# Proving [Subsets, interior, open ball]

• Design
In summary: A) is a subset of int(B) because every element of int(A) is also in int(B).You've shown that every element of int(A) is also in int(B). Use the definition of a subset. int(A) is a subset of int(B) because every element of int(A) is also in int(B).
Design

## Homework Statement

Prove that if A is a subset of B then int(A) is a subset of int(B).
int(A) = interior of A
int(B) = interior of B

## The Attempt at a Solution

Take some y E int(a) , this implies that B(r,y) is a subset of A.
Given that A is a subset of B, we know that B(r,y) is a subset of B.
Now take some z E int(b), this implies that B(r,z) is a subset of B.

I got this much but I don't understand how even if I show that B(r,y) is in B(r,z), this shows that it is an int(A) is a subset of int(B) or am I totally on the wrong track?

thank you

What does it mean precisely for z to be in int(B)? Is this condition satisfied by y in int(A)?

fzero said:
What does it mean precisely for z to be in int(B)? Is this condition satisfied by y in int(A)?

It means that there exist a B(r,z) in B. Don't understand what you mean here

Design said:
It means that there exist a B(r,z) in B. Don't understand what you mean here

OK, in your original post you showed that for y in int(A), there is a B(r,y) in B. Is there a y that is not in int(B)?

fzero said:
OK, in your original post you showed that for y in int(A), there is a B(r,y) in B. Is there a y that is not in int(B)?

No there is no y that is not that is not in int(B) since A is a subset of B.

Design said:
No there is no y that is not that is not in int(B) since A is a subset of B.

Well I'd be a bit more careful here. $$A \subset B$$ is obviously important, but the crucial condition for an object y to be in int(B) is that there is a B(r,y) in B. You have all of the results that you need to finish the proof, you just need to put them in the right order.

fzero said:
Well I'd be a bit more careful here. $$A \subset B$$ is obviously important, but the crucial condition for an object y to be in int(B) is that there is a B(r,y) in B. You have all of the results that you need to finish the proof, you just need to put them in the right order.

Thanks I think I understand what you mean.

How should I piece together the last part about how int(A) is a subset of int(B).
Should i say since the ball came from int(A) it must follow that int(A) is a subset of int(B)?

Design said:
Thanks I think I understand what you mean.

How should I piece together the last part about how int(A) is a subset of int(B).
Should i say since the ball came from int(A) it must follow that int(A) is a subset of int(B)?

You've shown that every element of int(A) is also in int(B). Use the definition of a subset.

## 1. What is the definition of a subset?

A subset is a set that contains only elements that are also contained in another set. In other words, all elements of the subset must also be elements of the larger set.

## 2. How do you prove that one set is a subset of another set?

To prove that one set is a subset of another set, you can show that all elements of the first set are also elements of the second set. This can be done by either listing out all the elements of both sets and comparing them, or using logical arguments and mathematical notation to show the relationship between the two sets.

## 3. What is the interior of a set?

The interior of a set is the largest open set that is contained within the original set. In other words, it is the set of all points within the original set that do not touch the boundary of the set.

## 4. How can you prove that a set is open?

To prove that a set is open, you can show that for any point in the set, there exists an open ball (a set of points within a certain distance from the given point) that is also contained within the set. This can be done using mathematical notation and logical arguments.

## 5. Can a closed set also be open?

No, a closed set cannot also be open. By definition, a closed set is a set that contains all its boundary points, while an open set does not contain any of its boundary points. Therefore, a set cannot be both closed and open at the same time.

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