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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?


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?

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The name of the result in the second part of your post is Tietze extension theorem. – Martin Sleziak Oct 23 '12 at 11:19
@Martin The link is what exactly i explained in last paragraph. The link is about extension of $f:X\rightarrow \mathbb{R}$, but the theorem above is about extension of $f:\mathbb{R} \rightarrow X$. Are these same? – Katlus Oct 23 '12 at 11:24
@Katlus Dugundji's generalization of Tietze's theorem covers the vector-valued case. The original paper was published in 1951 on Pacific J. Math. If you can enter a campus library, you can download it from projecteuclid. – Siminore Oct 23 '12 at 11:42
@Martin Edited. – Katlus Oct 23 '12 at 11:46
What about $E=\{0,1,2,\dots\}\cup\{1/2,1/3,\dots\}$ and $a=0$ as a counterexample for the first part of your question? – Martin Sleziak Oct 23 '12 at 11:46
up vote 2 down vote accepted

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.

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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 $X$.

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Frankly, i replaced it since the statement i posted seemed relatively easier than Tietze extension theorem, but now i think the statement i posted is rather false. Thank you – Katlus Oct 23 '12 at 11:58
I am not sure that the result you've posted is false. – Martin Sleziak Oct 23 '12 at 12:06
Ok, so now we have a counterexample to that result in sheesh's post. (And it was not that difficult.) – Martin Sleziak Oct 23 '12 at 12:20

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