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Let $E$ be a subset of $\mathbb{R}$ and let $f$ be a continuous function defined on $E$. Is it true that $f$ can always be extended to a function $\tilde{f}$ defined on $\mathbb{R}$, which is still continuous on $E$? I know that we cannot ask $\tilde{f}$ be to continuous on all $\mathbb{R}$, which is shown by the following example: Does every continuous map from $\mathbb{Q}$ to $\mathbb{Q}$ extends continuously as a map from $\mathbb{R}$ to $\mathbb{R}$?

Thank you in advance!

Edit: I wanted ask if $\tilde{f}$ could be continuous at every point of $E$. I hope it's clearer phrased this way!

Edit2: From comments. For example, if $E=\{0\}$, then $f$ with $f(0)=0$ is continuous. If we define $\tilde{f}(x)=1$ for $x\in\mathbb{R}\backslash\{0\}$ (and $\tilde{f}(x)=f(x)$ for $x\in E$), then the restriction of $\tilde{f}$ to $E$ is continouous but $\tilde{f}$ is not continuous at $0$.

Jiu
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  • Continuous at each point of $E$ or continuous when restricted to $E$ (w.r.t. the topology of $E$)? – geetha290krm Jun 27 '22 at 05:00
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    @copper.hat Your function does extend to a function which is continuous at each point of $E$. – geetha290krm Jun 27 '22 at 05:02
  • @geetha290krm continuous at every point of $E$ as a function on $\mathbb{R}$. – Jiu Jun 27 '22 at 05:10
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    @copper.hat Define $f(0)=0$. That extends your function to $\mathbb R$ and it is continuous at each point of $E$. – geetha290krm Jun 27 '22 at 05:25
  • @geetha290krm Sorry, I misread. – copper.hat Jun 27 '22 at 05:29
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    Define $f(x)$ to be $-{1 \over 12}$ for $x \notin E$. – copper.hat Jun 27 '22 at 05:29
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    @copper.hat Your last comment answers the OP's question, but I believe OP might want you to express your extended $f$ using a tilde over the $f$ for the extension function to keep it distinct from the original $f$ defined only on $E.$ [I agree this is a very trivial thing to point out.] – coffeemath Jun 27 '22 at 05:34
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    @coffeemath I missed my 5 mins. window. – copper.hat Jun 27 '22 at 05:38
  • @copper.hat I have difficulty understanding your answer with -1/12... For example, when $f$ is defined on $E={0}$ with $f(0)=0$, then this extension is not continuous on $0$ anymore. But does it mean that no extension could be continuous at $0$? – Jiu Jun 27 '22 at 05:41
  • @Jiu In your posted question, you only asked that the extended function be continuous on the set $E$ (the set on which your starting function was defined, and on which it was assumed continuous). – coffeemath Jun 27 '22 at 05:43
  • @coffeemath Maybe I should have written at every point of $E$. – Jiu Jun 27 '22 at 05:45
  • What does it mean to say a function is continuous on a singleton set such as ${0}$? – coffeemath Jun 27 '22 at 05:45
  • What I mean to ask in my last comment is: If in your set=up the subset $E$ of the reals is a single real number, say $0,$ then what does your initial assumption that your function is continuous on $E$ mean? – coffeemath Jun 27 '22 at 05:48
  • @coffeemath There is a unique structure of topological space on a singleton. And continuity is defined accordingly. Every function defined on a singleton is continuous. – Jiu Jun 27 '22 at 05:48
  • Let us continue this discussion in chat. – coffeemath Jun 27 '22 at 05:48

1 Answers1

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Given $f:E \to \mathbb R $ continuous on $E$, define $\tilde f:\mathbb R \to \mathbb R $ as follows:

  1. For $\, x \in E\, $, let $\, \tilde f(x):=f(x)$.

  2. If $x \in \overline{E}\setminus E$ and $\displaystyle \limsup_{E \ni t \to x} f(t)\in {\pm \infty}\, , \,$ let $\tilde f(x):=0$.

  3. If $x \in \overline{E}\setminus E$ and $\displaystyle \limsup_{E \ni t \to x} f(t)\in \mathbb R \, , \,$ let $\displaystyle \tilde f(x):= \limsup_{E \ni t \to x} f(t)$.

  4. If $x \in \mathbb R \setminus \overline{E}\, , \, $ let $\, d(x,E):=\inf_{t \in E} |x-t|$, and define $$ \tilde f(x):= \inf \{f(t) \, : \, t\in E, \, \, \,\; |t-x| \le 2d(x,E)\}\,.$$

It remains to check continuity of $\tilde f$ in points of $E$.

Given $z \in E$ and $\epsilon>0$, we know there exists $\delta>0$ such that $$\forall t \in E \cap(z-\delta,z+\delta), \quad |f(t)-f(z)| <\epsilon/2 \,.$$ Now suppose that $x \in \mathbb R \setminus {E}$ satisfies $|x-z|<\delta/3$. Then there are two possibilities:

(a) If $x\in \overline{E}$, then we must have $\displaystyle \limsup_{E \ni t \to x} f(t) \in [f(z)-\epsilon/2, f(z)+\epsilon/2]$, so in particular $$|\tilde f(x)-f(z)|<\epsilon \,. $$

(b) If $x\notin \overline{E}$, then $0<d(x,E) < \delta/3$, so all $t\in E$ such that $|t-x| \le 2d(x,E)< 2\delta/3$ satisfy $|t-z| <\delta$. Therefore $$|\tilde f(x)-f(z)|\le \epsilon/2 <\epsilon \,. $$

Yuval Peres
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