# Proof that $R^{[1,4]}$ is infinite dimensional. Is it correct?

Let $$R^{[1,4]}$$ denote the set of all function $$f:[1,4] \to R$$. I was attempting to prove that $$R^{[1,4]}$$ is infinite dimensional using the following theorem. Can someone check if the proof is correct?

Theorem 1: A vector space $$V$$ is infinite dimensional if and only if there is a sequence of vectors $$v_1,v_2,. . .$$ such that for all $$m \in Z^+$$, $$v_1,. . .,v_m$$ is linearly independent.

The following is my attempt at proving that $$R^{[1,4]}$$ is infinite dimensional.

Proof: Let $$f_1,. . .,f_n$$ be a list of vectors in $$R^{[1,4]}$$ such that each function is defined as follows. $$f_j(x) = x^{j+1}$$.

Then the list $$f_1,. . .,f_n$$ is linearly independent. To see this, let $$a_1,. . .,a_n \in R$$. It suffices to show that $$f_j$$ is not in the span of the previous vectors $$f_1,. . .,f_{j-1}$$. Because $$x \in [1,4]$$ and is not a fixed number, $$a_1x^2 +. . .+a_{j-1}x^j \ne x^{j+1}$$. Every time we would have to pick a different set of scalars to make that equation true. And there is no single choice of scalars that makes this equation true.

Hence, we can have a linearly independent list of arbitrary length. By Theorem 1 we conclude that $$R^{[1,4]}$$ is infinite dimensional.

Is this proof correct?

• Why not invoke the fundamental theorem of algebra? If $a_1f_1+\dots+a_jf_j-f_{j+1}=0$, then all the coefficients are zero. Contradiction. Commented Jul 9, 2022 at 8:51
• Actually you don't need FTA, just the factor theorem. I added it as an answer. Commented Jul 9, 2022 at 9:01

Writing “Every time we would have to pick a different set of scalars to make that equation true. And there is no single choice of scalars that makes this equation true.” doesn't seem to be a good justification.

You can start from the equality$$x^{j+1}=a_1x^2+\cdots+a_{j-1}x^j$$and differentiate both sides $$j+1$$ times. The LHS becomes a non-zero number, whereas the RHS becomes $$0$$.

• Thanks a lot. That’s a great solution! Yeah I was feeling that those sentences don’t justify the argument. Commented Jul 9, 2022 at 8:50

By the factor theorem, often included as part of the fundamental theorem of algebra, an $$n$$-th degree polynomial can have at most $$n$$ roots.

Since your polynomial would have infinitely many roots, all the coefficients have to be zero.

• That’s also a great solution. Thanks a lot. Commented Jul 9, 2022 at 9:01
• Thank you. You got most of it. Commented Jul 9, 2022 at 9:02

Yes. $$U=\Bbb{R}^{[0,1]}$$ is an infinite dimensional vector space over $$\Bbb{R}$$. Infact $$\dim(\Bbb{R}^{[0, 1]})\ge \mathfrak{c}$$.

For a vector space $$V$$ over $$F$$ , $$|V|=|F|^{\dim(V) }$$

$$|\Bbb{N}|=\mathfrak{a}$$

$$|[0,1]|=|\Bbb{R}|=2^{\mathfrak{a}}=\mathfrak{c}$$

$$|\Bbb{R}^{[0,1]}|=\mathfrak{c}^{\mathfrak{c}}=(2^{\mathfrak{a}})^{\mathfrak{c}}=2^{\mathfrak{c}}$$

From $$|\Bbb{R}^{[0,1]}|=|\Bbb{R}|^{\dim(U)}$$ , we have

$$2^{\mathfrak{c}}=\mathfrak{c}^{\dim(U) }$$

This shows that $$\dim(U) \ge \mathfrak{c}$$

Hence $$\dim(U) =\dim(\Bbb{R}^{[0,1]})$$ is uncountable.

Note: $$|A|:=$$ cardinality of $$A$$

• Infact $\dim(\Bbb{R}^{[0, 1]}) =\mathfrak{c}$. Commented Jul 9, 2022 at 9:58