# Prove continuity of a function from a subset

Let $$A\subseteq B \subseteq \mathbb R$$ let $f: B\rightarrow\mathbb R$ and let $g$ be the restriction of $f$ on $A$ (ie $g(x)=f(x)$ for $x\in A$)

a)If $f$ is continuous at $c\in A$, show that $g$ is continuous at $c$.

b)Show that if $g$ is continuous at $c$ then it is not necessary that $f$ is continuous at $c$ (by example)

My attempt:

a) $\mid x-c\mid\leq \delta$ implies $\mid f(x)-f(c)\mid \leq \epsilon$ by definition $g(x)=f(x)$ for $x\in A$ we have $\mid g(x)-g(c)\mid \leq \epsilon$

QED

b) I got stuck.

I am new to analysis so please be specific in the answers. Thanks.

For b,let $f$ be sign function, $B=(-1,1)$, $f$ is discontinuous at $c=0$, $A=(0,1)$ then $g=1$ on $A$, hence continuous. Note we exclude the the discontinuity point of $f$ from $A$.

Let $\epsilon > 0$. Then, because $f$ is continuous, there exists such a $\delta>0$ that for $x\in B$, $|x-c|<\delta$ implies $|f(x)-f(x)| < \epsilon$. This means that for each $x\in A$, we know that $x\in B$ (because $A\subseteq B$, and therefore $|x-c|<\delta$ implies $|g(x)-g(x)| = |f(x) - f(b)| < \epsilon$ by the definition of $g$.

For your second answer, you need to create a function which is continuous on some $A$ but not on its superset $B$. In order to find it, it is always a good idea to make your life as simple as possible, which you can do by:

1. Taking a simple set $B$, for example, taking an interval ($B=[0,1]$ or $(0,1)$), or just taking $B=\mathbb R$
2. Making $A$ as simple as possible: taking an interval for $A$, or maybe even a singleton $A=\{0\}$ will make it easy.
3. Choosing simple functions for your $f$, such as constant functions or piecewise constant functions, or taking functions for which you know are continuous almost everywhere, like $1/x$.

Possible examples you get include:

1. $A=\{0\}$: In this case, any function $f$, defined on $B$ (whatever that $B$ is), which is not continuous on $B$, is a counterexample, since any function $g:A\mathbb R$ is continuous.
2. $A=(0,1)$: Taking $B=[0,1)$ is enough to find the function $f(x) = \begin{cases}1 & x>0\\0 & x=0\end{cases}$ which is not continous on $B$ but is continuous on $A$. A more sophisticated example would be $$f(x) = \begin{cases}\frac1x & x>0\\c & x=0\end{cases}$$ where $c$ is any real number.
• Thanks, it is much clearer now Nov 13 '14 at 2:43