3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
5
$\begingroup$

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 ${\bf{R}}$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Yes, that is right. $\endgroup$ – user284331 Jan 25 '18 at 5:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.