enough, there are no points $x$ which work, so the part after the such that is vacuously true.