I am trying to show:

Algebraic dimension of infinite-dimensional Banach Space is uncountable.

By algebraic dimension it is meant that the cardinality of the Hamel Basis of the space.
Suppose we defined $V$ to be an infinitely dimensional normed linear space. So far I found out the followings:

  • Interior of any proper subspace of $V$ is empty
  • Every proper closed subspace of $V$ is nowhere dense
  • According to Baire's Theorem, $V$ can not be written as a union of countable union of nowhere dense closed sets.
  • By the way of contradiction, I assume there is a countable Hamel Basis for $V$.
  • Using the Hamel Basis, I need to construct closed sets such that their union gives me the whole space $V$.
  • Suppose $\{ x_n \}_{n \in \mathbb{N} }$ is the Hamel Basis and suppose $(x_1,x_2,...,x_n)$ be the subspace generated linearly by $x_1,x_2,...,x_n$.

Trouble Now all i need to show is the subspace $(x_1,x_2,...,x_n)$ is closed. I pick an element from the closure of that subspace and argue it can be written as an linear combination of $x_1,x_2,...,x_n$. I use bunch of triangle inequalities, cauchy sequences but i feel i am lost. Would please help me to complete this proof?


2 Answers 2


On a finite-dimensional ($\mathbb{R}$ or $\mathbb{C}$) vector space, all norms are equivalent.

The finite-dimensional subspace $\operatorname{span} \{ x_1,\dotsc, x_n\}$ is complete if we endow it with the norm induced by the Euclidean norm on $\mathbb{R}^n$ (or $\mathbb{C}^n$) via the isomorphism $(\alpha_1,\dotsc,\alpha_n) \mapsto \sum\limits_{k=1}^n \alpha_k\cdot x_k$. Hence it is complete in the norm induced from $V$.

A complete subset of a metric space is closed.

  • $\begingroup$ So are you saying that $span \{ x_1,...,x_n \}$ is complete because $\mathbb{R}^n$ is complete? $\endgroup$
    – iamvegan
    Jan 12, 2014 at 17:46
  • 2
    $\begingroup$ Sort of. Because $\mathbb{R}^n$ is complete, and the norms induced by the ambient space and the one induced via an isomorphism to $\mathbb{R}^n$ are equivalent. Both facts together yield the completeness of $\operatorname{span} \{x_1,\dotsc,x_n\}$. $\endgroup$ Jan 12, 2014 at 18:27
  • $\begingroup$ Okay i just looked up the definition of equivalent norms. I saw that convergence is preserved in equivalent norms. Now i understand what you mean. $\endgroup$
    – iamvegan
    Jan 12, 2014 at 18:48
  • $\begingroup$ Quick and easy proof, +1! $\endgroup$ Jan 25 at 23:08


Every infinite-dimensional Banach space $X$ contains a vector space that is algebraically isomorphic to $\ell^\infty$.

From the above lemma, your claim (algebraic dimension of $\infty$ dimensional Banach spaces is at least $\mathfrak c$) follows.

  • $\begingroup$ And the proof of that lemma is ... ? (I don't see how proving that is any easier than the question here itself.) $\endgroup$ Sep 6, 2021 at 5:03
  • $\begingroup$ It is easier (in my opinion) actually. Gotta rush somewhere now, but I will post a proof sometime later. $\endgroup$ Sep 6, 2021 at 5:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.