It seems like all the proofs I've seen of the theorem that monotonic functions can only have jump discontinuity at $x=a$ use $f(a)$. But how can we use this in the case that f is not defined in this point? For example the function $f(x)= \frac{x²-1}{x-1}$ if I don't decide on a value for $f(1)$ and keep it undefined, is that not removable discontinuity?
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2$\begingroup$ A function $f: E\to \mathbb R$ with $E\subset \mathbb R$ is said to be continuous at a point $x\in E$ if for any sequence of points $x_n\in E$ with $x_n\to x$, we have $f(x_n) \to f(x)$. A continuous function means $f$ is continuous at all points $x\in E$. This is the convention used when people talk of continuous functions. Thus $f(x) = (x^2-1)/(x-1)$ is continuous everywhere. The fact that $f$ is not defined at $x=1$ is irrelevant. In other words, if you want to claim $f$ is discontinuous at $x$, you better actually have $x\in \mathrm{dom}\,f$ $\endgroup$– AndrewJun 6 at 23:41
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$\begingroup$ From Wikipedia on the classification of discontinuities, it seems that some don't consider your $f$ (with $f(1)$ left undefined) to even have removable discontinuity. $\endgroup$– peterwhyJun 6 at 23:45
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$\begingroup$ Thank you so much!! $\endgroup$– user1188938Jun 7 at 0:05
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3 Answers
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If you don't choose a value for $f(1)$ for the function $f(x) = \frac{x^2-1}{x-1}$, then there is no real issue: we can only discuss continuity of a function within its domain.
If $f(1)$ is not defined, then $1$ is not in the domain of $f$, so there is no real "point" to discussing whether $f$ is continuous at $1$, because it is not defined there to begin with.
More broadly, if $f : D \to \mathbb{R}$ is a function of domain $D \subseteq \mathbb{R}$ then we only can discuss continuity at those $x$ which live in $D$, i.e. those $x$ for which $f(x)$ has a defined value.
The statement "monotone functions cannot have removable discontinuities" is concerned with discontinuities within their domain. That is, it would try to instead deal with a function looking like $$ f(x) = \begin{cases} \frac{x^2-1}{x-1} , & x \ne 1 \\ 4, & x = 1 \end{cases} $$ where $1$ is actually given a defined value.
The function $f$ you gave at the outset, just defined by the rule $f(x) = \frac{x^2-1}{x-1}$, is not defined at $1$, and so $1$ is not in its domain and it is therefore pointless to discuss continuity there. You might as well discuss the continuity of $\sqrt x$ at $-1$ (when interpreted as a function $\sqrt{\cdot} : [0,\infty) \to [0, \infty)$), it would be an equally pointless conversation.
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We can only reasonably talk about properties of a function at a point if said point is in the domain of the function. When you define $f(x) = \frac{x^2-1}{x-1}$, we must also specify the domain on which it is defined. If you say that $f \colon \mathbb R \setminus \left\{1\right\} \to \mathbb R$ (that is, $f$ is defined on all the reals except at $1$), then it doesn't make sense to say that $f$ has a removable discontinuity at $1$, since $f$ isn't defined there.
(This does seem to be a possible deviation from convention here however; see https://mathworld.wolfram.com/RemovableDiscontinuity.html).
A function must specify a value at all points in its domain. That's the definition of a function. So if you say we "leave $f(1)$ undefined", that either means excluding $1$ from the domain, in which case speaking of continuity at that point doesn't make sense; or we must include $1$ and thus specify a (finite) value for it.
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Emphasizing that the function $\frac{x^2-1}{x-1}$ is not defined in $1$ is formally true, but glosses over the fact that this is not the point. Even if we considered a function like brought below $$f(x) = \begin{cases} \frac{x^2-1}{x-1} , & x \ne 1 \\ a, & x = 1 \end{cases}$$
for some $a$, we will find, that monotone contradicts removable discontinuity: if $a=2$, then we lost discontinuity and function simply become $x+1$, and if $a\ne 2$ then we lost monotone property.
Monotonic function, defined on real interval, have very interesting property: it has limits from the right and from the left at every point of its domain.
But removable discontinuity we call case when function has limit in given point, but it is not equal to function value, i.e. it's strictly less or strictly more.
Joining these two facts gives, that function lost monotone property, if it have removable discontinuity.
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$\begingroup$ Constructive critique is welcome, especially as answer is correct. $\endgroup$– zkutchJun 7 at 0:10