# Continuity

• November 28th 2010, 09:31 AM
kathrynmath
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 |x-c|<delta and |g(x)-g(c)|<epsilon

Now I get stuck
• November 28th 2010, 11:37 AM
nimon
"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.
• November 28th 2010, 12:51 PM
kathrynmath
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 |x-c|<delta, it follows that |f(x)-f(c)|<epsilon.
• November 28th 2010, 01:32 PM
kathrynmath
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.
• November 28th 2010, 01:35 PM
Plato
Lets suppose that for each open set $O\subset \mathbb{R}$ it is the case that $f^{-1}(O)$ is also open.
Given a point $p\in\mathbb{R}$ and $\varepsilon > 0$ then $O=\left(f(p)- \varepsilon ,f(p)+ \varepsilon \right)$ is an open set.
So $Q=f^{-1}(O)$ is an open set that contains $p$.
Hence, $\left( {\exists \delta > 0} \right)$ such that $(p-\delta,p+\delta)\subset Q$.
But $f(p-\delta,p+\delta)\subset O$.
Which by the $\varepsilon/\delta$ definition $f$ is continuous.