Hi! im new to this site, and id like to start off by asking for help with this problem ive been struggling with for the past hour or so.
Problem: Let A be such that it is a subset of R (real numbers). Prove that A is open if and only if it does not contain its boundary points.
[Also let A(c) be the complement of A]
I started off by letting A be open. I said then that for each x in A, there exists an e-neighborhood V of x such that V is a subset of A.
So, for every x in A, there exists a V(x) such that
p in V(x) --> p is in A for some p in R -from definition of a subset
p not in A --> p is not in V(x)
p in A(c) --> p is not in V(x)
p in V(x) --> p not in A(c)
So as you see, im not really sure where im going with this or even if its right.
It seems to me that A being open implies that there is a neighborhood of x that yes, contains a point of A but not a point of A(c) [from the "p in A(c) --> p is not in V(x)" line]. However, i dont know what this means (like, does this just tell me p is not a boundary pt?) and i think im getting confused and i need to start over.
I think that I need to get to a statement such that an arbitrary boundary point p is not in G, but i dont know how
Any hints or clarification would be VERY appreciated. Thank you