Number of Polynomials and continuous functions on $\mathbb R$

Question

The question is what is the cardinalyty of

$1)$ $\mathcal P=$The set of all polynomials with real co-efficient and

$2)$ $\mathcal R=$The set of all real valued continuous functions

Now a polynomial's standard form is $a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+....+a_0$ where $x$ is the variable and $a_i\in \mathbb R \forall i.$

So for a fixed $n$, the cardinality of such polynomials is $|\mathbb R^n|=c.$

The set of all polynomials is bijective with the set $\bigcup_{n=0}^{\infty}\mathbb R^n.$ Hence the cardinality is $$\left|\bigcup_{n=0}^{\infty}\mathbb R^n\right|\\=\sum_{n=1}^{\infty}1.c\\=|\mathbb N|\cdot c\\=c$$

Am I correct here.

But every continuous function can be wrtten as the limit of a sequence of polynomials. So, $\mathcal R=\bar {\mathcal P}$ So we can say that $$\mathcal R\supset \mathcal P\\\implies |\mathcal R|\ge |\mathcal P|=c.$$

Now what do I do to find an equality$?$

Thanks.

Answer 1

Consider the set $A$ of polynomials with rational coefficients. This is a countable set.

A continuous function $f:[0,1]\to \mathbb R$ is the uniform limit of polynomials with rational coefficients by the Stone-Weierstrass Theorem. Thus, such a function can be identified with a sequence $\{p_n\}_{n\in \mathbb N} \subset A$ such that $\|p_n-f\|_\infty<1/n$. Hence, $f$ can be identified with a subset of $A$. In order to show that the identification is injective we have to show that we can recover $f$ uniquely from a subset $\{q_i:i\in \mathbb N\}$ of $A$, where $q_i$ is in $1-1$ correspondence with some sequence $p_n$ converging to $f$ as above. For each $n_0\in \mathbb N$, if $i_0\in \mathbb N$ is sufficiently large, we have $\{1,\dots,i_0\} \supset \{1,\dots,n_0\}$, and therefore $\|p_n-f\|_{\infty} <1/n_0$ for all $n\geq n_0$ and $\|q_i-f\|_{\infty}<1/n_0$ for all $i> i_0$. Thus $q_i$ converges uniformly to $f$.

Thus $f$ can be identified with an element of $\mathcal P(A)$, and this shows that $C([0,1])$ has cardinality at most $c=2^{\mathbb N}$.

It is not hard to generalize this to continuous functions $f:\mathbb R\to \mathbb R$, e.g. you can use a sequence $p_n$ of polynomials such that $p_n$ approximates $f$ very well in the interval $[-n,n]$.

Answer 2

Yes, you are correct that the set $\mathcal{P}$ of polynomials with real coefficients has cardinality $\mathfrak{c}.$

The set $\mathcal{R}$ of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$ also has cardinality $\mathfrak{c}.$ One easy way to see this is to observe that any continuous function from $\mathbb{R}$ to $\mathbb{R}$ is uniquely determined by its values on the set $\mathbb{Q}$ of rational numbers (this is true since $\mathbb{Q}$ is dense in $\mathbb{R}).$ It follows that the function $i\colon\mathcal{R}\to {}^\mathbb{Q}\mathbb{R}$ defined by setting $i(f)$ equal to the restriction of $f$ to $\mathbb{Q}$ is an injection. $({}^\mathbb{Q}\mathbb{R}$ here denotes the set of all functions from $\mathbb{Q}$ to $\mathbb{R}.)$

But $\mathbb{Q}$ is countable, so the cardinality of ${}^\mathbb{Q}\mathbb{R}$ equals the cardinality of ${}^\mathbb{N}\mathbb{R},$ which is $\mathfrak{c}.$ It follows that the cardinality of $\mathcal{R}$ is less than or equal to $\mathfrak{c}.$

Now $\mathcal{P}\subseteq\mathcal{R},$ and you already showed that the cardinality of $\mathcal{P}$ is $\mathfrak{c},$ so we have that the cardinality of $\mathcal{R}$ is greater than or equal to $\mathfrak{c}.$

We conclude that the cardinality of $\mathcal{R}$ is equal to $\mathfrak{c}.$

Answer 3

Establish the following facts:

1. $\mathcal R$ and $\mathcal P$ have at least the cardinality of continuum. (Consider the constants)

2. Any continuous function on $\mathbb R$ is completely determined by its values on the rational numbers, and hence can have at most the cardinality of $\mathfrak c$.

3. $\mathcal P \subseteq \mathcal R$