Prove that the set of $\mathcal{L}$-terms has size $\max\{\aleph_0, \#\text{ of constant sym plus the $\#$ function sym}\}$.

Question

Let $\mathcal{L}$ be a first-order language. Prove that the set of $\mathcal{L}$-terms has size $\max\{\aleph_0,\text{ the number of constant symbols plus the number of function symbols}\}$.

I know that the set of $\mathcal{L}$-terms is at least $\aleph_0$ because there are countably many variables. I also know that, if the number of constants and function symbols is at most countable then there is at most countable many terms.

I am not sure how to deal with the following case: suppose there are $\beta$ many function and constant symbols where $\beta$ is an uncountable ordinal. Would this mean that there would be at least $\aleph_0^\beta$ many terms? I believe $\aleph_0^\beta$ is greater than $\beta$, so I must be getting something wrong.

Any suggestions would be appreciated.

Answer 1

Your exponentiation is all wrong. You need to think in terms of cardinals, not ordinals, and recall that terms are finite strings from the alphabet formed under certain rules.

HINT: Recall that for every infinite set $X$, we have $|X|=|X\times X|$, and therefore $|X|=|\bigcup_{n\in\Bbb N}X^n|$.

Answer 2

Hint: each term has only a finite length. For any $n$, $\lvert X^n\rvert\leq \lvert X\rvert + \aleph_0$.