# Prove: If p has a neighborhood contained in A, then p is in the interior of A.

• Hodgey8806
In summary, the conversation discusses the definition of a neighborhood and its relationship to the interior of a set. The definition states that a neighborhood is an open subset of X. The conversation ends with a discussion on how to properly prove that p is in the union of all open sets.
Hodgey8806

## Homework Statement

If p has a neighborhood contained in A, then p is in the interior of A.

## Homework Equations

Int A = $\bigcup${C$\subseteq$X:C$\subseteq$A and C is open in X}
By the books definition, a neighborhood is open.

## The Attempt at a Solution

Let C'$\subseteq$A be a neighborhood of p contained in A
Then p$\in$$\bigcup${C$\subseteq$X:C$\subseteq$A and C is open in X}
p$\in$Int A

Seems too simple, but I'm going through the book's definition of "neighborhood of p"--and open subset of X.

Last edited:
You are thinking along the right lines. But C' as you chose it leads you nowhere. You have not proved that p is in any of C.

Should it state something like:
Let p$\in$A,
There exists an open set C'$\subseteq$A s.t. p$\in$C'
Thus it's in the union of all open sets.

Would that fix the issue? Thanks!

That works.

Excellent! Thank you for the help!

You just need to put things in a logical sequence without any gaps. First there is a neighborhood V. It contains some open set C'. Which must necessarily be one of all those open sets. So p is in the union. Q. e. d. If you are not explicit about every small detail, your professor might become unhappy :)

Lol, my professor is a stickler--but she is an excellent teacher! So she's only a stickler because she has covered the material. So hopefully I catch these things before a test---though we have a quiz coming up!

If you have time, do you mind helping me with my other post? I'm not quite sure I can stop where I did--though it makes sense to me!

## What does the statement "If p has a neighborhood contained in A, then p is in the interior of A" mean?

The statement means that if a point p is within a neighborhood of a set A, then p is also inside the interior of A. In other words, p is surrounded by points that are all contained within A.

## What is a neighborhood of a point?

A neighborhood of a point is a set of points that are close to the given point. The distance between the points in the neighborhood and the given point is usually less than a certain value, known as the radius of the neighborhood.

## How do you prove that a point is in the interior of a set?

To prove that a point p is in the interior of a set A, you must show that there is a neighborhood of p that is completely contained within A. This can be done by finding a radius for the neighborhood that is small enough so that all points within that radius are also inside A.

## Can a point be in the interior of a set if it is not in the set itself?

No, a point can only be in the interior of a set if it is also in the set itself. If a point is not in the set, then it cannot have a neighborhood that is completely contained within the set.

## Does this statement only apply to Euclidean spaces?

No, this statement applies to any topological space. In Euclidean spaces, neighborhoods are defined by distance, but in other topological spaces, neighborhoods can be defined in different ways. However, the concept of a point being in the interior of a set remains the same.

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