1. ## Continuity open sets

How are these equivalent to the regular epsilon-delta definition of continuity?

(i) Let $\displaystyle f: M \to N$. The pre-image of each closed set in $\displaystyle N$ is closed in $\displaystyle M$.

(ii) The pre-image of each open set in $\displaystyle N$ is open in $\displaystyle M$.

2. Originally Posted by shilz222
How are these equivalent to the regular epsilon-delta definition of continuity?
(i) Let $\displaystyle f: M \to N$. The pre-image of each closed set in $\displaystyle N$ is closed in $\displaystyle M$.
(ii) The pre-image of each open set in $\displaystyle N$ is open in $\displaystyle M$.
In a general space they are not equivalent. But they are in a metric space and in particular $\displaystyle \Re^1$.
Note that $\displaystyle \left| {x - x_0 } \right| < \delta \quad \Leftrightarrow \quad x \in \left( {x_0 - \delta ,x_0 + \delta } \right)$.
In $\displaystyle \Re^1$ the basic open set is an open inteveral.
$\displaystyle \left( {x_0 - \delta ,x_0 + \delta } \right)$ is the a basic open inteveral, in fact it is a ball centered at $\displaystyle x_0$ with radius $\displaystyle \delta$.

Can you work out the details?

3. That book shows that the 2 are equivalent by using sequences. They use the definition of sequence: $\displaystyle d(x,p) < \delta \implies f( f(x), f(p)) < \epsilon$.

I haven't covered balls yet, but I think now that I follow that the book is doing. It seems more intuitive with sequences.

4. Oh I got it. You have to use the rule that the complement of an open set is closed right?

5. Originally Posted by shilz222
Oh I got it. You have to use the rule that the complement of an open set is closed right?
Yes, that will work.