TITLE: Does a function have to be "continuous" at a point to be "defined" at the point?
QUESTION [9 upvotes]: I did search for whether this question was already answered but couldn't find any.
Does a function have to be "continuous" at a point to be "defined" at the point?
For example take the simple function $f(x) = {1 \over x}$; obviously it is not continuous at $x = 0$. However it does have the $-$ and $+$ limits because $\lim_{x \to 0-} f(x) = -\infty $ and $\lim_{x \to 0+} f(x) = +\infty$.
Would you then say that $f(x) = {1 \over x}$ is "defined" at $x = 0$ or not? Please justify your answer.
I ask because at http://en.wikipedia.org/wiki/Continuous_function#Examples there is the text: 

the function $f(x) = \frac {2x-1} {x+2}$ is defined for all real numbers $x \neq -2$ and is continuous at every such point. The question of continuity at $x = -2$ does not arise, since $x = -2$ is not in the domain of $f$. 

and the caption of the associated graph reads:

The function is not defined for $x = -2$.

Do I interpret this to mean that a function can not be defined at a point of discontinuity or merely that the function is intentionally not defined at the point of discontinuity only to achieve the status of being continuous (for whatever purpose) over the entire domain it is defined on?

REPLY [2 votes]: Surely, this is not a precise mathematical question - or, at least, as a precise mathematical question it is rather trivial: there are lots of discontinuous functions that are perfectly well-defined everywhere, if you want them to.
However, there is a deeper perspective coming from algebraic geometry and functional analysis, in which one works not with a single function, but rather with a ring $A$ and thinks of it as a ring of functions. Then one just defines the "points where these functions are defined" to correspond to maximal (or prime, if you really want them) ideals of that ring. This is what people call spectrum of a ring and Gelfand spectrum of a Banach algebra (let's stick to Banach algebras for the sake of simplicity, instead of polynormed algebras or even worse). The Gelfand spectrum of a Banach algebra carries a natural compact Hausdorff topology, generated by the functions in the algebra, and this topology is just designed to make these functions continuous. However, this construction depends only on $A$ being a Banach algebra, so if originally it was also an algebra of functions defined on some space, you may find that the spectrum is much richer than that space. In this case the spectrum provides the natural compactification, and in some cases it may be so large and complicated that it actually looks a little frightening at first sight (in particular, the spectra of $\ell^\infty$ or $L^\infty$ certainly do), but it is a really, really useful thing.
An alternative, but closely related, point of view is to fix a ring $A$ and also a field (or ring) $\mathbb{K}$, for your purpose usually $\mathbb{R}$ or $\mathbb{C}$, and define a $\mathbb{K}$-point of $A$ to be a ring homomorphism $A \to \mathbb{K}$ (see functor of points). If $\mathbb{K}$ carries a topology, then so does this space of $\mathbb{K}$-points, and again, this topology is designed to make the functions continuous. In ths case of complex Banach algebras the space of $\mathbb{C}$-points is the same as the Gelfand spectrum.