# Proof that vector space $C([a,b])$ of all functions from $[a,b]\to \mathbb{R}$ is infinite dimensional?

Let me first express how I understand an infinite dimensional vector space.

A vector space V is infinite dimensional if $\forall n\in \mathbb{N} (\{e_1,\dots , e_n\} \text{ linearly independent }\Rightarrow \{e_1 , \dots , e_{n+1}\}\text{ linearly independent })$.

Having the above definition in mind, I try to prove that the vector space $C([a,b])$ of all functions from $[a,b]\to \mathbb{R}$ is infinite dimensional, with basis the set of polynomials.

Attempt at a proof: Let $n\in \mathbb{N}$ be arbitrary, and assume that $\{x , x^2 , \dots , x^n\}$ is linearly independent. My goal is to show that $\{x,x^2,\dots , x^{n+1}\}$ is linearly independent. To do that, assume that $\sum_{k=1}^{n+1}b_kx^k=0$, and now I need to show that for all $k$, $b_k=0$. My feeling is that this should proceed by contradiction, namely I should assume that there exists $k$ such that $b_k\neq 0$...

I am missing the step from here on, how do I proceed. Also is the definition of infinite dimensionality correct? Is this how one proves that a vector space is infinite dimensional?

• The polynomials do not generate every function in your vector space. Jan 10 '18 at 16:21
• The problem is with the interpretation of the definition. $\text{If we know}$ (not assumed) that $\{v_1,...,v_n\}$ is l.i set, we $\text{can obtain}$ $v_{n+1}$ such that $\{v_1,...,v_n,v_{n+1}\}$ is l.i set Jan 10 '18 at 16:25
• If $V$ is just a vector space there's no such thing as "$e_n$" in $V$. Jan 10 '18 at 16:49
• The question as stated doesn't make sense.
– user370967
Jan 10 '18 at 17:29
• You can but you didn't. If you'd said what you were doing fine. But you didn't say anything about what the notation meant. Jan 11 '18 at 13:42

Your supposed definition of infinite dimensional vector space doesn't make sense. A vector space $V$ is infinite dimensional if no finite subset of $V$ generates $V$. This is equivalent to the assertion that $V$ has an infinite subset which is linearly independent.

The space $\mathcal{C}\bigl([a,b]\bigr)$ is infinite dimensional because the set $\{1,x,x^2,x^3,\ldots\}$ is linearly independent.

• I am puzzled by your answer. How do you know the set $X=\mathcal{C}([a,b])$ is not uncountably infinite instead of countably infinite as you assumed? Also how one show that the set $\{1, x, x^2, x^3, ...\}$ spans $X$? Further, shouldn't we be talking about $\mathcal{C}^0([a, b])$, instead of $\mathcal{C}^1((a,b))$. In either case, the basis elements proposed belong all to $\mathcal{C}^\infty((a,b))$. Sep 23 '18 at 3:10
• I am puzzled by your answer. How do you know the set $X=\mathcal{C}([a,b])$ is not uncountably infinite instead of countably infinite as you assumed? Also how could one show that the set $\{1, x, x^2, x^3, ...\}$ spans $X$? Further, shouldn't we be talking about $\mathcal{C}^0([a, b])$ (the set of continuous functions $f:I\rightarrow\Bbb R$, instead of $\mathcal{C}^1((a,b))$, note the open set. In either case, the basis elements proposed belong all to $\mathcal{C}^\infty((a,b))$. Or am I getting the notation wrong? (For some reason I couldn't edit my previous comment) Sep 23 '18 at 3:18
• @minimax All I wrote was that the set $\{1,x,x^2,x^3,\ldots\}$ is linearly independent. I never claimed that it spans the whole space (which is good, since it doesn't). Sep 23 '18 at 7:05

First of all you forgot to include $1$ in your set $\{x , x^2 , \dots , x^n\}$.

Also the assumption that $\{1, x , x^2 , \dots , x^n\}$. is linearly independent is not necessary to show that to show that $\{1, x , x^2 , \dots , x^{n+1}\}$ is linearly independent.

Let $C_1+C_2x+...+C_{n+2}x^{n+1}=0$.

Plog $x=0$ and you get $C_1=0$. Now differentiate and evaluate at $x=0$ and you will get $C_2=0$.

Continue and you will get all the coefficients equal $0$

Therefore you have an infinite set {${1, x , x^2 , ... , x^n, ...}$} of linearly independent continuous functions.

• "Plog $x=0$ and you get $C_{1}=0$. Now differentiate and evaluate at $x=0$ and you will get $C_{2}=0$ " What do you mean? It is correct if I consider $f (x) = \lbrace 1, x, x^{2}, x^{3}, ... \rbrace$, then as $f (x)$ is continuous when $x \rightarrow 0$, do I get $C_ {2} = 0$ ? @Mohammad Riazi-Kermani Feb 25 at 14:27

A vector space is infinite-dimensional if there does not exist a finite set that spans it -- is generally taken as the definition. More precisely, one defines a vector space to be finite-dimensional if there is a finite set that spans it and then an infinite-dimensional space is then merely defined to be a space that is not finite-dimensional.

EDIT- In response to your comment, the span of a set of vectors is the set of all linear combinations of those vectors. So, for instance the vector space of complex numbers over the field of real numbers is finite-dimensional because it is spanned by the finite set $$\{1,i\}$$

• Care to elaborate on the notion of span? Does it mean that the function $L_e :\mathbb{R}^n \to V$: $x \mapsto \sum_i x_i e_i$ is surjective? (for a finite dimensional vector space)
– EEEB
Jan 10 '18 at 16:31
• @EEEB Sure. See my edit. Jan 10 '18 at 16:37
• @EEEB Also, your basis will not always be countable and so you will not always be able to enumerate your basis vectors that way Jan 10 '18 at 17:02

Your definition at the moment doesn't quite make sense, since you haven't specified what the $e_n$'s are. It's also incorrect - what happens if $e_1 = 0$? Since there's quite a few different equivalent definitions, I will try and state several here.

To clarify notation, fix a vector space $V.$ We say a finite subset $F = \{e_1,\dots,e_n\}\subset V$ is linearly independent if for any choice of scalars $\lambda_1,\dots,\lambda_n,$ we have,

$$\sum_{i=1}^n \lambda_ie_i = 0 \quad \iff \quad \lambda_1=\dots=\lambda_n = 0.$$

A general subset $F \subset V$ is linearly independent if every finite subset $F_0 \subset F$ is linearly independent. This is equivalent to the assertion that for all $n \in \mathbb N,$ $x_1,\dots, x_n \in F$ and $\lambda_1,\dots,\lambda_n$ scalars, $\sum_{i=1}^n \lambda_ie_i = 0$ if and only if each $\lambda_i=0.$

Also a finite subset $F = \{e_1,\dots,e_n\} \subset V$ spans $V$ if for all $x \in V,$ there are scalars $\lambda_1,\dots,\lambda_n$ such that $x = \sum_{i=1}^n \lambda_ie_i.$ Similarly a general subset $F \subset V$ spans $V$ if for all $x \in V,$ there is $n \in \mathbb N,$ $x_1,\dots,x_n \in F$ and scalars $\lambda_1,\dots,\lambda_n$ such that $x = \sum_{i=1}^n\lambda_ix_i.$

Theorem The following are equivalent.

1. There exists an infinite subset $F \subset V$ such that $F$ is linearly independent.

2. There exists a countably infinite subset $F = \{e_n\}_{n \in \mathbb N} \subset V$ such that $F$ is linearly independent.

3. There exists a countably infinite subset $F = \{e_n\}_{n \in \mathbb N} \subset V$ such that for all $n \in \mathbb N,$ the subset $F_n = \{e_1,\dots,e_n\}$ is linearly independent.

4. There exists a countably infinite subset $F = \{e_n\}_{n \in \mathbb N} \subset V$ such that $e_1 \neq 0$ and for all $n \in \mathbb N$ (starting at $n=1$), $$\{e_1,\dots,e_n\} \text { is linearly independent } \implies \{e_1,\dots,e_{n+1}\} \text{ is linearly independent.}$$

5. There does not exist a finite subset $F \subset V$ spanning $V.$

6. (*) There exists a infinite subset $F \subset V$ which is both linearly independent and spanning.

If any of the above equivalent conditions holds, we say $V$ is infinite dimensional.

(*) Strictly speaking, the equivalence with (6) is not only non-trivial, but is equivalent the axiom of choice. It is commonly used however since it allows one to define the dimension of $V$ to be the cardinality of the basis $F,$ but I digress... Also on the note of choice, I will be using countable choice freely.

Proof: We will first show the series of implications,

$$(\mathrm{iv}) \implies (\mathrm{iii}) \implies (\mathrm{ii}) \implies (\mathrm{i}) \implies (\mathrm{iv}).$$

For $(\mathrm{iv}) \implies (\mathrm{iii})$ this is simply the principle of mathematical induction. We note $\{e_1\}$ is linearly independent provided $e_1 \neq 0,$ which establishes the base case.

For $(\mathrm{iii})\implies(\mathrm{ii}),$ we note that if $F \subset{e_n}_{n \in \mathbb N}$ is finite, then $n = \max F$ exists. So $F \subset \{e_1,\dots,e_n\}$ from which the implication follows. The implication $(\mathrm{ii})\implies (\mathrm{i})$ is trivial since $\{e_n\}_{n \in \mathbb N}$ is an infinite linearly independent subset. The final implication $(\mathrm{i}) \implies (\mathrm{iv})$ holds by noting said infinite linearly independent subset $F$ contains a countable subset $\{e_n\}_{n \in \mathbb N} \subset F.$ This is automatically linearly independent, so we can verify $(\mathrm{iv}).$

Finally $(\mathrm{i}) \iff (\mathrm{v})$ is equivalent to showing $\neg(\mathrm{i}) \iff \neg(\mathrm{v}),$ where no infinite linearly independent set exists if and only if a finite spanning set exists. This will require some results from finite dimensional linear algebra, which I will assume. The latter implication will also need a bit of choice also.

If a finite spanning set $F$ exists, then we know any set $G \subset V$ with $|G| > |F|$ (set cardinality) must be linearly dependent. This gives one direction.

For the converse, assume that any infinite subset $F \subset V$ is linearly independent. The strategy will be to start with a linearly independent set inductively add linearly independent vectors to our set until it spans the space. Indeed we will fix $x_1 \in V\setminus\{0\}$ and set $F_1 = \{x_1\}.$ Given $F_n = \{x_1,\dots,x_n\}$ is linearly independent, if $\mathrm{span}\ F_n \neq V$ then there exists $x_{n+1} \in V$ such that $F_{n+1} = \{x_1,\dots,x_{n+1}\}$ is linearly independent. So we continue inductively (note this uses dependent choice). If this process doesn't terminate, we will get an infinite subset $\{x_1,x_2,\dots\}$ which is linearly independent, contradicting our original assumption. So the process eventually terminates to give a finite spanning set.

For reasons mentioned above, I will skip the equivalence with $(\mathrm{vi}).$

For your specific problem I'll refer you to Mohammad Riazi-Kermani's answer, which gives an inductive proof (so it's an example of using $(\mathrm{iv})$).

• I feel like the equivalence of (i) and (v) is sort of the main point here - it would be worth elaborating on how that proof works (especially since I don't see how the proof of (v) implies (i) follows from finite dimensional linear algebra). You should also be careful about your claims regarding the axiom of choice here - note hat (i)$\Leftrightarrow$(vi) is equivalent to the axiom of choice, but you still use AC to show the equivalence of (i) implies (iv) (since you take a countable subset of an arbitrary infinite set) and the argument I would use for (v) implies (i) uses Zorn's lemma. Jan 10 '18 at 17:59
• @MiloBrandt Fair point. The reason I focused on the independence results was to clear confusion that the OP seemed to have, so the latter two were mostly intended for interest. I'll edit my answer later and mention the use of AC (somewhat) more precisely.
– ktoi
Jan 10 '18 at 18:17