This is the follow up question to this question (*)

According to page 2 of this link 1 and this link 2, $\mathbb{R^Z}$ (which is referred as $\mathbb{R^\infty}$ in link 1) has elements of the following form:

$$(y_k)_{k\mathbb{\in Z}}=(\dots y_{-1},y_0,y_{1}\dots)$$

Or more concisely:

$$y: \mathbb{Z}\rightarrow \mathbb{R}$$ where there exists $(y_k)$ that have infinite nonzero components

The author in link 1 then begin a heuristic argument showing how one can naively attempt to construct a Hamel Basis for $\mathbb{R^Z}$, before concluding the construction does not end and then he mentioned it can be proved that the Hamel Basis of it is uncountable.

Further down in page 3 and 4, he uses Zorn's Lemma to show that all vector spaces have a maximal spanning set, hence a (Hamel) basis (assuming Axiom of Choice)

Now from (*), we found that $\mathbb{R^Z}$ is uncountable. Based on the author's heuristic argument, it make me suspect that the only set that can span $\mathbb{R^Z}$ need all nonzero elements in $\mathbb{R^Z}$. This sounds reasonable since such set is uncountable, which matches the requirement of the Hamel basis of $\mathbb{R^Z}$. Thus the question boils down to

What are the approaches that one should try to construct a proof of the statement in the title, or find counterexamples to it?


1 Answer 1


I don't think any Hamel basis is explicitly known. Anyway any two elements $(\dots,a,\dots,a,\dots)$ ($a$ everywhere) and $(\dots,b,\dots,b,\dots)$ are clearly linearly dependent.

  • 1
    $\begingroup$ it is consistent with the failure of the axiom of choice that there is none. Which explains why no such basis can be explicitly given. $\endgroup$
    – Asaf Karagila
    May 25, 2015 at 10:08

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