4
$\begingroup$

Let $A$ be an irrational, countable dense subset of $[0,1]$, e.g, $A=\{q\sqrt 2|q\in\mathbb Q\} \cap[0,1] $. Let $\{a_i\}_{i\in \mathbb N} =A$ be an enumeration of $A$. Let $f:\mathbb Q\cap[0,1] \to \mathbb R$ be defined by:

$$f(x)=\sum_{i\in\mathbb N} \frac{\varepsilon_i}{|x-a_i|}.$$

Can the sequence $\{\varepsilon_i\}_{i\in\mathbb N}$ be chosen such that $f$ is well-defined at every point, i.e., the sum is finite?

I came up with this example in trying to construct a continuous function on a dense subset of the unit interval that has no continuous (domain) extension to any neighbourhood (in the unit interval) of any point, but I don’t know how to make to ensure it is well-defined

$\endgroup$
2
  • $\begingroup$ I suppose you want $\epsilon_i>0$ for each $i$ as well. $\endgroup$ Mar 15, 2023 at 0:19
  • $\begingroup$ I am not sure why this is of relevance $\endgroup$ Mar 15, 2023 at 13:06

1 Answer 1

6
$\begingroup$

Yes. Enumerate $\mathbb{Q}\cap[0,1]$ as $\{q_i\}_{i\in\mathbb{N}}$. Now choose $\epsilon_i>0$ such that $\frac{\epsilon_i}{|q_j-a_i|}<1/2^i$ for all $j\leq i$ (which is possible since there are only finitely many such $j$ for each given $i$). Then for any fixed $j$, the terms in $f(q_j)$ will all be less than $1/2^i$ for all $i\geq j$, and so the sum will converge.

(Note, though, that the convergence will not be uniform, so there is no reason to expect $f$ to be continuous. In fact, such an $f$ can never be continuous, since it must approach $\infty$ as you approach any point of $A$, so it is unbounded on every open interval. To get a function that is continuous, you can use a similar construction but using summands that have a jump discontinuity at each $a_i$ rather than blowing up to $\infty$, so that each summand will be bounded and the sum can converge uniformly.)

$\endgroup$
8
  • $\begingroup$ Thanks for your answer. I can’t see why it’s true that “[Then] for any fixed $j$, the terms in $f(q_j)$ will all be less than $1/2^i$ for all $i\geq j$”. Shouldn’t this be $i\le j$? $\endgroup$ Mar 15, 2023 at 1:26
  • $\begingroup$ I thought about the jump discontinuity, for example by replacing the summands by $\mathcal H(x-a_i) \varepsilon_i$, using Heaviside functions, but I would have to think through this example though $\endgroup$ Mar 15, 2023 at 1:34
  • $\begingroup$ The fact that it approaches infinity shouldn’t take away from its continuity per se, $f(x) = \frac{1}{x-\sqrt 2}$ is continuous on the rationals and approaches infinity as you approach $\sqrt 2$ $\endgroup$ Mar 15, 2023 at 1:37
  • 1
    $\begingroup$ No, it is for $i\geq j$. The variables $i$ and $j$ play the same role as they did in the previous sentence (where it was $j\leq i$), just now we are thinking about $j$ being fixed instead of $i$ being fixed. $\endgroup$ Mar 15, 2023 at 2:01
  • 1
    $\begingroup$ Well, each individual Heaviside function is continuous on the rationals, so you just have to make sure the sum converges uniformly to conclude it is continuous on the rationals. $\endgroup$ Mar 15, 2023 at 12:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .