Set of Polynomials over Infinite Set has Same Cardinality

Theorem

Let $S$ be a set of infinite cardinality $\kappa$.

Let $S \sqbrk x$ be the set of polynomial forms over $S$ in the indeterminate $x$.

Then $S \sqbrk x$ has cardinality $\kappa$.

Proof

Since $S \sqbrk x$ contains a copy of $S$ as constant polynomials, we have an injection $S \to S \sqbrk x$.

We define an injection from $S \sqbrk x$ to the set $\FF$ of finite sequences over $S$ as follows:

Each polynomial in $f \in S \sqbrk x$ is of the form:

$f = a_0 + a_1 x + a_2 x^2 + \dotsb + a_n x^n$

where $a_n$ is non-zero and each $a_i$ is in $S$.

We send each polynomial $f$ to the sequence of its coefficients $\sequence {a_0, \dotsc, a_n}$.

By the definition of equality of polynomials, this is injective.

Now the set of finite sequences over $S$ is a countable union of sets of cardinality $\kappa$.

From Cardinality of Infinite Union of Infinite Sets, $\FF$ has cardinality $\kappa$.

Therefore there is a bijection $\FF \leftrightarrow S$.

Composing this with the injection $S \sqbrk x \to \FF$, we have an injection $S \sqbrk x \to S$.

So by the Cantor-Bernstein-SchrÃ¶der Theorem there is a bijection $S \sqbrk x \leftrightarrow S$.

Hence, we have:

$\card {S \sqbrk x} = \card S = \kappa$

$\blacksquare$