Is the neighborhood of a point...?

[Trenters4325]

Is the neighborhood of a point in a one-dimensional topological space just a line segment?
Likewise, is the neighborhood of a point in a two-dimensional topological space just a "filled" circle and is the neighborhood of a point in a three-dimensional topological space just "filled" sphere?

 

[JakeD]

Is the neighborhood of a point in a one-dimensional topological space just a line segment?
Likewise, is the neighborhood of a point in a two-dimensional topological space just a "filled" circle and is the neighborhood of a point in a three-dimensional topological space just "filled" sphere?

In a general topological space the concepts of dimension, line segment, circle and sphere may not even be defined or have any relation to the topology. So the answer is no in general. But if the topology is defined using those concepts, such as for the Euclidean topology for \(\displaystyle \mathbb{R}^3\), the answer is of course yes.

The concept of neighborhood can always be defined for any topological space. One simple definition is this: a neighborhood of a point x is any open set containing x. Then "neighborhood of x" abbreviated nbd of x is just a shorthand for "an open set containing x."

See "http://en.wikipedia.org/wiki/Neighbourhood_(mathematics)" for alternative definitions.

 

[Trenters4325]

But if the topology is defined using those concepts, such as for the Euclidean topology for \(\displaystyle \mathbb{R}^3\), the answer is of course yes.

I'm confused. How can a filled circle be an open set when the points on the edge do not have any "wiggle" room?

 

[JakeD]

But if the topology is defined using those concepts, such as for the Euclidean topology for \(\displaystyle \mathbb{R}^3\), the answer is of course yes.

I'm confused. How can a filled circle be an open set when the points on the edge do not have any "wiggle" room?

Are you taking about this line from the Wikipedia page?

• Intuitively speaking, a neighbourhood of a point is a set containing the point where you can "wiggle" or "move" the point a bit without leaving the set.

When we have a neighborhood of x, the only point being wiggled is x. It stays within the neighborhood when wiggled a bit. The neighborhood itself and in particular the points on the edge do not move.

Another note: you have to be careful about whether a neighborhood is required to be an open set. A "filled circle" with the edge or boundary included is never an open set under the Euclidean topology, but it may be a neighborhood depending on the definition being used. As the Wikipedia page says

• Note that the neighbourhood V need not be an open set itself. If V is open it is called an open neighbourhood. Some authors require that neighbourhoods be open; be careful to note conventions.

When you said filled circle in your first post, I interpreted that to mean without the boundary. Much confusion is possible here!

 

[Trenters4325]

If X is a topological space and p is a point in X, a neighbourhood of p is a set V, which contains an open set U containing p.

Note that the neighbourhood V need not be an open set itself. If V is open it is called an open neighbourhood. Some authors require that neighbourhoods be open; be careful to note conventions.

How can a neighborhood contain an open set but not be an open set itself?

Would the following be an example of such a case?

Let x be the origin in R3, and let the neighborhood include a sphere around the origin as well as the x-y plane.

P.S. Do you use TeXaid? If so, what translator do you set it to when using it with this forum?

 

[JakeD]

If X is a topological space and p is a point in X, a neighbourhood of p is a set V, which contains an open set U containing p.

Note that the neighbourhood V need not be an open set itself. If V is open it is called an open neighbourhood. Some authors require that neighbourhoods be open; be careful to note conventions.

How can a neighborhood contain an open set but not be an open set itself?

Would the following be an example of such a case?

Let x be the origin in R3, and let the neighborhood include a sphere around the origin as well as the x-y plane.

There's one little problem with your example: the entire plane is an open set. Consider the sphere with boundary, which is a closed set, and the sphere without boundary, which is open.

P.S. Do you use TeXaid? If so, what translator do you set it to when using it with this forum?

No. But did you try the LaTeX settings and run into problems?

 

[Trenters4325]

If X is a topological space and p is a point in X, a neighbourhood of p is a set V, which contains an open set U containing p.

Note that the neighbourhood V need not be an open set itself. If V is open it is called an open neighbourhood. Some authors require that neighbourhoods be open; be careful to note conventions.

Let x be the origin in R3, and let the neighborhood include a sphere around the origin as well as the x-y plane.

There's one little problem with your example: the entire plane is an open set. Consider the sphere with boundary, which is a closed set, and the sphere without boundary, which is open.

Why is that a problem? I was trying to give an example of a neighborhood that contained an open set but was not an open set itself.

But did you try the LaTeX settings and run into problems?

Do you mean the LaTeX translators? Yes, I tried AMS LaTeX and LaTeX 2.09 or later and neither of them worked. What do you use for your TeX and LaTeX needs?

 

[Trenters4325]

Ah, I understand everything now.

No need to answer my questions above except those about TeX and LaTeX.

 

[JakeD]

But did you try the LaTeX settings and run into problems?

Do you mean the LaTeX translators? Yes, I tried AMS LaTeX and LaTeX 2.09 or later and neither of them worked. What do you use for your TeX and LaTeX needs?

I type in the LaTeX.

I suggest you post a description of the problems you're encountering to this forum. Include a small example such as this. You want \(\displaystyle \L \int\ f(x)\ dx\) which is produced by the LaTeX code

[tex]\int\ f(x)\ dx.[/tex]

Show what TeXaid gives you.

 

[Trenters4325]

TeXaid is a program in which I can use buttons to get functions and operators that appear as they would in a post and I can manipulate them in while in that form. When I copy and paste it, the translator changes it into code which this forum then (should) return to its originial form. Thus, I cannot "show what TeXaid gives" for that a code because you cannot input code into TeXaid.

Also, what do you mean you type in LaTeX? Does that mean you didn't download any editor and that you just type in all of the code from memory?

Also, the translation works fine at the SOS forum, so I do not think that they can help me.

 

[JakeD]

TeXaid is a program in which I can use buttons to get functions and operators that appear as they would in a post and I can manipulate them in while in that form. When I copy and paste it, the translator changes it into code which this forum then (should) return to its originial form. Thus, I cannot "show what TeXaid gives" for that a code because you cannot input code into TeXaid.

What TeXaid gives means what code you get when you cut and paste.

Also, what do you mean you type in LaTeX? Does that mean you didn't download any editor and that you just type in all of the code from memory?

Yup. Probably most posters who use LaTeX on the math help sites do that.

Also, the translation works fine at the SOS forum ...

Good.

so I do not think that they can help me.

That conclusion may not be true. There is overlap between the regulars on the different math help sites.

 

[pka]

Is the neighborhood of a point in a one-dimensional topological space just a line segment?
Likewise, is the neighborhood of a point in a two-dimensional topological space just a "filled" circle and is the neighborhood of a point in a three-dimensional topological space just "filled" sphere?

I have read though the above discussion and find needless confusion there!

Definition: The statement that N is a neighborhood of a point p means that there is an open set such that \(\displaystyle p \in O \subseteq N\).

You see that the notion of neighborhood depends upon on the meaning of open set. So you need to know about basic open sets. These change. Are you working in a basic Euclidian metric space, a Moore space, a Hausdorff space, a co-finite topology or what? You see, the idea if open set changes with the type of space. It is very dangerous to use Wikipedia or MathWorld for mathematical definitions in advanced courses such as topology. Use only your text material for definitions.

Now in basic Euclidian metric spaces basic open sets are simple.
In R<SUP>1</SUP> they are open intervals (a,b); in R<SUP>2</SUP> they are interiors of a circle; in R<SUP>3</SUP> they are interiors of spheres. Thus in these spaces a neighborhood of a point is a set that contains one of theses open sets about the point.

To answer your question about a line segment: a line segment is a neighborhood of each of its points except for its endpoints. Any open set is a neighborhood of all of its points.

 

[Trenters4325]

Are Euclidean metric spaces just R^1,R^2,...,R^n i.e. spaces of n dimensions in which each direction is orthogonal?

 

[pka]

Are Euclidean metric spaces just R^1,R^2,...,R^n i.e. spaces of n dimensions in which each direction is orthogonal?

I do not know what that means.
But metric means measure. We use the distance formula.
That is d(x,y) is the distance from x to y.
Do you mean that the axis are perpendicular? If so, then yes that is true.

 

[Trenters4325]

Yes, thats what I mean.

So the term Euclidean space applies to any space in which the the distance formula applies including Rs greater than 3 where "distance" doesn't really make physical sense. Interesting.

 

[pka]

So the term Euclidean space applies to any space in which the the distance formula applies including Rs greater than 3 where "distance" doesn't really make physical sense.

Well no, not always!
If the distance is the Euclidean Distance then yes.
But there are other distance formulae.
The Euclidean Distance is derived from the Pythagorean Theorem.

 

[Trenters4325]

OK, forget about distance. Is this correct:

The term "Euclidean space" applies to any n-dimensional coordinate system (R^n) in which each of the dimensions (or their axes) are orthogonal?

 

[pka]

OK, forget about distance. Is this correct:

The term "Euclidean space" applies to any n-dimensional coordinate system (R^n) in which each of the dimensions (or their axes) are orthogonal?

YES