Hamel bases

See Vaidyanathan (2017).

Definition 348.1(Definition).

A Hamel basis for a vector space $E$ over $\mathbb{R}$ or $\mathbb{C}$ is a set $\mathcal{B}\subseteq E$ such that every element $x\in E$ can be expressed uniquely as a (finite) linear combination of elements in $\mathcal{B}$.

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Definition 348.2(Definition).

An infinite set of vectors is said to be linearly independent if every finite subset is linearly independent.

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Proposition 348.3(Proposition).

For a subset $\mathcal{B}\subseteq E$, TFAE:

1. $\mathcal{B}$ is a Hamel basis for $E$;
2. $\mathcal{B}$ is a maximal linearly independent set;
3. $\mathcal{B}$ is a minimal spanning set.

We have seen previously that every finite dimensional vector space has a basis. With the extension of the notion of basis to infinite dimensional spaces with Hamel bases, this holds for all vector spaces:

Theorem 348.4(Theorem).

Every vector space has a basis.

Also see Cor 169.3.

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${\mathbb{R}}^{\mathbb{N}}$ does not have a countable basis

Lemma 348.5(Lemma).

A countable product of completely metrizable spaces is completely metrizable.

Proof.

Given a metric space $(Y,d)$, the metric $d^{\prime }(x,y)=\min \{1,d(x,y)\}$ is also complete, and induces the same topology $Y$ as $d$.

Now, suppose that $(X_i,{\rho }_i{)}_{i=1}^{\infty }$ is a sequence of metric spaces. For sequences $\mathbf{x}=\{x_i\}$ and $\mathbf{y}=\{y_i\}$, define

$$\rho (\mathbf{x},\mathbf{y})=\sum _{i=1}^{\infty }\frac{1}{2^i}{\rho }_i(x_i,y_i).$$

It can be shown, in the style of Exr 15.2, that $\rho$ metrizes the product topology on

$$X=\prod _{i=1}^{\infty }{X}_i.$$

If $(X_i,d_i)$ is a sequence of complete metric spaces, each $(X_i,{\rho }_i)$ is also complete and the metric $\rho$ as above is complete as well (see Rmk 15.3). Thus, the product of completely metrizable spaces is also completely metrizable.□

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Exercise 348.6(Exercise).

Show that ${\mathbb{R}}^{\mathbb{N}}$, the set of all functions $\mathbb{N}\to \mathbb{R}$, does not have a countable basis.

Proof.

Suppose ${\mathbb{R}}^{\mathbb{N}}$ has a countable basis $\mathcal{B}=\{w_i{\}}_{i=1}^{\infty }$. Equip ${\mathbb{R}}^{\mathbb{N}}$ with the product topology $\tau$. By Lem 5, $V:=({\mathbb{R}}^{\mathbb{N}},\tau )$ is completely metrizable. $V$ inherits the countable basis $\mathcal{B}$. Denote by $V_i$ the subspace of $V$ spanned by $\{w_1,\dots ,w_i\}$. It is clear from the topology of $V$ that each $V_i$ is nowhere dense¹. Additionally, each $V_i$, being homeomorphic to ${\mathbb{R}}^i$, is closed. Since every $v\in {\mathbb{R}}^{\mathbb{N}}$ can be expressed as a finite linear combination of $\{w_i{\}}_{i=1}^{\infty }$, we have

$$V=\bigcup _{i=1}^{\infty }{V}_i,$$

contradicting the Baire category theorem (Cor 168.9).□

Proof.

Taken from Schipperus (2017)
Here is a diagonalization argument. Let $\{f_i{\}}_{i=1}^{\infty }$ be a countable set we find $g\notin \text{span}\{f_i\}$. Construct $g$ as follows.

Consider the vector

$$(f_1(1),f_2(2)).$$

Define $(g(0),g(1))$ not to be a linear multiple of this vector.
Now look at the vectors

$$\begin{array}{r}(f_0(2),f_0(3),f_0(4)),\\ (f_1(2),f_1(3),f_1(4)).\end{array}$$

Define $(g(2),g(3),g(4))$ to not be a linear combination of these vectors. Now look at the vectors

$$\begin{array}{r}(f_0(5),f_0(6),f_0(7),f_0(8)),\\ (f_1(5),f_1(6),f_1(7),f_1(8)),\\ (f_2(5),f_2(6),f_2(7),f_2(8)).\end{array}$$

Define $(g(5),g(6),g(7),g(8))$ not in the span of these three vectors, and so on. It is clear that $g$ is not in the span of $\{f_i\}$.□

Footnotes

1. any open neighborhood in ${\mathbb{R}}^{\mathbb{N}}$ leaves infinitely many coordinates free; no finite-dimensional subspace can contain such a neighborhood. ↩

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References

Schipperus, R. (2017). Answer to “Proof of Uncountable Basis for $\mathbb{N} \to \mathbb{R}$ over $\mathbb{R}$.” In Mathematics Stack Exchange. https://math.stackexchange.com/a/2140854/1677240

Vaidyanathan, D. P. (2017). MTH 503: Functional Analysis.