High School Continuity of ln(x) function

[eneacasucci]

it is correct to say that if we consider the whole of R as the domain, the function ln(x)is not continuous, whereas if we consider the domain of the function as the domain, then it is continuous?

 

[PeroK]

it is correct to say that if we consider the whole of R as the domain, the function ln(x)is not continuous, whereas if we consider the domain of the function as the domain, then it is continuous?

Not quite. Continuity can only be defined where the function is defined. I.e. on its domain. Outside of its domain, where the function is not defined, it has no properties.

 

[fresh_42]

it is correct to say that if we consider the whole of R as the domain, the function ln(x)is not continuous, whereas if we consider the domain of the function as the domain, then it is continuous?

Yes. However, it has a flaw. Continuity is a local property, something that happens at a certain location. We only say that a function is continuous if it is everywhere continuous for reasons of simplicity and laziness to write everywhere every time. So continuity at a point $x<0$ makes little sense. However, continuity, i.e. discontinuity in this case at $x=0$ is an issue. We could e.g. say that it is right-continuous.

Not quite. Continuity can only be defined where the function is defined. I.e. on its domain. Outside of its domain, where the function is not defined, it has no properties.

Not quite. We say indeed that $x \longmapsto 1/x$ is discontinuous at $x=0$ although it is not defined. You would abolish any singularities by that definition!

 

[eneacasucci]

Not quite. Continuity can only be defined where the function is defined. I.e. on its domain. Outside of its domain, where the function is not defined, it has no properties.

So to make it discontinuous we should explicitly define a piecewise function like: f(x)=ln(x) for x>0 and f(x)=0 for x≤0? It is not enough to say "consider ln(x) over whole R" then, right?

 

[fresh_42]

So to make it discontinuous we should explicitly define a piecewise function like: f(x)=ln(x) for x>0 and f(x)=0 for x≤0? It is not enough to say "consider ln(x) over whole R" then, right?

@PeroK's answer was a bit sloppy, I think. It is simply arduous to talk about points that are undefined and not bordering the domain. That rules out the negative numbers in that case, but not necessarily $0.$ We need (open) neighborhoods of function values so that the term continuity makes sense. No function values, no neighborhoods. We don't have function values at $\log(-|x|)$ but we have function values at $\log n^{-1}$ so that it's legitimate to ask what happens at the limit.

 

[PeroK]

@PeroK's answer was a bit sloppy, I think. It is simply arduous to talk about points that are undefined and not bordering the domain. That rules out the negative numbers in that case, but not necessarily $0.$ We need (open) neighborhoods of function values so that the term continuity makes sense. No function values, no neighborhoods. We don't have function values at $\log(-|x|)$ but we have function values at $\log n^{-1}$ so that it's legitimate to ask what happens at the limit.

You can ask what happens at the limit. In this case, the function tends to $-\infty$.

The definition of continuity at a point is clear. The point must be in the domain to be considered. You can either treat points outside the domain as undefined. Or, treat them all as discontinuities. There's nothing in the definition of continuity that allows you to choose specific points outside the domain over others.

 

[Infrared]

The definition of continuity at a point is clear. The point must be in the domain to be considered. You can either treat points outside the domain as undefined. Or, treat them all as discontinuities. There's nothing in the definition of continuity that allows you to choose specific points outside the domain over others.

While this is the definition I was taught and think makes the most sense, the other perspective is also out there. From https://en.wikipedia.org/wiki/Continuous_function#Real_functions

"A partial function is discontinuous at a point if the point belongs to the topological closure of its domain, and either the point does not belong to the domain of the function or the function is not continuous at the point."

 

[WWGD]

If you want to make it continuous on the left you should define it as $0$ for $x\leq 1$, or as $ln(\epsilon)$ for some $\epsilon >0$. Basically, it cannot be redefined at $0$ to be made continuous.

 

[PeroK]

While this is the definition I was taught and think makes the most sense, the other perspective is also out there. From https://en.wikipedia.org/wiki/Continuous_function#Real_functions

"A partial function is discontinuous at a point if the point belongs to the topological closure of its domain, and either the point does not belong to the domain of the function or the function is not continuous at the point."

One of the problems with that approach comes when working with disconnected sets. An alternative condition for a set $X$ to be disconnected is that there exists a continuous function from $X$ onto the two-point set $\{0, 1\}$. If we take $X = (-\infty, 0) \cup (0, \infty)$, then we can show it is disconnected by considering the function $f(x) = 0 \ (x < 0)$ and $f(x) = 1 \ (x > 0)$.

With the alternative definition of continuity, this function $f$ is not continuous and we lose an important characterisation of disconnectedness - or, at least, muddy the waters. In fact, the alternative definition of continuity tends to confuse the concept of a continuous function with that of a function defined on a disconnected domain. Or, even, a function simply restricted to an open subset.

In fact, that alternative definition appears to make every function defined on an open subset of $\mathbb R$ not continuous. So, the set of continuous functions defined on $(0, 1)$ becomes the empty set!

 

[PeroK]

Not quite. We say indeed that $x \longmapsto 1/x$ is discontinuous at $x=0$ although it is not defined.

Not by the standard epsilon-delta definition (*). Technically, $1/x$ is a continuous function on a disconnected domain.

You would abolish any singularities by that definition!

A singularity is not the same as a discontinuity.

 

[WWGD]

$1/x$ cannot be redefined at $0$ to be made continuous there. Just what is a singularity, if not a point where a function is discontinuous?

 

[PeroK]

$1/x$ cannot be redefined at $0$ to be made continuous there. Just what is a singularity, if not a point where a function is discontinuous?

A singularity is a point where the function cannot be defined. The gamma factor is singular at $v = c$:
$$\gamma(v) = \frac{1}{\sqrt{1- \frac {v^2}{c^2}}}$$That's a continuous function of $v$ on the domain $(-c, c)$. The definition of continuity does not demand that the function be bounded. A continuous function can tend to $\pm \infty$.