Countable basis for function space

Question

Show that the space of functions $f:\Bbb{N}\to\Bbb{R}$ does not have a countable basis.

If the domain were a finite set instead of $\Bbb{N}$, then the set of functions that takes the value $1$ at a single point and vanishes elsewhere forms a basis. Now that the domain is countably infinite we get countable infinitely many functions $f_n, n\in\Bbb{N},$ defined by $f_n(m)=\delta_{mn}$. These are clearly linearly independent, but don't span the entire space. For example, the constant function $f(n)=1$ is not in their span.

But, we can extend that set of functions to a bigger countable collection. Can anyone give some hint about how to start the problem?

It is possible to explicitly describe an uncountable set of linearly independent $f$ in this space. — Angina Seng, May 24 '17 at 21:50
You mean a basis ${\phi_n}{n \in \mathbb{N}}$ such that any function $f : \mathbb{N} \to \mathbb{R}$ is a finite linear combination of the $\phi_n$. Suppose this is the case, remove the $\phi_n$ that are linear combination of the $\phi_m,m < n$ and look at $g(k) = \sum{n=0}^\infty a_n \phi_n(k)$. Can we always choose $a_n$ non-zero such that the series converges for every $k$ ? — reuns, May 24 '17 at 21:51
Your functions are just sequences in $\mathbb R.$ Consider the collection of all sequences of $0's$ and $1's.$ — Matematleta, May 24 '17 at 22:36
What kind of space and what kind of basis? (Vector space, or a topological space?) It seems from the comments everyone is assuming a vector space. Hint. Assume there was a countable basis, and "diagonalize" to construct a function which cannot be expressed as a (finite) linear combination of the elements of the basis. You can make this function approach infinity faster than any of the functions of the presumed basis. — Mirko, May 24 '17 at 23:43
@user1952009 The cardinality of $\mathbb{R}^\mathbb{N}$ is $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}=\mathfrak c$, same as the cardinality of $\mathbb R$. — Mirko, May 24 '17 at 23:54
@LordSharktheUnknown Could you please be more explicit as to how to "explicitly describe an uncountable set of linearly independent $f$ in this space"? — Mirko, May 24 '17 at 23:57
I think it does have a countable basis. Consider the set the set of functions { fi(x) : fi(x)=1 (for x=i) & 0 otherwise } (i belongs to N). — Prajwal Samal, Jun 03 '18 at 05:11
@Matematleta, could you give a separate answer to your comment? I have tried to use your comment but I failed. Could you show please why $\mathbb{R}^{\omega}$ has not countable basis? BTW, I have done the following: Suppose it has countable basis ${e_k}_k$ then we form the new vector say $(x_1,x_2,\dots)$ taking 1st coordinate from $e_1$, 2nd coordinate from $e_2$ and etc. Let's take the new vector $(y_1,y_2,\dots)$ such that $y_i\neq x_i$. Is this reasoning correct? — RFZ, Nov 14 '19 at 22:06
@Mirko, could you give a more detailed answer what do you mean by "diagonalize"? I have spent some time trying to solve it but no results yet. — RFZ, Nov 15 '19 at 00:38
@PrajwalSamal With the set of functions that you propose, one would only be able to express as a (finite) linear combination each function $f$ that has only finitely many $n$ with $f(n)\neq0$ — Mirko, Nov 15 '19 at 00:45
@ZFR I added an answer, hope it helps, please let me know if you have questions. — Mirko, Nov 15 '19 at 04:45

Mirko · Answer 1 · 2019-11-15T04:35:38.467

2

$\def\R{\Bbb R} \def\N{\Bbb N} \def\F{\mathcal F} \def\B{\mathcal B}$ Suppose $\B=\{b_n:n\in\N\}$ were a countable basis for the space $\mathcal F$ of all functions $f:\N\to\R$. That is, each $b_n\in\F$ and each $f\in\F$ is a linear combination with real coefficients of finitely many functions in $\B$. For convenience, assume $0\notin\N=\{1,2,3,...\}$.

For each $n$ let $M_n=n\cdot\sum_{k=1}^n|b_k(n)|$. Define $g\in\F$ by $g(n)=M_n+1$. (This is what I refer to as "diagonalize", in my comment.) We will show that $g\notin span(\B)$.

Suppose $g=a_1b_{n_1}+\cdots+a_jb_{n_j}$ for some $j\in\N$, some $a_1,...,a_j\in\R$, and some $n_1<\dots<n_j$. Let $A=\max\{|a_1|,\dots,|a_j|\}$. Take $n>\max\{n_j,A\}$, and so then $g=a_1b_{n_1}+\cdots+a_jb_{n_j}+0b_n$. But then $|g(n)|\le|a_1b_{n_1}(n)|+\cdots+|a_jb_{n_j}(n)|+|0b_n(n)|\le$ $A\cdot\bigl(|b_{n_1}(n)|+\cdots+|b_{n_j}(n)|+|b_n(n)|\bigr)\le$ $A\cdot\sum_{k=1}^{n}|b_k(n)|\le$ $n\cdot\sum_{k=1}^n|b_k(n)|=M_n<M_n+1=g(n)$, a contradiction.

edited Nov 15 '19 at 04:35

answered Nov 15 '19 at 04:21

Mirko

13,789

Dear Mirko! This is a really nice idea. I have never seen something before. That's why i have tried it couple hours and was not able to solve it. But could you explain what is the meaning of those $M_n$? Moreover, i got the proof which you provided but could you give some motivation how did you come up with this, please? I would like to know the basic idea behind the proof! – RFZ Nov 15 '19 at 14:12
1

@ZFR There is a limit as to what you could express as a linear combination using only the first $n$ many $b_k$, and only coefficients smaller by abs value than $n$. $M_n$ captures this limit at the $n$-th coordinate. Once limits are established, you diagonalize, exceed it,a contradiction. My original hint was incomplete, I had to think about details in vector basis context. I have seen this idea constructing a "scale" of functions in $\omega^{\omega}$,in M.E.Rudin's Lectures on Set Theoretic Topology,also Eric van Douwen chapter in Handbook Set-Theoretic Topology, "tower","splitting" cardinals – Mirko Nov 15 '19 at 15:31
I am very thankful for your reply. One more question: why each $M_n$ is multiple of $n$ with this sum? What if take $M_n$ just to be that sum? I mean without factor $n$. – RFZ Nov 15 '19 at 17:34
@ZFR Without factor $n$ you would only be taking care of linear combinations with coefficients of abs value at most $1$. With $n$, as $n\to\infty$, you allow not only the use of a larger set of basic functions (from $b_1$ to $b_n$), but also the use of larger coefficients. In this way you eventually take care of every possible linear combination. Indeed,every linear combination uses only finitely many basic functions, up to some $b_{n_j}$ and some finite coefficient going with each basic function. So $A$ is max of abs value of coefficients, we then take $n>A$ and then that $M_n$ is big enough. – Mirko Nov 15 '19 at 22:40
@ZFR If someone wrote the above comment to me I would have been totally lost/perplexed. At any rate,I tried. But another way to explain,just look at that long sequence of inequalities,and try to determine which step would not have worked, if we didn't incorporate $n$ in the definition of $M_n$. How about $A\cdot\sum_{k=1}^{n}|b_k(n)|\le n\cdot\sum_{k=1}^{n}|b_k(n)|$ (which is essentially saying/using that $A\le n$)? This is the step BTW, that I had to stop,and think about,before I wrote my answer,since it seems to be a necessary new element of the proof,compared to what I had seen about scales – Mirko Nov 15 '19 at 22:50
@ZFR I no longer understand what I did and why it works, you are on your own :) It just came to me. The more you read/see, the more things like this might come to you too. The best way (I have practised it in the past), sit down and try to reproduce the proof (write it down), without looking at my answer. You are allowed to study it as much as you wish in advance, and then give yourself a few hours (or one to three days), to forget it, and then try to reproduce it. Whatever you forgot, you will have to think and fix it on your own. Then you will understand it (better yet, your own way). – Mirko Nov 15 '19 at 23:03

score 2 · Answer 2 · answered Dec 24 '19 at 13:42

Vandermonde determinants to the rescue.

For each $x\in\Bbb{R}$ define the function $f_x:\Bbb{N}\to\Bbb{R}$ by the rule $f_x(n)=x^n$.

Claim. The set $\{f_x\mid x\in\Bbb{R}\}$ is linearly independent.

Proof. Assuming contrariwise that some non-trivial linear combination of $f_{x_i}, i=1,2,\ldots,n$, $x_i\neq x_j$ whenever $i\neq j$, is the constant function zero, i.e. for all $k\in\Bbb{N}$ $$\sum_{i=1}^nc_if_{x_i}(k)=0.\qquad(*)$$ The identity $(*)$ must hold in particular, when $k=0,1,2,\ldots,n-1$. Then the resulting linear system of equations on the unknowns $c_i$ should also have a non-trivial solution. But, the matrix of the system is a Vandermonde matrix with distinct columns, so its determinant is non-zero. Implying that all zeros is the only solution. A contradiction.

Given that we displayed an explicit uncountable linearly independent collection, no countably infinite basis is possible.

score -1 · Answer 3 · edited Jul 01 '23 at 15:03

For a given hypothetical countable basis $(f_{i})_{i \in \mathbb{N}}$. Define $f : \mathbb{N} \rightarrow \mathbb{R}$ as such : $$f(n \in \mathbb{N}) = n \cdot \sup_{i \leq n} f_i(n)$$ Let $a_0, ..., a_{k \in \mathbb{N}}$ be real numbers such that $f = \sum_{0 \leq i \leq k} a_i \cdot f_i$ .

Prove that for all $n \geq k$ : $$ 0 \leq f(n) \leq \left ( \frac{k}{n} \sum_{0 \leq i \leq k} |a_i| \right ) f(n)$$

Conclude.

Countable basis for function space

3 Answers3