Topology: is this an open cover of an unbounded subspace of a metric space?

[fraggle]

Homework Statement

Suppose A is an unbounded subspace of a metric space (X,d) (where d is the metric on X).
Fix a point b in A let B(b,k)={a in X s.t d(b,a)<k where k>0 is a natural number}.

Let A^B(b,k) denote the intersection of the subspace A with the set B(b,k).

Then the collection {A^B(b,k)} over all k in the natural numbers (i.e keeps going on and on) is an open cover of A in X.

Homework Equations

I believe that this is an open cover. If it is then the proof I'm doing will work.
If somebody can prove me wrong could they please suggest an open cover?
Thank you very much

The Attempt at a Solution

since each B(b,k) is a basis element for the metric topology on X B(b,k) is open in X and thus A^B(b,k) is open in A. Since there are infinitely many balls they will cover A. Therefore {A^B(b,k)} is an open cover of A as a subspace of X.

The only possible problem I see is that A is unbounded. Perhaps this collection does not cover all elements of A?

Thank you for the help

 

[Office_Shredder]

Even if A is unbounded, for any specific point in a, you have that d(a,b)=R for some real number. You can find k>R, and a is contained in B(b,k)

You might be interested in noting that b doesn't even need to be in A, just in X

 

[fraggle]

thanks, that makes sense