We know that, the function $f: \mathbb{R}→ \mathbb{R}$ such that $f(x) = x$ is uniformly continuous on $\mathbb{R}$.

Now, my Question is, if we define the function $g:\mathbb{Q} →\mathbb{R}$ such that $f(x) = x$ for all $x ∈ Q$ then is function $g$ is uniformly continuous on $\mathbb{Q}$?

My attempt: as we know, continuity is domain based property. So to check continuity of $g$, we must consider the point of domain first!!. Now let $q ∈\mathbb{Q}$ then as we know, rationals are dense in $\mathbb{R}$ so, $q$ must be limit point of domain and not an isolated point of domain. So $g$ is continuous at $q$ if and only if, $$\lim_{x\to q} g(x) = g(q)$$ But is $$\lim_{x\to q+} g(x)=\lim_{x\to q-} g(x)$$? (I stuck here)

As between two rationals there is an irrational number and function is not defined their. So is left hand limit and right hand limit of $g$ as $x$ tends to $q$ are equal? Is $g$ is continuous at $q$ ? and what about uniform continuity? Please help me.

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    $\begingroup$ Start with the definition of uniform continuity, please. $\endgroup$ – Professor Vector Jan 25 '18 at 5:10
  • $\begingroup$ Sir, to be uniform continuous function must be continuous there! So is $g$ is continuous on $\mathbb{Q}$ ? $\endgroup$ – Akash Patalwanshi Jan 25 '18 at 5:11
  • $\begingroup$ Start with the definition of uniform continuity, please. $\endgroup$ – Professor Vector Jan 25 '18 at 5:14
  • $\begingroup$ Please anyone help me $\endgroup$ – Akash Patalwanshi Jan 25 '18 at 5:16

If you have something like for all $x,y\in{\bf{R}}$, if $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon$.

Then, of course, for all $x,y\in S$, if $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon$, (or put it in this way: $|g(x)-g(y)|<\epsilon$.)

Here $S$ is any subset of ${\bf{R}}$.

  • $\begingroup$ But here $g$ is not $f$. $g$ is not defined on $\mathbb{Q^c}$. So how left hand limit of $g$ is equal to right and limit of $g$ (at $q$) $\endgroup$ – Akash Patalwanshi Jan 25 '18 at 5:19
  • $\begingroup$ You see, actually you can replace the $f$ by $g$, do you know what I mean? $\endgroup$ – user284331 Jan 25 '18 at 5:20
  • $\begingroup$ No. How we can? $g$ is restriction of $f$ to $\mathbb{Q}$ $\endgroup$ – Akash Patalwanshi Jan 25 '18 at 5:22
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    $\begingroup$ There is no connectedness in issue for uniform continuity. As long as the expression like, for all $x,y\in\text{Domain of the function}$, if $|x-y|<\delta$, then... $\endgroup$ – user284331 Jan 25 '18 at 5:26
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    $\begingroup$ Yes, that is right. $\endgroup$ – user284331 Jan 25 '18 at 5:38

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