# Function proof

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• Feb 14th 2010, 08:40 PM
jzellt
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.
• Feb 14th 2010, 08:59 PM
Jhevon
Quote:

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?
• Feb 15th 2010, 12:15 PM
jzellt
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 PM
JoachimAgrell
Does it involve sequences? I mean (given f:X->R, a in the set of limit points of X), did you define limits like :

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

if, and only if, for all sequence $(x_n)_{n\in\mathbb{N}}$ such that $x_n\in X\backslash{\{a\}}\forall n\in\mathbb{N}$ and $\lim_{n\to \infty}x_n=a$ we have $\lim_{n\to\infty}f(x_n)=L$?
• Feb 15th 2010, 06:42 PM
mabruka
Maybe his definition of continuity is

"f is continuous in a iff $\lim_{x\to a}{f(x)}=f(a)$ "
• Feb 15th 2010, 06:53 PM
Drexel28
Quote:

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

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

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