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Theorem #. Let $A \subset \Bbb{R}$ be a subset of $\Bbb{R}$ whose interior $U = \operatorname{int}(A)$ is not bounded above, i.e., $\sup U = \infty$. Also assume that $(c_n)$ is an increasing sequence of real numbers such that $c_n \to \infty$ and $c_{n+1} - c_{n} \to 0$ as $n \to \infty$. Then the set

$$ D = \{ x \in \Bbb{R} : x + c_{n} \in A \text{ for infinitely many } n \} $$

is dense in $\Bbb{R}$.

Proof. It is enough to prove that for any open interval $I$, the intersection $I \cap D$ is non-empty. Indeed, let $I = (a, b)$ be an open interval, $N$ be a positive integer and consider the open set

$$ V = \bigcup_{n=N+1}^{\infty} (c_n + I) = \bigcup_{n=N+1}^{\infty} (c_n + a, c_n + b). $$

By the conditions on $(c_n)$, we find that $V$ contains an open interval of the form $(r, \infty)$. In particular, $V \cap U$ is non-empty. That is, there exists $n > N$ such that $(c_{n} + a, c_{n} + b) \cap U$ is non-empty. Choose an open interval contained in this intersection and denote it as

$$(c_{n} + a', c_{n} + b')$$

It is clear that we may choose $a'$ and $b'$ in such a way that in fact we have

$$ [c_{n}+a', c_{n}+b'] \subset (c_{n} + a, c_{n} + b) \cap U. $$

Summarizing, for any open interval $I$ and a positive integer $N$, we found an open subinterval $I'$ and $n > N$ such that $\operatorname{cl}(c_{n} + I') \subset (c_{n} + I) \cap A$.

Now let $I_1$ be any open interval, and repeatedly apply the previous result to obtain a sequence $(I_k, n_k)$ of interval-integer pairs such that

$$ I_1 \supset \overline{I_2} \supset I_2 \supset \overline{I_3} \supset I_3 \supset \cdots $$

and $c_{n_k} + I_k \subset A$. Then by the nested interval theorem, the intersection

$$ K = \bigcap_{k=1}^{\infty} I_k $$

is non-empty. Let $x \in K$. Then $x \in I_k$ for all $k$ and thus we have $x + c_{n_k} \in A$ for all $k$. Therefore $x \in D$ and $I \cap D \neq \varnothing$, proving that $D$ is dense in $\Bbb{R}$.

A function $f : \Bbb{R} \to \Bbb{R}$ is called one-sided continuous if for each $x \in \Bbb{R}$, either $f$ is right-continuous at $x$ or is left-continuous at $x$.

Corollary #. (Croft's Lemma) Let $(c_n)$ be as in Theorem #label_thm_main, and $f : \Bbb{R} \to \Bbb{R}$ be a one-sided continuous function such that

$$ \lim_{n\to\infty} f(x + c_n) = \ell, \quad \forall x \in \Bbb{R}. $$

Then we also have

$$ \lim_{x\to\infty} f(x) = \ell. $$

Proof. If the statement fails to hold, then there exists $\epsilon > 0$ such that the set

$$A = \{ x \in \Bbb{R} : |f(x) - \ell | > \epsilon \}$$

is unbounded above. Then it is easy to find that $\operatorname{int}(A)$ is also unbounded above, and by the Theorem #label_thm_main, and we obtain a contradiction.

Corollary #. Let $(a_n)$ be a sequence of real numbers such that $a_{[r^n]} \to \ell$ as $n\to\infty$ for any $r > 1$. Then $a_n \to \ell$ as $n \to \infty$.

Proof. Let $f(x) = a_{[\exp(\exp(x))]}$. Clearly $f$ is a one-sided continuous function. Then the condition in the problem amounts to say that $f(x + \log n) \to \alpha$ for any $x \in \Bbb{R}$. Since $c_n = \log n$ satisfies the condition of the theorem, we obtain the conclusion.