Consider a function $f(x)$ that has no jump, infinite, or removable discontinuities in the middle anywhere — but maybe the domain is limited:
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Would the endpoint $[a,$ be considered continuous?
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What about the endpoint $(a,$?
I ask because I often see "a function is continuous if we can draw it without lifting up the pencil" but I didn't know to what extent this applies to the endpoints and whether or not it matters if the points themselves are defined.
Best Answer
The function is continuous iff it is continuous at each point of the domain, so we need only consider points in the domain.
Hence, if the domain is of the form $(a,...$, the end point $a$ is of no concern since it is not part of the domain. For example, $f(x) = {1 \over x}$ is continuous on $(0,\infty)$. The point $0$ is not an issue since it is not part of the domain.
If the domain has the form $[a,...$ then if $f$ is continuous at $x=a$ then $f$ must have values close to $f(a)$ for $x$ close to $a$ (and in the domain). In particular, $f$ must be defined at $x=a$.
One characterization is for all sequences $x_n \to a$ (with $x_n $ in the domain) we must have $f(x_n) \to f(a)$. This is equivalent to drawing the curve without lifting.
Illustrations:
The function $f(x) = -1 $ for $x \in [-1,0]$ and $f(x) = 1$ for $x \in (1,2]$, with domain $[-1,0] \cup (1,2]$ is continuous everywhere.
The function $f(x) = -1 $ for $x \in [-1,0]$ and $f(x) = 1$ for $x \in (0,2]$, with domain $[-1,2]$ is continuous everywhere except at $x=0$.
The function that is $0$ everywhere except at $x=0$ where $f(0) = 1$ is continuous everywhere except at $x=0$.