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Can a Hamel basis of $C[a,b]$ be dense in $C[a,b]$? Also, can ALL the elements in C[a,b] be expressed as linear combination of the Hamel basis? by Hamel Basis $\{e_\alpha\}$ I just mean a maximal linear independent set of elements in $X$. Therefore, in general, It's not necessary to have $\text{span}\{e_\alpha\} = X$.

Ma Joad
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    Editing to remove half your question after someone has answered is not very nice: it makes their answer look stupid, and removes the value of their answer to future readers. – John Hughes Dec 28 '17 at 13:17
  • John! I just add the missing part back! Just give me time! Be paient! Come on! I am just adding more details to make me less stupid and help others to understand! Just understand me! I am not deleting any parts of the question I have put before! – Ma Joad Dec 28 '17 at 13:25
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    A maximal linearly independent set is a basis: if $B$ is a linearly independent set and $v\in X\setminus\operatorname{span}(B)$, then $B\cup{v}$ is linearly independent, so $B$ is not maximal. – egreg Dec 28 '17 at 13:27
  • How stupid I am...... – Ma Joad Dec 28 '17 at 13:39
  • You can make edits without saving the intermediate state. I arrived and looked at this in its broken state, checked the original question, and didn't like what appeared to have happened. Usually we edit questions only to fix typos or to address details raised in the comments. (e.g. a comment like, "Does $C[a, b]$ denote the space of all continuous functions on the interval $[a.b]?$ might make you add that clarification to your question.) – John Hughes Dec 28 '17 at 13:40
  • "How stupid I am .... " Nope: just learning. We're happy to support that in general. – John Hughes Dec 28 '17 at 13:42
  • See also this old question. –  Dec 28 '17 at 18:21

3 Answers3

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Yes; in fact

Theorem 1. If $X$ is an infinite-dimensional separable Banach space then there exists a Hamel basis which is dense in $X$.

Note The assumption that $X$ is separable is not needed; see below. We include this version just because the proof is more elementary.

Lemma If $X$ is a Banach space and $Y$ is a proper subspace then $Y$ has empty interior.

Proof: If $B(x,r)\subset Y$ then $B(0,r) = B(x,r) - x\subset Y$, hence $Y=X$.

Proof of the theorem: Say $B_1,B_2,\dots$ is a base for the topology of $X$. The lemma shows we can recursively choose $x_n\in B_n$ so that $x_{n+1}$ is not in the span of $x_1,\dots, x_n$. So $\{x_1,\dots\}$ is independent, hence it is a subset of some basis. QED.

Theorem 2. Same as Theorem 1 except $X$ need not be separable.

Proof. Just a transfinite version of the previous proof.

Say the dimension is $\kappa$. Say $B$ is a Hamel basis for $X$ of cardinality $\kappa$. Then there is a base for the topology of cardinality $\kappa$, consisting of all $B(x,r)$ where $x$ is a rational linear combination of (finitely many) elements of $B$ and $r>0$ is rational. Say $(B_\alpha)_{\alpha<\kappa}$ is such a base (the index $\alpha$ running over all ordinals less than $\kappa$.)

Now suppose $\alpha<\kappa$ and $(x_\beta)_{\beta<\alpha}$ is an independent set with $x_\beta\in B_\beta$ for all $\beta<\alpha$. Since $\alpha<\kappa$ the span of $(x_\beta)_{\beta<\alpha}$ cannot be $X$, so there exists $x_\alpha\in B_\alpha$ which is not in this span. So by induction we obtain an independent dense set $(x_\alpha)_{\alpha<\kappa}$.

David C. Ullrich
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To construct a clear example, take a Hamel basis and renormalize it so that all elements have norm equal to one. Then the zero function is at distance 1 from the basis (and any element with norm other than 1 will have positive distance to the basis).

As for your second question: yes, that is in the definition of basis.

Martin Argerami
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  • Thank you for answering hte first question. However, by Hamel Basis ${e_\alpha}$ I just mean a maximal linear independent set. In general, It's not necessary to have $span {e_\alpha} = X$ – Ma Joad Dec 28 '17 at 13:18
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    Can't you just take an element $e \in X \setminus \operatorname{span}{e_\alpha}$ which is then linearly independent and thus the chosen Hamel Basis wouldn't be a maximal independent set? –  Dec 28 '17 at 13:27
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    @txokmdlq: I think you need to learn your vector spaces. A maximal linearly independent set does span the space, as Paul K. says. – Martin Argerami Dec 28 '17 at 13:57
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A separable Banach space $X$ with $\dim X\ge\aleph_0$ (and thus $\dim X=\beth_1$) always admits a dense Hamel basis.

Since $X$ is separable, it has a dense subspace $V$ such that $\dim V=\aleph_0$. This subspace has a basis $\mathcal B$ and $\overline{\operatorname{span}_{\Bbb Q} \mathcal B}=X$. $\mathcal B$ may be extended to a Hamel basis $\mathcal B\cup \mathcal W$ by some set such that $\#\mathcal W=\beth_1$. The idea is to use the elements of $\mathcal W$ to approximate the elements of the countable set $\operatorname{span}_{\Bbb Q}\mathcal B$. Since $\aleph_0\cdot\aleph_0=\aleph_0<\beth_1$, there is an injective function $\Psi:\Bbb N\times\operatorname{span}_{\Bbb Q}\mathcal B\hookrightarrow \mathcal W$. Let's call $\Phi(n,v)=v+\frac{1}{2^n\lVert \Psi(n,v)\rVert}\Psi(n,v)$ for $n\in\Bbb N$ and $v\in\operatorname{span}_{\Bbb Q}\mathcal B$.

Now, I claim that $\mathcal X=\mathcal B\cup\{\Phi(n,v)\,:\,n\in\Bbb N,\ v\in\operatorname{span}_{\Bbb Q}\mathcal B\}$ is a linearly independent set/sequence. In fact, if $b_i\in\mathcal B$ and $$\alpha_1 b_1+\cdots+\alpha_m b_m+\beta_1\Phi(n_1,v_1)+\cdots+\beta_k\Phi(n_k,v_k)=0$$ then $$\underbrace{\sum_i\alpha_ib_i+\sum_j\beta_j v_j}_{\in\ \operatorname{span}\mathcal B}+\underbrace{\sum_j\frac{\beta_j}{2^{n_j}\lVert\Psi(n_j,v_j)\rVert}\Psi(n_j,v_j)}_{\in\ \operatorname{span}\mathcal W}=0$$

By $\operatorname{span}\mathcal W$ and $\operatorname{span}\mathcal B$ being in direct sum, the former implies $$\begin{cases}\sum_j\frac{\beta_j}{2^{n_j}\lVert\Psi(n_j,v_j)\rVert}\Psi(n_j,v_j)=0\\\sum_i\alpha_ib_i+\sum_j\beta_j v_j=0\end{cases}\iff\begin{cases}\forall j,\ \beta_j=0\\\sum_i\alpha_ib_i+\underbrace{\sum_j\beta_j v_j}_{=0}=0\end{cases}\iff\\\iff\begin{cases}\forall j,\ \beta_j=0\\\forall i,\ \alpha_i=0\end{cases}$$

It is clear that $\lim_{n\to\infty}\Phi(n,v)=v$, and thus $\overline{\mathcal X}\supseteq\operatorname{span}_{\Bbb Q}\mathcal B$, making it a dense subset. Finally, if you so desire, $\mathcal X$ may be completed to a Hamel basis.