How to prove that $max(\aleph_{0}, card(X)) = max(\aleph_{0}, card(L(X)))$?

Question

I struggle with the following problem. Let $X$ be a set of elementary sentences and $L(X)$ be the smallest elementary language in which we can express all the sentences from $X$. How to prove that $max(\aleph_{0}, card(X)) = max(\aleph_{0}, card(L(X)))$?

I’m aware that for any elementary language L, $card(L) = card(Form(L)) = card(Sent(L)) = max(\aleph_{0},card(L))$, but I have no idea how to use it in a sensible way in establishing the desired equality.

Thanks for help!

Answer 1

$L(X)$ is the first-order language generated by the symbols appearing in sentences in $X$. Let's denote this set of symbols by $\Sigma$. Then $|L(X)| = \max(\aleph_0,|\Sigma|)$.

Now each first-order sentence contains only finitely many symbols, so $|\Sigma| \leq |X|\cdot \aleph_0 = \max(\aleph_0,|X|)$. We have $$|L(X)| = \max(\aleph_0,|\Sigma|) \leq \max(\aleph_0,\max(\aleph_0,|X|)) = \max(\aleph_0,|X|).$$

Also, any first-order language is infinite, so $\max(\aleph_0,|L(X)|) = |L(X)| \leq \max(\aleph_0,|X|)$.

For the other direction, $X\subseteq L(X)$, so $|X|\leq |L(X)|$, and hence $\max(\aleph_0,|X|) \leq \max(\aleph_0,|L(X)|)$.