butterfly/kb
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity
a246fdfd34
kb/data/en.wikipedia.org/wiki/Classification_of_discontinuities-2.md

1257 lines
15 KiB
Markdown
Raw Blame History

This file contains ambiguous Unicode characters

This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

---
title: "Classification of discontinuities"
chunk: 3/4
source: "https://en.wikipedia.org/wiki/Classification_of_discontinuities"
category: "reference"
tags: "science, encyclopedia"
date_saved: "2026-05-05T09:08:17.602531+00:00"
instance: "kb-cron"
---
1
C
(
x
)
=
{
1
x
∈
C
0
x
∈
[
0
,
1
]
∖
C
.
{\displaystyle \mathbf {1} _{\mathcal {C}}(x)={\begin{cases}1&x\in {\mathcal {C}}\\0&x\in [0,1]\setminus {\mathcal {C}}.\end{cases}}}
One way to construct the Cantor set
C
{\displaystyle {\mathcal {C}}}
is given by
C
:=
⋂
n
=
0
∞
C
n
{\textstyle {\mathcal {C}}:=\bigcap _{n=0}^{\infty }C_{n}}
where the sets
C
n
{\displaystyle C_{n}}
are obtained by recurrence according to
C
n
=
C
n
−
1
3
∪
(
2
3
+
C
n
−
1
3
)
for
n
≥
1
,
and
C
0
=
[
0
,
1
]
.
{\displaystyle C_{n}={\frac {C_{n-1}}{3}}\cup \left({\frac {2}{3}}+{\frac {C_{n-1}}{3}}\right){\text{ for }}n\geq 1,{\text{ and }}C_{0}=[0,1].}
In view of the discontinuities of the function
1
C
(
x
)
,
{\displaystyle \mathbf {1} _{\mathcal {C}}(x),}
let's assume a point
x
0
∉
C
.
{\displaystyle x_{0}\not \in {\mathcal {C}}.}
Therefore there exists a set
C
n
,
{\displaystyle C_{n},}
used in the formulation of
C
{\displaystyle {\mathcal {C}}}
, which does not contain
x
0
.
{\displaystyle x_{0}.}
That is,
x
0
{\displaystyle x_{0}}
belongs to one of the open intervals which were removed in the construction of
C
n
.
{\displaystyle C_{n}.}
This way,
x
0
{\displaystyle x_{0}}
has a neighbourhood with no points of
C
.
{\displaystyle {\mathcal {C}}.}
(In another way, the same conclusion follows taking into account that
C
{\displaystyle {\mathcal {C}}}
is a closed set and so its complementary with respect to
[
0
,
1
]
{\displaystyle [0,1]}
is open). Therefore
1
C
{\displaystyle \mathbf {1} _{\mathcal {C}}}
only assumes the value zero in some neighbourhood of
x
0
.
{\displaystyle x_{0}.}
Hence
1
C
{\displaystyle \mathbf {1} _{\mathcal {C}}}
is continuous at
x
0
.
{\displaystyle x_{0}.}
This means that the set
D
{\displaystyle D}
of all discontinuities of
1
C
{\displaystyle \mathbf {1} _{\mathcal {C}}}
on the interval
[
0
,
1
]
{\displaystyle [0,1]}
is a subset of
C
.
{\displaystyle {\mathcal {C}}.}
Since
C
{\displaystyle {\mathcal {C}}}
is an uncountable set with null Lebesgue measure, also
D
{\displaystyle D}
is a null Lebesgue measure set and so in the regard of Lebesgue-Vitali theorem
1
C
{\displaystyle \mathbf {1} _{\mathcal {C}}}
is a Riemann integrable function.
More precisely one has
D
=
C
.
{\displaystyle D={\mathcal {C}}.}
In fact, since
C
{\displaystyle {\mathcal {C}}}
is a nonwhere dense set, if
x
0
∈
C
{\displaystyle x_{0}\in {\mathcal {C}}}
then no neighbourhood
(
x
0
−
ε
,
x
0
+
ε
)
{\displaystyle \left(x_{0}-\varepsilon ,x_{0}+\varepsilon \right)}
of
x
0
,
{\displaystyle x_{0},}
can be contained in
C
.
{\displaystyle {\mathcal {C}}.}
This way, any neighbourhood of
x
0
∈
C
{\displaystyle x_{0}\in {\mathcal {C}}}
contains points of
C
{\displaystyle {\mathcal {C}}}
and points which are not of
C
.
{\displaystyle {\mathcal {C}}.}
In terms of the function
1
C
{\displaystyle \mathbf {1} _{\mathcal {C}}}
this means that both
lim
x
→
x
0
−
1
C
(
x
)
{\textstyle \lim _{x\to x_{0}^{-}}\mathbf {1} _{\mathcal {C}}(x)}
and
lim
x
→
x
0
+
1
C
(
x
)
{\textstyle \lim _{x\to x_{0}^{+}}1_{\mathcal {C}}(x)}
do not exist. That is,
D
=
E
1
,
{\displaystyle D=E_{1},}
where by
E
1
,
{\displaystyle E_{1},}
as before, we denote the set of all essential discontinuities of first kind of the function
1
C
.
{\displaystyle \mathbf {1} _{\mathcal {C}}.}
Clearly
∫
0
1
1
C
(
x
)
d
x
=
0.
{\textstyle \int _{0}^{1}\mathbf {1} _{\mathcal {C}}(x)dx=0.}
== Discontinuities of derivatives ==
Let
I
⊆
R
{\displaystyle I\subseteq \mathbb {R} }
an open interval, let
F
:
I
→
R
{\displaystyle F:I\to \mathbb {R} }
be differentiable on
I
,
{\displaystyle I,}
and let
f
:
I
→
R
{\displaystyle f:I\to \mathbb {R} }
be the derivative of
F
.
{\displaystyle F.}
That is,
F
′
(
x
)
=
f
(
x
)
{\displaystyle F'(x)=f(x)}
for every
x
∈
I
{\displaystyle x\in I}
.
According to Darboux's theorem, the derivative function
f
:
I
→
R
{\displaystyle f:I\to \mathbb {R} }
satisfies the intermediate value property.
The function
f
{\displaystyle f}
can, of course, be continuous on the interval
I
,
{\displaystyle I,}
in which case Bolzano's theorem also applies. Recall that Bolzano's theorem asserts that every continuous function satisfies the intermediate value property.
On the other hand, the converse is false: Darboux's theorem does not assume
f
{\displaystyle f}
to be continuous and the intermediate value property does not imply
f
{\displaystyle f}
is continuous on
I
.
{\displaystyle I.}
Darboux's theorem does, however, have an immediate consequence on the type of discontinuities that
f
{\displaystyle f}
can have. In fact, if
x
0
∈
I
{\displaystyle x_{0}\in I}
is a point of discontinuity of
f
{\displaystyle f}
, then necessarily
x
0
{\displaystyle x_{0}}
is an essential discontinuity of
f
{\displaystyle f}
.
This means in particular that the following two situations cannot occur:
Furthermore, two other situations have to be excluded (see John Klippert):
Observe that whenever one of the conditions (i), (ii), (iii), or (iv) is fulfilled for some
x
0
∈
I
{\displaystyle x_{0}\in I}
one can conclude that
f
{\displaystyle f}
fails to possess an antiderivative,
F
{\displaystyle F}
, on the interval
I
{\displaystyle I}
.
On the other hand, a new type of discontinuity with respect to any function
f
:
I
→
R
{\displaystyle f:I\to \mathbb {R} }
can be introduced: an essential discontinuity,
x
0
∈
I
{\displaystyle x_{0}\in I}
, of the function
f
{\displaystyle f}
, is said to be a fundamental essential discontinuity of
f
{\displaystyle f}
if
lim
x
→
x
0
−
f
(
x
)
≠
±
∞
{\displaystyle \lim _{x\to x_{0}^{-}}f(x)\neq \pm \infty }
and
lim
x
→
x
0
+
f
(
x
)
≠
±
∞
.
{\displaystyle \lim _{x\to x_{0}^{+}}f(x)\neq \pm \infty .}
Reference in New Issue View Git Blame Copy Permalink