Generalization of Tietze Extension Theorem I have been reading about Tietze extension theorem and i wondered if we could do it for any subset of $\mathbb{R}$, i.e ,
Let $A$ be a closed subset of a normal space $X$ and we have a continuous function $f: A \rightarrow C$ where $C$ is any subset of $\mathbb{R}$, is there any way we can make a continuous extension?
My intuition says that we cant because in the proof of the theorem we use the fact that we are working with intervals, but im not quite seeing how to create a counterexample, so any enlightenment would be appreciated.
Thanks.
 A: In the statement of the Tietze extension theorem you can in general replace $\Bbb R$ by any so-called AR (sometimes also called AE) space, absolute retracts. These have a quite general (but technical) characterisation, but among the main examples are all convex subsets of separable metric locally convex linear spaces, and this thus includes all intervals in $\Bbb R$ or disks in the plane and solid disks in higher dimensions. See Hu's book theory of retracts, among others, or books like van Mill's "infinite-dimensional topology (prerequisites and an introduction)", or Borsuk's original texts, also see Wikipedia for some basic info and references. 
A: The answer is no, in general. Let $A=\{0,1\}=C$, and $f: A \rightarrow C$ be the identity function, where $X=\Bbb R$. If $g$ were an extension of $f$, then the image of $g$ must be connected (since $\Bbb R$ is). On the other hand the image must be $C$ (since the image of $A$ is $C$, and the extension is only supposed to take values in $C$, according to my understanding of your question), which is not connected. 
A: If $A$ is a non-closed subset of $\Bbb R$, let $a$ be an element of the closure
of $A$ in $\Bbb R$ that is not an element of $A$. Then $x\mapsto 1/(x-a)$ is
a continuous function from $A$ to $\Bbb R$ which does not extend to a continuous
function from $\Bbb R$ to $\Bbb R$.
The same trick works if $\Bbb R$ is replaced by a metric space $X$.
This time take $x\mapsto d(x,a)^{-1}$ to be the function.
