# If A is contained in B, then the interior of A is contained in B.

• kpoltorak
In summary: That is correct. In summary, the given problem asks to prove that if A and B are subsets of a topological space X and A is a subset of B, then the interior of A is a proper subset of the interior of B. This can be proven using the fact that for any subset X of a topological space, the interior of X is equal to X itself if and only if X is an open set.
kpoltorak

## Homework Statement

Given that A and B are in a topology, show that if A is contained in B, then the interior of A is contained in B.

## Homework Equations

The interior of A:={a: there exists a neighborhood which is a subset of A}

## The Attempt at a Solution

I can prove that the interior of A is contained in OR EQUAL TO the interior of B but I am having trouble showing that they cannot be equal. I have tried setting int(A)=int(B) and working towards a contradiction to no avail.

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What's the interior of R and Q?
I can't think of any nonempty ones off the top of my head.

The interior of Q is empty, and I believe the interior of R would be R if the topology is in R.

There's no contradiction if a subset actually coincides with its superset. You've evidently proven the result.

Sorry, my original post was not entirely correct. I need to prove that the the interior of A is a PROPER subset of the interior of B.

So you're given that A is a proper subset of B? Is the lemma that given A a proper subset of B, int(A) is a proper subset of int(B)? Why isn't $$A=\text{int}(A), B = \bar{A}$$ a counterexample, where $$\bar{A}$$ is the closure of A? Serious question as I'm mainly a physicist.

Edit: I meant $$A=\text{int}(C), B = \bar{C}$$.

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Are you working in a topological space? If you are, it just suffices to show $$\cup \{ X \in \tau : X\subseteq B \mbox{ and } X \cap A = \emptyset \}$$ is nonempty.

If you're working in R^2 you can just play around with lengths to get proper bounds for epsilon.

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another similar counter example, what if B is the union of A with a discrete point not contained in any neighbourhood of A

then doesn't
int(A) = int(B)

can you write the question exactly as it was stated?

I believe I have found an answer. $$\forall{X}\in{\tau}:X=X^{\circ}$$
Where $$X^{\circ}$$ is the interior of X, and $$\tau$$ is a topological space. Doing some reading I found a definition of a topology which defines it as a collection of open sets, and then a definition which said that a set is open iff it is equal to its interior.

Can somebody please verify this is true?

The exact question:

Let $$X$$ be topological space. Prove that if $$A,B \subset{X}$$ and $$A\subset{B}$$ then
a) $$A^{\circ}\subset{B^{\circ}}$$
b) $$\bar{A} \subset \bar{B}$$

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Generally, in the literature $$\subset$$ is synonymous with $$\subseteq$$. $$\varsubsetneq \not= \subset$$

ZioX said:
Generally, in the literature $$\subset$$ is synonymous with $$\subseteq$$. $$\varsubsetneq \not= \subset$$

Apparently not in this particular piece of literature (my homework): http://www.stat.psu.edu/~arkady/403/HWKs/Assignment%205.pdf" See Problem C.

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Thank you everybody for your help, I believe I have enough information to solve the problem.

sorry, I shortcutted to A & B subsets of a topological space

as you point out if A & B are in the topology they are open and equal to their interior, thus it holds for subsets & their proper subsets

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## 1. What does it mean for A to be contained in B?

When we say that A is contained in B, it means that every element of A is also an element of B. In other words, A is a subset of B.

## 2. How is this different from saying that the interior of A is contained in B?

The interior of A refers to the set of all points within A that are not on the boundary of A. So, if the interior of A is contained in B, it means that all the points within A are also elements of B, and that the boundary points of A (if any) are not necessarily included in B.

## 3. Why is it important for the interior of A to be contained in B if A is contained in B?

It is important because the interior of A captures the essence of A, while the boundary points may not be as relevant. So, if the interior of A is contained in B, it ensures that B contains the most important and relevant parts of A.

## 4. Can the interior of A be contained in B without A being contained in B?

No, this is not possible. If the interior of A is contained in B, it means that all the elements of A are also elements of B. Therefore, A must also be contained in B.

## 5. Is this statement always true for any sets A and B?

No, this statement is not always true. In order for the interior of A to be contained in B, it is necessary for A to be a subset of B. However, there may be cases where A is not a subset of B, and therefore the statement would not hold true.

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