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

[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.

 

[ZioX]

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

 

[kpoltorak]

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

 

[fzero]

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

 

[kpoltorak]

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.

 

[fzero]

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}.

 

[ZioX]

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.

 

[lanedance]

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?

 

[kpoltorak]

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}

 

[ZioX]

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

 

[kpoltorak]

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.

 

[kpoltorak]

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

 

[lanedance]

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