# Topology, Int(A) is an open set

## Homework Statement

Question: Prove Int(A) is an open set, given Int(A) is the set of all interior pts of A where x is an interior pt of A if it is the centre of an open ball in A.

None

## The Attempt at a Solution

Attempted Soln: Suppose x is an element of Int(A).
Then there exists r > 0 such that B(x, r) is a subset of A.
Have tried to extend this to say there exists r > 0 such that B(x, r) is a subset of Int(A), but with no success.

Have also tried to prove C(Int(A)) contains all its limit pts and thus is closed. Then Int(A) would be open.

Just looking for a hint to get me on the right road. Thanks

Dick
Homework Helper
Take a point y in your B(x,r). Can you also show y is in Int(A)? Hint: consider |x-y|, can you say anything about that distance?

HallsofIvy
By the way, there are a number of different ways of approaching topology. You appear (since you say "there exists r > 0 such that B(x, r) is a subset of Int(A)") to be assuming a topology defined by a metric. I notice Dick refers to |x-y| rather than d(x,y), so he may be assuming you are in R, the real numbers with the usual topology. In more general topology an "open set" is simply a member of the "topology" and the "interior" of a set, A, is defined as the union of all open subsets of A. It would help if you would say explicitely what kind of topology you have, what definitions you are using. In any case, since you are definitely in a metric space, you can use the "triangle" inequality: that $d(x,y)\le d(x,z)+ d(z,y)$ (or $|x-y|\le |x-z|+ |z- y|$. That should be useful.