"Then |x-c|<delta and |g(x)-g(c)|<epsilon."
This just won't do. Write back with the proper definition of continuity and we'll carry on from there.
Let g be defined on all of R. If A is a subset of R, define the set g^-1(A) by
g^-1(A)={x in R : g(x) in A}.
Show that g is continuous iff g^-1(O) is open whenever O contained in R is an open set.
well g^-1(O) means g(O) is in A.
Let g be continuous and O be an open subset of R.
Then |x-c|<delta and |g(x)-g(c)|<epsilon
Now I get stuck
I tried to get started on both directions:
suppose g^-1(O) is open for each open subset O of R. Fix a real number x, and let epsilon be given. The neighborhood N centered at g(x) with radius epsilon is an open set, so the inverse image of N under g is an open set containing x. Thus there is a neighborhood V of x with radius delta such that V is a subset of g^-1(N)
If g is continuous at x, then there exists a delta such that whenever y is less than delta apart from x, g(y) is in a neighborhood of g(x) with radius epsilon.