Math Help - dimension

1. dimension

I have a homework problem that I can't figure out...

Is the space of all real valued continuous functions defined on the interval [0,1] finite dimensional?

2. Originally Posted by JackieJo
I have a homework problem that I can't figure out...

Is the space of all real valued continuous functions defined on the interval [0,1] finite dimensional?

No. Because the space of all real valued continous functions on $[0,1]$ contains the space of all real polynomials over $[0,1]$. Can you see why it makes it infinite dimensional?

3. because it has infinite linearly dependent elements? is that right?

4. Originally Posted by JackieJo
because it has infinite linearly dependent elements? is that right?
You need to be more formal. If $\mathcal{C}[0,1]$ (continous functions) were to be a finite dimensionsal vector space then every subspace would be finite dimensional also, with a smaller degree (this is a theorem). But then $\mathcal{P}[0,1]$ (polynomials functions) would be finite dimensional. But that is impossible. Why?

5. because P is a subspace of C and P is infinite so C has to also be infinite dimensional? Thanks by the way

6. Originally Posted by JackieJo
because P is a subspace of C and P is infinite so C has to also be infinite dimensional? Thanks by the way
Again, why is P[0,1] infinite? You need to be more formal. Here is how you show it cannot have a finite basis. Suppose that P[0,1] has a finite basis $\{ p_1(x),...,p_n(x)\}$ where $p_i(x)$ are polynomials. Then it would mean any polynomial in P[0,1] can be expressed in the form $a_1p_1(x)+...+a_np_n(x)$. Let $f(x)$ be a polynomial with a larger degree then all of the $p_i(x)$ polynomials. Then $f(x)=a_1p_1(x)+...+a_np_n(x)$ is impossible because polynomials can only be equal if their have the same coefficients and the same degree, but it is impossible since the degree of f(x) exceedes the RHS.

7. Thanks!

I think I basically understand, I just didn't know how to represent it....so that's what I needed, thanks!