Continuous extension of a real function Related;
Open set in $\mathbb{R}$ is a union of at most countable collection of disjoint segments
This is the theorem i need to prove;
"Let $E(\subset \mathbb{R})$ be closed subset and $f:E\rightarrow \mathbb{R}$ be a contiuous function. Then there exists a continuous function $g:\mathbb{R} \rightarrow \mathbb{R}$ such that $g(x)=f(x), \forall x\in E$."
I have tried hours to prove this, but couldn't. I found some solutions, but ridiculously all are wrong. Every solution states that "If $x\in E$ and $x$ is not an interior point of $E$, then $x$ is an endpoint of a segment of at most countable collection of disjoint segments.". However, this is indeed false! (Check Arthur's argument in the link above)
Wrong solution Q4.5;
http://www.math.ust.hk/~majhu/Math203/Rudin/Homework15.pdf
Just like the argument in this solution, i can see that $g$ is continuous on $E^c$ and $Int(E)$. But how do i show that $g$ is continuous on $E$?

This is a special case of the Tietze extension theorem. This is a standard result whose proof can be found in any decent topology text. A rather different proof can be found here.
