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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?

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  • $\begingroup$ 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. $\endgroup$
    – i can try
    Commented Jul 9, 2022 at 8:51
  • $\begingroup$ Actually you don't need FTA, just the factor theorem. I added it as an answer. $\endgroup$
    – i can try
    Commented Jul 9, 2022 at 9:01

3 Answers 3

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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 $0$.

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

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  • $\begingroup$ That’s also a great solution. Thanks a lot. $\endgroup$
    – Seeker
    Commented Jul 9, 2022 at 9:01
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    $\begingroup$ Thank you. You got most of it. $\endgroup$
    – i can try
    Commented Jul 9, 2022 at 9:02
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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 $A$

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

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