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.

Printable View

- Feb 14th 2010, 08:40 PMjzelltFunction 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.

- Feb 14th 2010, 08:59 PMJhevon
- Feb 15th 2010, 12:15 PMjzellt
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.. - Feb 15th 2010, 02:05 PMJoachimAgrell
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 $\displaystyle \lim_{n\to\infty}f(x_n)=L$? - Feb 15th 2010, 06:42 PMmabruka
Maybe his definition of continuity is

"f is continuous in a iff $\displaystyle \lim_{x\to a}{f(x)}=f(a)$ " - Feb 15th 2010, 06:53 PMDrexel28
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 $\displaystyle g':D\mapsto C$ 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.