Suppose \(f \colon \R^2 \to \R\) is a polynomial. That is,

\begin{equation*}
f(x,y) =
\sum_{j=0}^d
\sum_{k=0}^{d-j}
a_{jk}\,x^jy^k =
a_{0\,0} + a_{1\,0} \, x +
a_{0\,1} \, y+
a_{2\,0} \, x^2+
a_{1\,1} \, xy+
a_{0\,2} \, y^2+ \cdots +
a_{0\,d} \, y^d ,
\end{equation*}

for some

\(d \in \N\) (the degree) and

\(a_{jk} \in \R\text{.}\) We claim

\(f\) is continuous. Let

\(\bigl\{ (x_n,y_n) \bigr\}_{n=1}^\infty\) be a sequence in

\(\R^2\) that converges to

\((x,y) \in \R^2\text{.}\) We proved that this means

\(\lim_{n\to\infty} x_n = x\) and

\(\lim_{n\to\infty} y_n = y\text{.}\) By

Proposition 2.2.5,

\begin{equation*}
\lim_{n\to\infty}
f(x_n,y_n) =
\lim_{n\to\infty}
\sum_{j=0}^d
\sum_{k=0}^{d-j}
a_{jk} \, x_n^jy_n^k
=
\sum_{j=0}^d
\sum_{k=0}^{d-j}
a_{jk} \, x^jy^k
=
f(x,y) .
\end{equation*}

So \(f\) is continuous at \((x,y)\text{,}\) and as \((x,y)\) was arbitrary \(f\) is continuous everywhere. Similarly, a polynomial in \(n\) variables is continuous.