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Former math grad student, now a lawyer for the last $28$ years. Just doing math for fun in my spare time.

I was browsing questions here and my mind went on a tangent. The following questions occurred to me, which I strongly suspect are related:

Let $S = \{x_n \mid n \in \Bbb N$ }$ \subseteq \Bbb R$ be a set with no accumulation points. (A subset of $\Bbb R$ with no accumulation points has to be countable, because you can partition $\Bbb R$ into countably many intervals as fine as you like, and if $S$ is not countable, the pigeonhole principle assures us that one of those intervals has uncountably many elements of $S$. Consider that interval and repeat the process countably many times with intervals having diameters that go to $0$, and you'll end up with an accumulation point of $S$.) Let $f:S \to \Bbb R$ be arbitrary. Is it always possible to extend $f$ to a $C^\infty$ function on $\Bbb R$? To an analytic function on $\Bbb R$?

If instead $S \subseteq \Bbb C$ (where $S$ has no accumulation points) and $f:S \to \Bbb C$ is arbitrary, is it always possible to extend $f$ to an analytic function on $\Bbb C$?

When I was in grad school, I focused on algebra and mathematical logic so I really have little idea how one would approach this question.

Robert Shore
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  • Yes. As explained in the linked Q&As, for a given sequence $(z_n)$ of complex numbers without accumulation point and an arbitrary sequence $(w_n)$ of complex numbers there is an entire function $f$ satisfying $f(z_n) = w_n$ for all $n$. – Martin R Aug 26 '21 at 20:42
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    this follows from Weierstrass and Mittag Leffler theorems for example (giving always an entire function $f$ that extends the given $f$, while in the case when $x_n, f(x_n)$ are real, $f$ has real Taylor coefficients at zero so it restricts to an analytic function on the reals with real image too); by Weierstrass one constructs an entire nonzero function with $g(x_n)=0, g'(x_n) \ne 0$ and then one uses Mittag Leffler to get a meromorphic function $h$ with principal part $f(x_n)/(z-x_n)$ at each $x_n$ which are the poles; $f=gh$ will do and both constructions are conjugate invariant – Conrad Aug 26 '21 at 20:43
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    even better one can specify arbitrary values not only for $f(x_n)$ but for any finite number of derivatives at each $x_n$ - number of derivatives which may depend on $n$ also - and get an $f$ like that and similarly the construction is conjugation invariant so if everything is real, that $f$ will send the reals into the reals – Conrad Aug 26 '21 at 20:48
  • In line of your question I would like to mention the Weierestrass factorization theorem. Its n a particular complex analytic case https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem – Ali Taghavi Jun 28 '24 at 01:43
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    Your enthusiasm on mathematics is very admirable – Ali Taghavi Jun 28 '24 at 01:47

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Since i am a bit strapped for time here is a solution for the smooth case which does not work for analytic functions (but verbatim for subsets of $\mathbb{R}^d$): First note that $S$ is discrete and we can find for every element $x$ in $S$ a small ball around $x$ which contains no other element of $S$. Adjusting choices we can arrange that no two of these balls intersect.

To see this we use that in every closed Ball around $0$ of Radius $n$ there are only finitely many Elements of $S$ (by your argument). So in particular we can separate every element in $B_n$ from each other as described such that the boundaries of the Balls have positive distance from the boundary of $B_n$. Continuing for every $n$ we can thus separate all points as described.

Now it is well known that every ball in the reals admits a smooth bump function, i.e. a smooth function $b$ which takes the value 1 at a Point (for our balls we choose the unique $x\in S$ as this point) and which is uniformly 0 outside of the ball. Pick for every ball such a bump function and multiply it with the function value of $f$ at the point $x$ which is contained in the ball. Summing up all These functions you obtain a smooth function taking the desired values, i.e. extending $f$.

Note: Due to the identity Theorem for analytic functions there are no analytic bump functions.

Alexander Schmeding
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