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?

  • $\begingroup$ I'm not sure I understand the definition of your "norm" $\|x\|$. What are the $\gamma_\lambda$'s? Are they just $\gamma_\lambda = |x_\lambda|$? $\endgroup$
    – Ehsaan
    Apr 22, 2019 at 2:12
  • $\begingroup$ Oh I see, you just fix a set $\{\gamma_\lambda\}$. $\endgroup$
    – Ehsaan
    Apr 22, 2019 at 2:17
  • $\begingroup$ @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. $\endgroup$
    – yoshi
    Apr 22, 2019 at 2:31
  • $\begingroup$ So $\|e_{\alpha}-e_{\beta}\|= (\gamma_\alpha + \gamma_\beta)=2 \quad \text {(for all $\alpha \neq \beta$)}$? And so all $\gamma_\alpha = 1$? $\endgroup$ Apr 22, 2019 at 11:03
  • $\begingroup$ 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. $\endgroup$ Apr 22, 2019 at 11:18

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .