Dimensions and isomorphisms I am convinced that there is no isomorphism $\theta: \mathbb R^n\to\{\text{polynomials of degree}\leq n\}$ because $n+1$ numbers are required to represent the latter set whereas the former only has $n$. But how does one prove that no isomorphism can exist? (i.e. how can I turn this instinct into a mathematical argument)? Could I just say that the dimension of the latter is different from that of the former? But how is dimension defined for a set of polynomials? Thanks.
 A: The polynomials of degree $\le n$ form a vector space. The dimension of a vector space can be defined as the cardinality of a basis of the vector space (which is the same for every basis). In this case there's a particularly simple basis, consisting of the monomials. There are $n+1$ of these, and it's clear that you can't write any of them as a linear combination of the others and that you can write any polynomial as a linear combination of them. Thus the vector space of polynomials of degree $\le n$ has dimension $n+1$, and therefore can't be isomorphic to $\mathbb R^n$, which has dimension $n$.
A: Surely you are speaking of linear isomorphisms. In that case, as you already noticed, dimension is the key: two isomorphic vector spaces have the same dimension, thus $\mathbb{R}^n$ and $\{\text{polynomials of degree }\le n\}$ cannot be isomorphic because the first has dimension $n$ and the second $n+1$. To show this last assertion just observe that $\{1, x, \ldots x^n\}$ is a basis of the second space and it contains $n+1$ elements.
