If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A=\{e_1,\dots,e_n\}$, then, given a vector $v\in V$ we can find a new basis $E$ of $V$ such that $v\in E$; this is done as follows:

First we form the set $\{v\}\cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0=\{v\}$. If $e_1\not\in\text{span}(E_0)$, then we set $E_1=\{v\}\cup\{e_1\}$. Otherwise, $E_1=E_0$. Then, if $e_2\not\in\text{span}(E_1)$, we set $E_2=E_1\cup\{e_2\}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $\lambda_i,\mu_i$ such that $e=v+\sum\lambda_ie_i$, $e'=v+\sum\mu_ie_i$, then $e=e'+\sum(\lambda_i-\mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.

My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $v\in E$?

P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.

  • $\begingroup$ BTW. A property $P(x)$, which any set $x$ may or may not have, is called a property of finite character when $P(x)\iff \forall$ finite $y\subset x\,(P(y))$ holds for every $x$. For example, being a linearly independent subset of $V$ is a property of finite character. The Teichmuller-Tukey Lemma: If $P$ is of finite character and $P(\{x\})$ for some $x\in X$ then there is a $\subset$-maximal $Y\subset X$ such that $x\in Y$ and $P(Y).$ In the axiom system ZF, this "lemma" is equivalent to the axiom of choice. $\endgroup$ – DanielWainfleet Feb 28 '19 at 10:41

Yes (as long as $v \neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.


Let $A$ be a basis of $V$ (using AC) and $v \in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is $$ v = a + k_2a_2 + \cdots + k_na_n . $$

Then $B = A/\{a\} \cup \{v\}$ is a basis: it spans, and any finite subset is linearly independent.


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