# Counterexample of Existence of a continuous extension of a Continuous function

Till now, I have proved followings;

Suppose $X,Y$ are metric spaces and $E$ is dense in $X$ and $f:E\rightarrow Y$ is uniformly continuous. Then,

1. $Y=\mathbb{R}^k \Rightarrow \exists$ a continuous extension.

2. $Y$ is compact $\Rightarrow \exists$ a continuous extension.

3. $Y$ is complete $\Rightarrow \exists$ a continuous extension. (AC$_\omega$)

4. $E$ is countable & $Y$ is complete $\Rightarrow \exists$ a continuous extension.

What are true and what are false if $f$ is replaced by a 'continuous function', not uniformly?

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## 2 Answers

For example, with $X = [0,1]$ , $Y = [-1,1]$ and $E$ the rationals in $(0,1)$, take $f(x) = \cos(\pi/x)$.

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$X=\mathbb R$, $Y=[-\frac \pi2,\frac \pi2]\subseteq \mathbb R$, $E=\mathbb Q$, $f(x)=\arctan\left(\frac1{x-\sqrt 2}\right)$ kills them all

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