# Vector Space with Hamel Basis is not Separable when all basis elements are 2 apart

Consider a Hamel basis, $$\{e_{\lambda}\}_{\lambda \in \Lambda}$$, for an infinite dimensional linear vector space. I'm reading something that makes the following claim in passing:

Note that if for any $$\alpha, \beta \in \Lambda$$ we have $$\|e_{\alpha}-e_{\beta}\|=2 \quad (\alpha \neq \beta)$$ Then the vector space is not separable.

(Where any $$x$$ is given by $$x = \sum_{\lambda \in \Lambda}e_{\lambda}x_{\lambda}$$ and the norm is given by $$\|x\| = \sum_{\lambda \in \Lambda}\gamma_{\lambda}|x_{\lambda}|$$ where $$\gamma_{\lambda}$$ is an element in some collection of positive numbers $$\{\gamma_{\lambda}\}_{\lambda \in \Lambda}$$)

I wasn't able to show this for myself. I want to take a countable dense subset, choose a basis element not in this countable collection, and show that its "separate" from the dense set. In particular, I want to use a projection operator where I project to this "new" basis element -- but I don't think the projection operator is continuous though so this won't work. How else can I prove the claim?

• I'm not sure I understand the definition of your "norm" $\|x\|$. What are the $\gamma_\lambda$'s? Are they just $\gamma_\lambda = |x_\lambda|$? Apr 22, 2019 at 2:12
• Oh I see, you just fix a set $\{\gamma_\lambda\}$. Apr 22, 2019 at 2:17
• @Ehsaan yes! sorry if that wasn't clear. Fix a set of positive numbers. The sum will have 0 in all but a finite number of places. Maybe it was confusing because I wrote the basis and the coordinate backwards in the sum. Apr 22, 2019 at 2:31
• So $\|e_{\alpha}-e_{\beta}\|= (\gamma_\alpha + \gamma_\beta)=2 \quad \text {(for all$\alpha \neq \beta$)}$? And so all $\gamma_\alpha = 1$? Apr 22, 2019 at 11:03
• The space will be separable iff the basis is countable. That is possible in general, but may not be if you have other conditions, e.g. completeness. Apr 22, 2019 at 11:18

The fact that $$\|e_\alpha - e_\beta\| = 2$$ for all $$\alpha, \beta \in \Lambda$$ with $$\alpha \neq \beta$$ implies that the metric balls $$B(a_\alpha, 1), \alpha \in \Lambda$$ are pairwise disjoint (using the metric on $$X$$ defined by the norm, so $$d(x,y)=\|x-y\|$$). If $$D$$ is dense in $$X$$ (for the metric topology) it must intersect all open balls and as the balls are disjoint, all such intersection points will give different points of $$D$$.
Conclusion: If the vectors $$\{e_\alpha: \alpha \in \Lambda\}$$ obey the norm property then $$|D|\ge |\Lambda|$$ for any dense set $$D$$ of $$X$$. So $$X$$ can only be separable if $$\Lambda$$ is at most countable.
If the $$e_\alpha$$ form a countable Hamel base, it’s quite easy to see that the finite linear combinations with rational coefficients form a countable dense set in $$X$$. So it depends on the size of the Hamel basis: countable then separable, uncountable then $$X$$ is not separable.