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I have some confusions.

I was reading the answer here.

Suppose $f:\mathbb R\to\mathbb R$ is a continuous function. Let $x\in\mathbb R$. Then there is a sequence of rational numbers $(q_n)_{n=1}^\infty$ that converges to $x$. Continuity of $f$ means that $$\lim_{n\to\infty}f(q_n) = f(\lim_{n\to\infty}q_n)=f(x).$$ This means that the values of $f$ at rational numbers already determine $f$.

  1. What do we mean by "the values of $f$ at rational numbers already determine $f$". What is the necessity of this line?
  2. What is $f|_\mathbb{Q}$.

Please help me. Thanks.

emacs drives me nuts
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user1234
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    $\mathbb Q$ is dense, so if you know the values of a continuous map on $\mathbb Q$, then you also know them on $\mathbb R$ – AlvinL Jul 20 '22 at 06:28

1 Answers1

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$f|_\mathbb{Q}$ is the restriction of $f$ to $\mathbb Q$.

  1. means if $f$ and $g$ are two continuous functions $: \mathbb R \to \mathbb R$ such that $f(q)=g(q)$ for all $q \in \mathbb Q$ then $f(x)=g(x)$ for all $x \in \mathbb R$. (This is clear from the fourth line in your question applied to both $f$ and $g$: [$f(x)=\lim f(q_n)=\lim g(q_n)=g(x)$].).
Kavi Rama Murthy
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  • Thanks for your answer. But how the values of $f$ at rational numbers determine $f$? Does the fourth line imply that $f(q)$ exists for $x\in \mathbb{Q}$? – user1234 Jul 20 '22 at 06:27
  • I don't understand what you are asking. $f$ is defined on $\mathbb R$ so $f(q)$ exists for every real number $q$, hence also for every rational number $q$. @user1234 – Kavi Rama Murthy Jul 20 '22 at 06:30
  • I have edited my question. I am asking that why we need the 4th line? – user1234 Jul 20 '22 at 06:34
  • Do you have a proof without the fourth line? I have mentioned how that line is used in showing that $f(x)=g(x)$ for all real numbers $x$. @user1234 – Kavi Rama Murthy Jul 20 '22 at 06:38
  • Ok now I understood that your answer is used to show $\Phi$ is injection( in the main proof). Thanks. – user1234 Jul 20 '22 at 06:41