$$\{\left[\begin{array}{c}1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\end{array}\right]\}\$\$isa[[ConceptWiki/Hamelbasis\parallel Hamelbasis]]for\${\mathbb{R}}^2\$.$$

If $V$ is a vector space, then $V$ has a Hamel basis.

Let $E$ be the set of linearly independent subsets of $V$. Define a partial order on $E$, $⪯$ via inclusion:

Towards applying Zorn, let $C$ be a chain in $E$. Define

This is a linearly independent subset:

• Let $v_1,\dots ,v_n\in c$. Thus there exist $e_1,\dots ,e_n\in C$ where for all $j$ we have $v_j\in e_j$.
• Since $C$ is a chain, there exists some $J$ such that for all $j=1,\dots ,n$ we have $e_j⪯e_J$ (ie $e_j\subseteq e_J$). Thus $v_1,\dots ,v_n\in e_J$.

Now, since $e_J$ is a linearly independent subset, this implies that $v_1,\dots ,v_n$ are themselves linearly independent. Thus $c\in E$

Thus for all $e\in C$, we have $e⪯c$. ie, there is an upper bound of $C$.

By Zorn's lemma, $E$ has a maximal element $H$. I claim that $H$ spans $V$.

Suppose BWOC that $H$ does not span $V$. Then there exists some $v\in V$ such that $v$ cannot be written as a finite linear combination of elements in $H$. Then $H\cup \{v\}$ is linearly independent. But then $H\subseteq H\cup \{v\}$ and $H\prec H\cup \{v\}$ so $H$ is not maximal $⟹⟸$

Thus $H$ spans $V$.

Thus every vector space has a Hamel basis

Let $V$ be a normed vector space and $M\subset V$ a subspace. If $u:M\to \mathbb{C}$ is a linear map such that for all $t\in M$

(ie, we have a bounded linear functional), then there exists a continuous extension $U:V\to \mathbb{C}$ such that $U\in \mathcal{B}(V,\mathbb{C})$ and

(with the same $t$ as above)

If $V$ is a normed vector space and $M\subset V$ is a subspace and $u:M\to \mathbb{C}$ is linear such that $|u(t)|\le C||t||$ for all $t\in M$, and

• $x\notin M$
  Then there exists a function $u^{\prime }:M^{\prime }\to \mathbb{C}$ which is linear
• (Where $M^{\prime }=M+\mathbb{C}x=\{t+ax:t\in M,a\in \mathbb{C}\}$ )
  Such that $u^{\prime }{|}_M=u$ and for all $t^{\prime }\in M^{\prime }$ we have $|u^{\prime }(t^{\prime })|\le C||t^{\prime }||$

Let $E$ be the set of all continuous extensions of $u$:

ie, each element is a bounded linear functional on $N$ with the same bound $C$ as the original functional $u$.

Now, define the partial order $⪯$ on $E$ as follows:

Let $C$ be a chain in $E$ indexed by the set $I$:

Then for all $i_1,i_2\in I$ either $(v_{i_1},N_{i_1})⪯(v_{i_2},N_{i_2})$ or vice versa. Now, let

be the union of all such subspaces $N_i$. We can show that $N$ is indeed a subspace.

Let $x_1,x_2\in N$ and $a_1,a_2\in \mathbb{C}$. Then there exist $i_1,i_2\in I$ such that $x_1\in N_{i_1}$ and $x_2\in N_{i_2}$. Since $C$ is a chain, WLOG, $N_{i_1}\subset N_{i_2}$.

Then $x_1,x_2\in N_{i_2}⟹a_1{x}_1+a_2{x}_2\in N_{i_2}\subset N⟹N$ is a subspace of $V$.

Now we can define $v:N\to \mathbb{C}$ by

• if $t\in N_i$ then $v(t)=v_i(t)$

Is this well-defined? ie, if $t\in N_{i_1}\cap N_{i_2}$ does this mean $v_{i_1}(t)=v_{i_2}(t)$?

• Suppose $t\in N_{i_1}\cap N_{i_2}$ and $(v_{i_1},N_{i_1})⪯(v_{i_2},N_{i_2})$. Since $v_{i_1}{|}_{N_{i_1}}=v_{i_1}$ then $v_{i_1}(t)=v_{i_2}(t)$. Thus $(v_{i_2},N_{i_2})$ is an extension of $(v_{i_1},N_{i_1})$

Similarly, we can show that $v$ is linear and also the extension of any $v_i$.

So then for all $i\in I$, we have $(v_i,N_i)⪯(v,N)$. ie, $(v,N)$ is an upper bound for $C$.

By Zorn, $E$ has a maximal element $(U,N)$ and claim $N=V$. Suppose BWOC this is not the case. Then let $x\notin N$. Then since we can always extend functions on subspaces, there exists a continuous extension $v$ of $U$ for $N+\mathbb{C}x$ (ie, $(v,N+\mathbb{C}x)\in E$). But then we have $(U,N)⪯(v,N+\mathbb{C}x)$ so $(U,N)$ is not maximal $⟹⟸$

First, note that if $t^{\prime }\in M^{\prime }(=M+\mathbb{C}x)$ then there exists unique $t\in M$ and $a\in \mathbb{C}$ such that $t^{\prime }=t+ax$.

Thus, upon choosing $\lambda \in \mathbb{C}$, the map

is well-defined on $M^{\prime }$ and $u^{\prime }:M^{\prime }\to \mathbb{C}$ is linear.

WLOG, suppose $C=1$. We want to choose $\lambda \in \mathbb{C}$ such that

Once $\lambda$ is fixed, $u^{\prime }$ is our continuous extension. When $a=0$, this inequality holds regardless of $\lambda$ (because it holds on $M$). Thus we only need to choose $\lambda$ for $a\ne 0$. When $a\ne 0$, we have

Since $\frac{t}{-a}\in M$. We first show there exists some $\alpha \in \mathbb{R}$ such that

where $w(t)=\frac{u(t)-\overline{u(t)}}{2}=\mathrm{Re}(u(t))$. Now, note that

So we can choose an $\alpha$ between these two quantities such that for all $t\in M$ we have

Using the same process for the imaginary part of $u(t)$ and repeating the argument with $ix$, we can find a $\lambda$ such that the desired bound holds.

This defines our function

on all of $M+\mathbb{C}x$. Thus we are done.