Continuous extension / Closed set in $\mathbb{R}$ I have two questions.
First: Let $E$ be a unbounded closed set in $\mathbb{R}$ and $a\in E$. Suppose $\inf \{x\in E : a<x\} = a$. Here, i think it must be true that "$\exists \epsilon>0$ such that $(a,a+\epsilon) \subset E$". How do i prove this?
Secondly,
PMA, Rudin p.99 states;
It can be proven that "If $E$ is a closed set in $\mathbb{R}$ and $f:E\rightarrow X$ is a continuous function from $E$ to a metric space $X$, then $\exists$ continuous extensions $g:\mathbb{R} \rightarrow X$ such that $\forall x\in E, f(x)=g(x)$. However proof is not simple".
How exactly not simple?
Is this almost impossible to prove this with only basic concept of topology?
I googled it, but it seems like the theorem above is not even generally called 'continuous extension' since the proof for continuous extension on wikiproof shows that "There exists an extension of $f:E\rightarrow \mathbb{R}$ where $E\subset X$, a metric space".
Anyway, how do i prove that with relatively easy concepts?
 A: Take $E = X = \{0,1\}$, and $f(x) = x$. Then there is no continuous extension $g\colon \mathbb{R} \to X$ because $\mathbb{R}$ is connected while $X$ isn't.
A: This is not an answer, but it would be too long for a comment.
It is possible that I have an older copy of Rudin, but in my copy I read the following:

If $f$ is a real continuous function defined on a closed set $E\subseteq R^1$, prove that there
  exist continuous real functions $g$ on $R^1$ such that $g(x) =f(x)$ for all $x \in E$. (Such
  functions $g$ are called continuous extensions of $f$ from $E$ to $R^1$.) Show that the
  result becomes false if the word "closed" is omitted. Extend the result to vector-valued 
  functions. Hint: Let the graph of $g$ be a straight line on each of the segments 
  which constitute the complement of $E$ (compare Exercise 29, Chap. 2).
  The result remains true if $R^1$ is replaced by any metric space, but the proof is not
  so simple. 

There are two places where $R^1$ is used in this exercise: One of them is $E\subset R^1$, the second one is $R^1$ as the codomain.
I think that when the author writes that $R^1$ is replaced by any metric space he means changing $E\subset R^1$ to $E\subset X$, where $X$ is a metric space.
(And this is, indeed, special case of Tietze extension theorem.)
In your post you suggested extending $E\to X$ to $\mathbb R\to X$, instead; so you replaced the other occurrence of $R^1$ by $X$.
