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Decide if the following statements are true or false and justify yor answer.

(a) Every bounded function $f:\mathbb Q\cap [0,1]\to\mathbb R$ is continuous.

(b) Every continuous function $g:\mathbb Q\cap [0,1]\to\mathbb R$ is uniformly continuous.

(c) Let $h:\mathbb Q\cap [0,1]\to\mathbb R$. The following statements are equivalent:

i) There exists a continuous function $\overline{h}:[0,1]\to\mathbb R$, $\overline{h}|_{\mathbb Q\cap [0,1]}=h$.

ii) $h$ is uniformly continuous.

My solution:

(a) False, let $$f(x)=\begin{cases} 0 & \text{si}& x\leq 1/2\\ 1 & \text{si}& x> 1/2\end{cases}$$. (b) Is true, since $\mathbb Q\cap [0,1]$ is compact and $g$ is continuous.

(c) $i)\Rightarrow ii)$ Since $\overline{h}$ is continuous, then $h$ is continuous, and for (b) $h$ is uniformly continuous.

$i)\Leftarrow ii)$ Since $h$ is unif. cont. it can be extended to an uniformly continuous function.

  • 1
    If $\mathbb{Q} \cap [0,1]$ is given the subspace topology inherited from $\mathbb{R}$, it actually has no isolated points. E.g. $\frac{1}{2}- \frac{1}{n} \to \frac{1}{2}$ as $n \to \infty$. – 2017-02-01
  • 1
    $\mathbb Q\cap [0,1]$ is not compact ! – 2017-02-01
  • 0
    Also $\mathbb{Q}\cap [0,1]$ is not compact (not closed). – 2017-02-01
  • 0
    for (b) and (c) i think $\phi (x)=\frac{1}{x-\sqrt{2}/2}$ works – 2017-02-01
  • 0
    I see, I will think more about it. Thank you very much – 2017-02-01

0 Answers 0