Topology, defn of a nowhere dense set in a metric space

[rourky]

Homework Statement

Defn: A subset A of a metric space (X, d) is NOWHERE DENSE if its closure has empty interior.

Now I am told that this implies 1. A is nowhere dense iff closure of A does not contain any non-empty open set and 2. A is nowhere dense iff each non-empty open set has a non-empty open subset disjoint from A.

I have no problem with statement 1, but with regards to statement 2, my query is, is it implicit that each non-empty open set is a non-empty open set in X?

Homework Equations

The closure of A is the union of A and its limit pts

The Attempt at a Solution

I ask the question because from statement 1, there can be no open set in the closure of A, and thus in A or the set of its limit pts. The only other set where an open set could be seems to be X.

 

[Dick]

Yes. I would take 'open' to mean 'open in X'. After all, a nowhere dense set has to have someplace to be nowhere dense in.

 

[rourky]

Thanks Dick, that's the second time you've helped me out this week.