1. ## Function proof

Let g be defined on a set containing the range of a function f. If f is continuous at z0 and g is continuous at f(z0), then g(f(z)) is continuous at z0.

2. Originally Posted by jzellt
Let g be defined on a set containing the range of a function f. If f is continuous at z0 and g is continuous at f(z0), then g(f(z)) is continuous at z0.
what have you tried? what definition of continuity are you working with?

3. The definition of continuity we're unsing involves limits and is NOT the epsilon delta method.

I havn' really gotten anywhere because I havn't a clue how to do this..

4. Does it involve sequences? I mean (given f:X->R, a in the set of limit points of X), did you define limits like :

$\displaystyle \lim_{x\to a}{f(x)}=L$

if, and only if, for all sequence $\displaystyle (x_n)_{n\in\mathbb{N}}$ such that $\displaystyle x_n\in X\backslash{\{a\}}\forall n\in\mathbb{N}$ and $\displaystyle \lim_{n\to \infty}x_n=a$ we have $\displaystyle \lim_{n\to\infty}f(x_n)=L$?

5. Maybe his definition of continuity is

"f is continuous in a iff $\displaystyle \lim_{x\to a}{f(x)}=f(a)$ "

6. Originally Posted by jzellt
Let g be defined on a set containing the range of a function f. If f is continuous at z0 and g is continuous at f(z0), then g(f(z)) is continuous at z0.
This is a classic and easy to find on the internet theorem...but let met give you an outline of the proof.

Let $\displaystyle f:A\mapsto B$ and let $\displaystyle g:B\mapsto C$ where $\displaystyle f,g$ are both continuous. (we actually only have to have $\displaystyle g'\mapsto C$ where $\displaystyle g'|B=g$ but that is neither here nor there). Then $\displaystyle gf:A\mapsto C$ is also continuous.

Proof: Let $\displaystyle B_{\varepsilon}(g(f(c)))$ be arbitrary. Since $\displaystyle f(c)\in B$ and $\displaystyle g$ is continuous there exists some $\displaystyle B_{\delta}(f(c))$ such that $\displaystyle g\left(B_{\delta}(f(c)\right)\subseteq B_{\varepsilon}(g(f(c)))$. And since $\displaystyle f$ is continuous there exists some $\displaystyle B_{\sigma}(c)$ such that $\displaystyle f\left(B_{\sigma}(c)\right)\subseteq B_{\delta}(f(c))$ and so $\displaystyle g\left(f\left(B_{\sigma}(c)\right)\right)\subseteq g\left(B_{\delta}(f(c))\right)\subseteq B_{\varepsilon}(g(f(c)))$

Or much nicer, if you are using the more topological definition (which is equivalent to the regular in metric spaces) that $\displaystyle k:X\mapsto Y$ is continuous iff $\displaystyle k^{-1}(O)$ is open in $\displaystyle X$ whenever $\displaystyle O$ is open in $\displaystyle Y$ merely note that $\displaystyle (gf)^{-1}(O)=f^{-1}(g^{-1}(O))$ and $\displaystyle g^{-1}(O)$ is ope in $\displaystyle B$ and so $\displaystyle f^{-1}(g^{-1}(O))$ is open in $\displaystyle A$. Done.