# What kind of definition is this?

Gold Member
Summary:
Hi, PF, I've got a quote, and a philosophical question

Figure shown suggests that a function ##f(x)## can have local extreme values only at ##x## points of three special types:
(i) Critical points of ##f## (points ##x## in ##\mathfrak{D}(f)## where ##f'(x)=0##)
(ii) Singular points of ##f## (points ##x## in ##\mathfrak{D}(f)## where ##f'(x)## is not defined)
(iii) Endpoints of the domain of ##f## (points in ##\mathfrak{D}(f)## that do not belong to any open interval contained in ##\mathfrak{D}(f)##)

And now my "brainstorm": (iii) is not defined as "reductio ad absurdum"; not "principle of exclusion".
My question: what kind of argument makes use of? I'm just overthinking? Why does talk about closed intervals in such a "implicit" (suggested, but not communicated directly) way?
PS: I'm not sure of the LaTeX used, and insecure about "Preview" tool, so I will click "Post thread". Please check. Thanks!

Office_Shredder
Staff Emeritus
Gold Member
It talks about the endpoints in a weird way because the domain of f can be really weird. For example the cantor set contains no interior, so any function defined on it is continuous.

Edit: actually I'm not confident in this claim. But you can imagine some bizarre domains and functions where the endpoints don't just look like the endpoints of closed intervals.

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mcastillo356
anuttarasammyak
Gold Member
It seems just saying there is no point beyond end point in words of mathematics.

mcastillo356
Mark44
Mentor
And now my "brainstorm": (iii) is not defined as "reductio ad absurdum"; not "principle of exclusion".
I'm not sure that it's useful to categorize these parts of the definition.
In the figure you attached, points a and b are, respectively, a local maximum point and a local minimum point. At neither one is the derivative of the function equal to zero, and at neither is the derivative undefined (although you will need to use one-sided limits for each one).
Clearly each of these points has a y-value greater/smaller than all other points in the immediate vicinity, which makes them local extreme points.
My question: what kind of argument makes use of? I'm just overthinking?
This.

mcastillo356 and sysprog
Gold Member
Hi, PF

Office_Shredder, very inquisitive your message, but far from my background.
anuttarasammyak, a concise comment.
I'm not sure that it's useful to categorize these parts of the definition.
In the figure you attached, points a and b are, respectively, a local maximum point and a local minimum point. At neither one is the derivative of the function equal to zero, and at neither is the derivative undefined (although you will need to use one-sided limits for each one).
Clearly each of these points has a y-value greater/smaller than all other points in the immediate vicinity, which makes them local extreme points.
Still stucked, I've read it, but... Could explain further?
Questions:
Why ##a## and ##b## are, respectively, a local maximum point and a local minimum point?. It's clear what you say later, but I don't relate it with the sentence I want to understand.

jbriggs444
Homework Helper
Why ##a## and ##b## are, respectively, a local maximum point and a local minimum point?. It's clear what you say later, but I don't relate it with the sentence I want to understand.
##a## is a local maximum for the function because it fits the definition of a local maximum.

1. It is within the domain of the function.
2. There is a neighborhood around a within which all points that fall within the domain of the function have function values less than or equal to that of a.

Similarly, ##b## is a local minimum because it fits the definition of a local minimum.

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mcastillo356 and sysprog
anuttarasammyak
Gold Member
If you are wondering why they use open interval not closed interval, I see current standard mathematics prefer open interval as primary idea and introduce closed interval as its complementary.

pbuk and mcastillo356
Gold Member
jbriggs444, anuttarasammyak, PF, what is my question? That's the question. Someone said. I've been thinking... (don't scare): Why does my textbook, "Calculus", by Robert A. Adams, defines endpoints in terms of open intervals?
My opinion:
(i) endpoints appear, closed intervals appear
(ii) when I look for the meaning of a word, I will never find the word, they use another words
Does it make sense?

Mark44
Mentor
Why a and b are, respectively, a local maximum point and a local minimum point?
To add to what @jbriggs444 already said, look at the graph you attached. Are there any points near a that have larger y values? Are there any points near b that have smaller y values?
Here "near" should be interpreted to mean between a and x1, and between x6 and b, respectively.
My opinion:
(i) endpoints appear, closed intervals appear
With an open interval, every point is an interior point; i.e., every point has neighbors that also belong to the open interval on either side of that point. With a closed interval, the endpoints have neighboring points that are in the interval only on one side.
(ii) when I look for the meaning of a word, I will never find the word, they use another words
That's how definitions work -- most often the meaning of a word will be described in a string of words rather than by a single word. If you're looking for a one-word meaning, that would be a synonym, another word that means the same thing.

anuttarasammyak and mcastillo356
Gold Member
If I talk sincerely... I'll be direct: I'm going to publish the same topic at a Spanish math forum. Neigbourhood... that's topology... I'll be back (bad joke)
See you later, greetings

mathwonk
Homework Helper
2020 Award
remark: the cantor set is infinite, bounded and closed. thus it contains some of its own accumulation points, at which the value of a continuous function is determined by its values on the other points.

mcastillo356
Gold Member
Hi, mathwonk, PF
I've already published at Spanish math forum "Rincón Matemático" (Mathematical Corner, translated). My aim is to work it out, put things in common, solve my quests...
We are in touch

pbuk
Gold Member
If you are wondering why they use open interval not closed interval, I see current standard mathematics prefer open interval as primary idea and introduce closed interval as its complementary.
(Edited) No, the complement of an open interval is not necessarily even an interval.

Also consider the intervals ## [0, 1) ## which is neither closed nor open and ##(- \infty, + \infty)## which is both closed and open.

The definitions are clear (from MathWorld).
• An open interval is an interval that does not include its end points.
• A closed interval is an interval that includes all of its limit points.

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mcastillo356
anuttarasammyak
Gold Member
They define "open" first and then define "closed" making use of "open".

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pbuk
Gold Member
By definition, a subset
of a topological space
is called closed if its complement
is an open subset of
;
Yes but that is not the same as saying the 'idea' of an open interval is complementary to the 'idea' of a closed interval, and I have edited my post to provide examples which explain why.

mcastillo356
Gold Member
Well, let's read the definition of Local Extreme Values, previous to the types of them: now, the question again
Summary:: Pseudophilosophal question about a definition

Hi, PF, I've got a quote, and a philosophical question

Figure shown suggests that a function ##f(x)## can have local extreme values only at ##x## points of three special types:
(i) Critical points of ##f## (points ##x## in ##\mathfrak{D}(f)## where ##f'(x)=0##)
(ii) Singular points of ##f## (points ##x## in ##\mathfrak{D}(f)## where ##f'(x)## is not defined)
(iii) Endpoints of the domain of ##f## (points in ##\mathfrak{D}(f)## that do not belong to any open interval contained in ##\mathfrak{D}(f)##)
View attachment 291084
(iii) Endpoints of the domain of ##f## (points in ##\mathfrak{D}(f)## that do not belong to any open interval contained in ##\mathfrak{D}(f)##): What does it mean?
My opinion:
1- We are talking about points at the end of something: an interval
2- They do not belong to any open interval contained in ##\mathfrak{D}(f)##
3- Therefore, they are the extremes of a closed interval.

PS: I will probably edit, I have great doubts about this post. Just want to make my contribution, but still indecisive

Gold Member
Example

 ##f:(-2,-1]\cup \{0\}\cup [1,2)\to \mathbb{R}\qquad f(x)=x.## - 0 is an endpoint, and a local maximum and a local minimum

pbuk
Gold Member
(Edited) No, the complement of an open interval is not necessarily even an interval.

Also consider the intervals ## [0, 1) ## which is neither closed nor open and ##(- \infty, + \infty)## which is both closed and open.

The definitions are clear (from MathWorld).
• An open interval is an interval that does not include its end points.
• A closed interval is an interval that includes all of its limit points.

1- We are talking about points at the end of something: an interval
We are talking about points that exist at the end of a bounded interval, ## \pm \infty ## would be troublesome here.
2- They do not belong to any open interval contained in ##\mathfrak{D}(f)##
Yes
3- Therefore, they are the extremes of a closed interval.
Yes
 ##f:(-2,-1]\cup \{0\}\cup [1,2)\to \mathbb{R}\qquad f(x)=x.## - 0 is an endpoint, and a local maximum and a local minimum
Yes. ## 0 ## is also a singular point.

mcastillo356
Gold Member
##a## is a local maximum for the function because it fits the definition of a local maximum.

1. It is within the domain of the function.
2. There is a neighborhood around a within which all points that fall within the domain of the function have function values less than or equal to that of a.

Similarly, ##b## is a local minimum because it fits the definition of a local minimum.
Might the following capture the concept of neighborhood for the case of ##a##, always in a naive way?
$$\exists{\;h>0}\quad\mbox{such that if}\;x\in{(a-h,a+h)}\cap{\mbox{dom}(f)}\quad\mbox{then}\;f(x)\leq{f(a)}$$

anuttarasammyak
Gold Member
Summary:: Pseudophilosophal question about a definition

(iii) Endpoints of the domain of f (points in D(f) that do not belong to any open interval contained in D(f))
We may say it in the way that "The points that have no (open) Neighborhood.
https://en.wikipedia.org/wiki/Neighbourhood_(mathematics)

mcastillo356
Gold Member
$$\exists{\;h>0}\quad\mbox{such that if}\;x\in{(a-h,a+h)}\cap{\mbox{dom}(f)}\quad\mbox{then}\;f(x)\leq{f(a)}\Leftrightarrow$$
$$\exists{\;h>0}\quad\mbox{such that if}\;x\in{[a,a+h)}\cap{\mbox{dom}(f)}\quad\mbox{then}\;f(x)\leq{f(a)}$$
Sorry, it was incomplete. On the path of reaching some kind of approval from you. Not now; not have a training in Topology, just searching for a fix.
Love

Mark44
Mentor
$$\exists{\;h>0}\quad\mbox{such that if}\;x\in{(a-h,a+h)}\cap{\mbox{dom}(f)}\quad\mbox{then}\;f(x)\leq{f(a)}\Leftrightarrow$$
$$\exists{\;h>0}\quad\mbox{such that if}\;x\in{[a,a+h)}\cap{\mbox{dom}(f)}\quad\mbox{then}\;f(x)\leq{f(a)}$$
Sorry, it was incomplete. On the path of reaching some kind of approval from you. Not now; not have a training in Topology, just searching for a fix.
Love
Point a is apparently the left endpoint of the domain, so my preference would be the second statement. The first one isn't wrong, but needlessly mentions points in the interval (a - h, a).

mcastillo356
Gold Member
Point a is apparently the left endpoint of the domain, so my preference would be the second statement. The first one isn't wrong, but needlessly mentions points in the interval (a - h, a).
Love, Physics Forums, quests solved.