B Topological neigborhoods that are not intervals in the real line?

[mcastillo356]

TL;DR Summary

I have some clues, but not a definitive answer to the title question

Hi, PF

$U=]2−\sqrt{2},2+\sqrt{2}[\cup{7}$ is a neighborhood of $2$ because there exists an open set $G$ such that $2\in{G}\subseteq{U}$; in this case, it might be $G=E(2,\sqrt{2})=]2−\sqrt{2},2+\sqrt{2}[$, which is an open set, for it is an open interval, and provided it satisfies

$$2\in G\subseteq {U}\Longleftrightarrow {2\in ]2-\sqrt{2},2+\sqrt{2}[}\,\subseteq {]2-\sqrt{2},2+\sqrt{2}[}\cup \{7\}$$

$U$ is not an interval since there are numbers in $U$ with real numbers between them that do not belong to $U$, ie, $2,7\in U$ but $\forall\,x\in [2+\sqrt{2},7[$ we see this fact: $x\notin{U}$

It is some neigborhood because if $(X,T)$ is a topological space...

Formally we say topological space to the ordered pair $(X,T)$ formed by a set $X$ and a topology $T$ over $X$, ie, a collection of subsets of $X$ that meet all three of the following properties:

1. The empty set and $X$ belong to $T$.

$$\emptyset\in T,\,X\in T$$

2. The intersection of any finite subcollection of sets of $T$ is in $T$.

$$(O_1\in{T},O_2\in{T})\Rightarrow{(O_1\cap O_2\in{T})}$$

3. The union of any subcollection of sets of $T$ is in $T$.

$$\forall{S}\subset{T},\,\bigcup_{O\in{S}}{\,O\in{T}}$$

Sets belonging to the topology $T$ are called open sets, or simply opens of $(X,T)$.

If $(X,T)$ is a topological space and $p$ is a point belonging to $X$, a neighborhood of $p$ is a set $V$ in which is contained an open set $U$ that has as its element the point $p$.

$$p\in U\subseteq{V}$$

Well, we are in the real line. Why it is not an interval?

An open interval, like $]2−\sqrt{2},2+\sqrt{2}[$, for example, is an open; therefore is a neighborhood of each of its points, not only of $2$. This is used to say that $]a,b[$ is an interval but I don't think it works to justify it is a neighborhood. For example, let's say

$$[a,b]=\{x\in{\mathbb R}|a\leq x \leq b\}$$

then between the two extremes there is no missing number, that's why it's an interval: however, it is not a neghborhood of $a$ neither $b$, since there is no $G$ that meets $a\in {G}\subset{[a,b]}$ nor is there an open $B$ that satisfies $b\in B\subset{[a,b]}$.

$U=]2-\sqrt{2},2+\sqrt{2}[\cup\{7\}$ is a neigborhood of $2$ since there is an open $G$ such that $2\in {G}\subseteq {U}$; in this case, it might be $G=E(2,\sqrt{2})=]2-\sqrt{2},2+\sqrt{2}[$, which is an open, since it is an open interval, and it also meets

$$2\in G\subseteq{U}\Longleftrightarrow {2\in ]2-\sqrt{2},2+\sqrt{2}[}\,\subseteq\,]2-\sqrt{2},2+\sqrt{2}[\cup \{7\}$$

Doubt:

$U$ is not an interval because there are reals in $U$ with numbers between them that do not belong to $U$, for example, $2,7\in U$ but $\forall\,x\in [2+\sqrt{2},7[$ we have $x\notin U$?

Greetings

Marcos

 

[martinbn]

...
Post without preview, love and peace

We can tell. About the rest of the post the answer is yes.

 

[fresh_42]

Neighborhoods around a point $x$ are usually considered to be open. So we need an open set $U$ that contains $x$. An open set of the real line with the standard Euclidean topology is a union of open sets $U_\iota.$ We may consider only those sets $U_\iota$ which contain $x$ since the others are not what we call a neighborhood. The union of open sets that all contain a common point $x$ is an open interval.

Open neighborhoods are also commonly regarded as being small to make sense of the word local. E.g. a linear approximation by a derivative is only an approximation around $x$ and not far away. $\mathbb{R}$ is open and contains $x$ but we wouldn't call it a neighborhood, only an open set. Continuity, differentiability, and Taylor series are all local phenomena.

 

[martinbn]

Neighborhoods around a point $x$ are usually considered to be open.

But he said what definition he uses, and it doesn't assume that a neighborhood is open.

 

[fresh_42]

But he said what definition he uses, and it doesn't assume that a neighborhood is open.

Yes, but this is not how it is used. It would make $\{x\}$ a neighborhood of $x$ which doesn't make much sense.

I primarily wrote my answer to avoid confusion among all readers that are neither you and me, nor the OP.

 

[martinbn]

Yes, but this is not how it is used. It would make $\{x\}$ a neighborhood of $x$ which doesn't make much sense.

Only if it is open. His preferred definitions says that a neighborhood is a set containing an open sets, which contains the point.

 

[fresh_42]

Only if it is open. His preferred definitions says that a neighborhood is a set containing an open sets, which contains the point.

In which case you wouldn't need the deviation of arbitrary sets. One could as well directly refer to the open interval. Not open neighborhoods are possible but senseless.

 

[martinbn]

In which case you wouldn't need the deviation of arbitrary sets. One could as well directly refer to the open interval. Not open neighborhoods are possible but senseless.

It doesn't matter. His question is about neighborhoods in the sense he uses. It is not about what is the best definition.

 

[fresh_42]

It doesn't matter. His question is about neighborhoods in the sense he uses. It is not about what is the best definition.

I wanted to avoid confusion among third-party readers, especially as post #1 is very hard to read with all the missing LaTeX tags and syntax errors. I did this because you already addressed and answered his question and he acknowledged it. That's why I used the word usually so often. My answer was supposed to be additional, not alternative information.

The important term of locality in calculus requires open sets. Closed boundaries are the exception and usually require a specific treatment.

 

[WWGD]

Only if it is open. His preferred definitions says that a neighborhood is a set containing an open sets, which contains the point.

There may be a Topology in which these sets are open. Certainly not the standard one, though.

 

[martinbn]

There may be a Topology in which these sets are open. Certainly not the standard one, though.

Definitely, for example the topology where all subsets are open.

 

[WWGD]

Definitely, for example the topology where all subsets are open.

I believe if such sets are seen as a basis, there's a minimal topology where they're open, but I don't think it's the Discrete one.

 

[mcastillo356]

Hi, PF

Thanks for all the answers. I've posted a question that's hard for me, ie, topology, which I fear a lot. I will answer as soon as possible, and probably the answer will consist in more questions. There goes the first question:

But he said what definition he uses, and it doesn't assume that a neighborhood is open.

The concept of neighborhood in Topology is tightly related with the idea of "open". A neighborhood of a point is any set containing an open to which belongs the point.

Neighborhoods that are not intervals can be found in the real line:

$$U=]2-\sqrt{2},2+\sqrt{2}[\cup \{7\}[$$ is a neighborhood of $2$ because there exists an open $G$ such that $2\in G\subseteq {U}$ ; in this case it might be
$$G=E(2,\sqrt{2})=]2-\sqrt{2},2+\sqrt{2}[$$, which is an open, since it is an open interval, and eventually meets $$2\in{G}\subseteq{U}\Longleftrightarrow{2\in{]2-\sqrt{2},2+\sqrt{2}[}}\subseteq{]2-\sqrt{2},2+\sqrt{2}[\cup{\{7\}}}$$.
It is not an interval due to the infinite numbers in $U$ that do not belong to $U$, for example, $$2,7\in{U}$$, but $$\forall{\,x\in [2+\sqrt{2},7[}$$ we have $x\notin{U}$

Marcos

 

[Gavran]

It is not an interval due to the infinite numbers in $U$ that do not belong to $U$,

This is a contradictory statement (the post #13).

$U$ is not an interval because there are reals in $U$ with numbers between them that do not belong to $U$, for example, $2,7\in U$ but $\forall\,x\in [2+\sqrt{2},7[$ we have $x\notin U$?

This is a correct statement (the post #1). One more thing, you have to change $\forall\,x\in [2+\sqrt{2},7[$ into $\forall\,x\in (2+\sqrt{2},7)$.
From your statement in the post #1 it is obvious that there are some points between $2-\sqrt{2}$ and $7$, which are not included into $U$. In other words, there is “a gap” between $2+\sqrt{2}$ and $7$. By definition, $U$ can not be an interval.

 

[mcastillo356]

Hi, dear PF, @Gavran, brilliant your post, but I've decided not to quote yours, and get with some textbook on Topology. I won't either ask the Forum for suggestions about any required reading. I confess I've been talking in my posts this thread like a parrot that copies answers from a Spanish maths forum, ie, without nearly a bit of understanding. My next step will be to turn to the original source. Paradoxically, I will ask Rincón Matemático for English written reference book.

The contains will have to include topology of the real line.

I've tried hard not to ramble, but take this post as a draft, although I'm going to publish it. The reason is that from this moment I start from void, I mean, from zero. The target is the only thing I know right now: introduction to topology.

Hope this is all understandable. If it is not, I beg for questions, or whatever.



Marcos, at 21/01, 2 am

 

[WWGD]

Hi, dear PF, @Gavran, brilliant your post, but I've decided not to quote yours, and get with some textbook on Topology. I won't either ask the Forum for suggestions about any required reading. I confess I've been talking in my posts this thread like a parrot that copies answers from a Spanish maths forum, ie, without nearly a bit of understanding. My next step will be to turn to the original source. Paradoxically, I will ask Rincón Matemático for English written reference book.

The contains will have to include topology of the real line.

I've tried hard not to ramble, but take this post as a draft, although I'm going to publish it. The reason is that from this moment I start from void, I mean, from zero. The target is the only thing I know right now: introduction to topology.

Hope this is all understandable. If it is not, I beg for questions, or whatever.



Marcos, at 21/01, 2 am

Make sure el Runcon Matemático is an open set !!.