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Find a function which has limit only in advanced marked points$(a_1,a_2,....a_n)$.

Here is my example.

$g(x) = \begin{cases} 1, & \text{if}~x\in\mathbb{Q} \\ 0, & \text{if}~x\in\Bbb{R}\backslash \Bbb{Q} \end{cases}$

$f(x) = (x-a_1)(x-a_2)...(x-a_n)g(x)$

But the problem I can't solve is that,is there a function which has limit only in advanced given $a_1,a_2...a_n....$ infinite points.

K.defaoite
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Hrant Baloyan
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3 Answers3

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An equivalent way of asking this question is whether there exists a function which is continuous only on a given sequence.

Such a function may fail to exists, for example when your sequence is dense in $\mathbb{R}$ (e.g. $\mathbb{Q}$). Because the set of continuity points must be a $G_\delta$ subset (and therefore cannot be dense and countable).

On the other hand, if your sequence has no limit points, then one can easily construct such a function by dividing $\mathbb{R}$ into intervals, each containing a single element of the sequence, and then define the function on the each interval to be continuous only on that element (for example as you suggested in your question).

MOMO
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    Why are both questions equivalent? What if the limit exists at some points $a_i$ but the function is not continuous at that points? – jjagmath Apr 01 '21 at 09:55
  • @jjagmath . I agree. The Q asks whether there exists $f$ such that $\lim_{y\to x}f(y)$ exists iff $x\in{a_n:n\in\Bbb N}$, not that $\lim_{y\to x}f(y)=f(x)$. – DanielWainfleet Apr 04 '21 at 03:10
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If I understand correctly your statement, then as stated this is not possible. You need some further hypothesis on the sequence $(a_n)$.

Otherwise take $a_n$ to be an enumeration of the rationals. If $f$ satisfies your condition then by assigning at each $a_n$ the limit value to $f(a_n)$ you obtain a function which is continuous on the rationals and discontinuous on irrationals. This is not possibly by Baire's category theorem.

When the sequence has no accumulation points there are several ways. You could e.g. use Weierstrass' Factorization Theorem which states that given any sequence $(a_n)_{n\geq 1}$ of complex numbers, with $|a_n|\to \infty$, there is an entire function $h(z)$ whose zeroes are exactly the elements of the sequence (see e.g. the link). Then your argument taking your $g$ and $f(z)=h(z)g(z)$ still goes through. But this may violate your requirement of 'not using series'?

H. H. Rugh
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  • The trouble with this A is that the Q asks that $\lim_{x\to a_n}f(x)$ exists but not that $f$ is necessarily continuous at $a_n$. Some other answers also have this oversight. – DanielWainfleet Apr 04 '21 at 17:10
  • Thanks, but I am aware of this. I explicitly state that BY ASSIGNING the limit value at each $a_n$, then you obtain a function which is continuous at $a_n$ and discontinuous everywhere else. And for this there are counterexamples (as also stated by @MOMO). – H. H. Rugh Apr 04 '21 at 17:38
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Assume that there is an order preserving bijection between $a_n$ and N. Define $$ f(x)=\frac{1}{n}, \quad x \in[a_n, a_{n+1}). $$

Then the function is discontinuous at every $a_n$ but has a right limit as $x \to a_n$. If, however, you want double-sided limits then take a small $\delta$ and modify as $f(x)=\frac{1}{n}$ for x $\in(a_n-\delta, a_{n}+\delta)$ and $f(x)=i$ inbeween (imaginary). But in this case, you are going outside ot the real numbers. From the perspective of the real numbers, the function is undefined between the intervals.

user48672
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    This only work if the sequence $(a_n)$ is strictly increasing. – jjagmath Mar 30 '21 at 21:17
  • @ jjagmath. Not true. Think before you downvote. Do you know the definition of a limit? – user48672 Mar 30 '21 at 21:18
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    @user48672 He is correct, it is not a function otherwise. – Dole Mar 30 '21 at 21:22
  • @Dole A function is a mapping between 2 sets. So the pre-image set must be specified. I can take $R \to R$ but also $R - > C$. What is wanted to point out is that the question underspecified. – user48672 Mar 30 '21 at 21:31
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    I think their point is that, if $a_2<a_4<a_3<a_5$, then your function is defined so that $f(x)=1/2=1/4$ for $a_4<x<a_3$ – Joe Mar 30 '21 at 21:37
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    As in, although the points $a_n$ are countable, there may not exist an order preserving bijection with the set of points and $\mathbb{N}$ that allows $f(x)$ to be defined piecewise on disjoint intervals $[a_n,a_{n+1})$ – Joe Mar 30 '21 at 21:44
  • The definition holds, as long as the intervals are disjoint, which is the case in my definition. so in any case $f(x)=1/2$ for $[a_2, a_4)$ and f(x)=1/4 for $[a_4, a_3)$ and $f(x)=1/3$ for $[a_3, a_5)$. – user48672 Mar 30 '21 at 22:38
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    Then I think you should start your answer by saying “If there is an order preserving bijection between ${a_n}$ and $\mathbb{N}$, then define $f(x)=$...”, since otherwise you cannot definite the intervals $[a_n, a_{n+1})$ to be disjoint. – Joe Mar 30 '21 at 22:53