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

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 $\mathfrak{c}.$
