
Continuity
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 xc<delta and g(x)g(c)<epsilon
Now I get stuck

"Then xc<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.

A function f:A>R is continuous at a point c in A if, for all epsilon epsilon>0, there exists a delta>0 such that whenever xc<delta, it follows that f(x)f(c)<epsilon.

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.

Lets suppose that for each open set $\displaystyle O\subset \mathbb{R}$ it is the case that $\displaystyle f^{1}(O)$ is also open.
Given a point $\displaystyle p\in\mathbb{R}$ and $\displaystyle \varepsilon > 0$ then $\displaystyle O=\left(f(p) \varepsilon ,f(p)+ \varepsilon \right)$ is an open set.
So $\displaystyle Q=f^{1}(O)$ is an open set that contains $\displaystyle p$.
Hence, $\displaystyle \left( {\exists \delta > 0} \right)$ such that $\displaystyle (p\delta,p+\delta)\subset Q$.
But $\displaystyle f(p\delta,p+\delta)\subset O$.
Which by the $\displaystyle \varepsilon/\delta$ definition $\displaystyle f$ is continuous.