Suppose $f:E→\mathbb R$ is continuous at $x_o$, and $x_o\in F\subset E$. Define $g:F→\mathbb R$ by $g(x)=f(x)$ for all $x\in F$. Prove that $g$ is continuous at $x_o$. Show by example that the continuity of $g$ at $x_o$ need not imply the continuity of $f$ at $x_o$.
I'm not sure if I have to use a theorem or if this proof is just supposed to be that simple. $f$ has been defined to be continuous at $x_o$. So, it makes sense that $g$ is equal to $f$ on a smaller domain. So, if we restrict $f$ to the same domain as $g$ it would still be continuous. We could write, $f:F→\mathbb R$ is continuous at $x_o$, but that's just $g$, so $g(x)$ is continuous at $x_o$.
Or maybe this is supposed to be more difficult and I'm doing it wrong.
Useful theorem:
Let $f:E→\mathbb R$ with $x_o\in E$ and $x_o$ an accumulation point of $E$. Then, the following are equivalent:
1) $f$ is continuous at $x_o$.
2) $f$ has a limit at $x_o$ and $\lim_{x→x_o}f(x)=f(x_o)$.
3) For every sequence $\{x_n\}_{n=1}^\infty$ converging to $x_o$ with $x_n\in E$ for each $n$, $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x_o)$.
UPDATE: Here's what I've put together so far:
Assuming $f$ is continuous at $x_o$, then by the useful theorem, for any sequence $\{x_n\}_{n=1}^\infty$ converging to $x_o$, $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x_o)$. Let $\{x_m\}_{m=1}^\infty$ be a sequence for $x_m\in F$. If $F\subset E$, then $\{x_m\}_{m=1}^\infty$ is a subsequence of $\{x_n\}_{n=1}^\infty$. Since $\{x_n\}_{n=1}^\infty$ converges to $x_o$, by another theorem, all subsequences of a converging sequence have to converge to the same to the same value, so $\{x_m\}_{m=1}^\infty$ converges to $x_o$. By the definition of $g$ the domain is $F$, so $\{g(x_n)\}_{n=1}^\infty$ converges to $g(x_o)$ implying $g$ is continuous at $x_o$. $\square$