Producing infinite family of transcendental numbers

Question

Weierstrass proved the result [Lindemann-Weierstrass theorem] that if $a_1, \cdots, a_n$ are reals linearly independent over the rationals, then $e^{a_1}, \cdots, e^{a_n}$ are algebraically independent.

I would like to know if the result holds for infinitely many numbers. Explicitely, if $\{a_1, a_2, \cdots \}$ is an infinite family of real numbers such that every finite subset is linearly independent over $\mathbb Q$, then is it true that every finite subset of $\{ e^{a_1}, e^{a_2}, \cdots \}$ is algebraically independent over $\mathbb Q$?

I'd be happy to know any other result in that spirit.

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Edit: Can one explicitly write an infinite family of real numbers linearly independent over the rationals?

(Sorry, this was the question I had originally in mind. Thanks anon for pointing that out).

Answer 1

If $\Gamma$ is a finite subset of $\{e^{a_1},e^{a_2}\cdots\}$, then $\Lambda=\{\log\gamma : \gamma\in\Gamma\}$ is a finite subset of $\{a_1,a_2,\cdots\}$. The former is thus linearly independent over $\mathbb{Q}$ by hypothesis, and by the LW theorem you cite this implies that $\{e^{\lambda}:\lambda\in\Lambda\}=\Gamma$ is algebraically independent over $\mathbb{Q}$. Does this answer your question?

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Update in response to OPs edit: Yes, one can explicitly construct an example. Via Dubuque:

$$\rm \Lambda=\{\, \log p: ~ p ~~ prime \,\} $$

is an example of an infinite set linearly independent over the rationals. The fundamental theorem of arithmetic is key to understanding why this is independent. However, the $\Gamma$ associated to this $\Lambda$ (defined as before), i.e. the primes, are not even linearly independent, let alone algebraically, over the rationals. Is your question actually for an explicit $\mathbb{Q}$-lin. ind. $\Lambda$ with $\mathbb{Q}$-alg. ind. $\Gamma$?