Is it always possible to extend a linearly independent set to a basis in infinite dimensional vector space?

I was proving with the following argument: If S is a linearly independent set, if it spans the vector space then done else keep on adding the elements such that the resultant set is also linearly independent, till it spans the vector space . But the problem is how can we guarantee that the process will stop?


2 Answers 2


Let $V$ be a vector space, $S\subseteq V$ a linearly independent subset and $\mathcal{A}=\{T\subseteq V: S\subseteq T \text{and $T$ is linearly independent}\}$. It is easy to see that any chain on $\mathcal{A}$ has an upper bound on $\mathcal{A}$ (we can take the union). Then, it follows from Zorn's lemma that $\mathcal{A}$ has a maximal element $R$. If $\langle R\rangle\neq V$ then we can consider $R\cup\{v\}$ for some $v\notin \langle R\rangle$ and we obtain an element of $\mathcal{A}$ which is greater than a maximal element. The contradiction comes from our assumption that $\langle R\rangle\neq V$. So, we must have $\langle R\rangle = V$.

  • 1
    $\begingroup$ Nice, just that it'd be a good idea, perhaps, to remark to the OP that it's necessary to show that $\;R\cup\{v\}\;$ (set union, not sum!) is lin. ind. to get our contradiction...+1 $\endgroup$
    – DonAntonio
    Dec 16, 2013 at 16:59
  • $\begingroup$ Something I've always wondered about is the following: how do we know that if $R$ is linearly independent and $v \notin \langle R \rangle$, then $R \cup \{v\}$ is independent? This seems to be a feature of modules specifically over fields; for instance, if we're talking $\mathbb{Z}$-modules, observe that $\{2\}$ is a linearly independent subset of $\mathbb{Z}$, and that $1 \notin 2\mathbb{Z}$, but that $\{1,2\}$ is not linearly independent. $\endgroup$ Sep 27, 2016 at 17:18
  • $\begingroup$ For fields the proof goes as follows: Take a linear combination of elements of $R$ and $v$ which is $0$. If the coefficient of $v$ is $0$ then it was a linear combination of $R$ which is $0$ and hence all the coefficients are null. In the case where the coefficient of $v$ is not zero you can divide by it and get a linear combination of elements of $R$ which is equal to $v$, which is a contradiction. $\endgroup$
    – Quimey
    Oct 31, 2016 at 17:27
  • $\begingroup$ Also from this, you can see that you don't need the condition of the base ring to be a field. It is enough to have that $\lambda v \in \langle R \rangle$ implies $v \in \langle R \rangle$. $\endgroup$
    – Quimey
    Oct 31, 2016 at 17:29

Let be $V$ a vector space and $W\leq V$ a subspace of $V$ with basis $Y$. If we consider the quotient space $V/W$ , by Zorn's lemma, we can obtain a basis of $V/W$, denoted $\overline{S}$.

If $v\in V$, $\overline{v}=\beta_1\overline{\alpha_1}+\cdots+\beta_k\overline{\alpha_k}$ where $\alpha_i\in S$, then $v-(\beta_1\alpha_1+\cdots+\beta_k\alpha_k)\in W$ so $v=\beta_1\alpha_1+\cdots+\beta_k\alpha_k+\gamma_1\zeta_1+\cdots+\gamma_r\zeta_r$, where $\zeta_i\in Y$. Thus, $S$ (yes, without bar) and $Y$ generate $V$.

Finally, whit the same notation above, if $\{{\alpha_1,\cdots,\alpha_k}\}\subseteq S$ and $\{{\zeta_1,\cdots,\zeta_r}\}\subseteq Y$ the equation $$\beta_1\alpha_1+\cdots+\beta_k\alpha_k+\gamma_1\zeta_1+\cdots+\gamma_r\zeta_r=0$$ implies that each escalar is zero. Indeed, the equation implies that $$\beta_1\alpha_1+\cdots+\beta_k\alpha_k\in W,$$ then $\beta_1\overline{\alpha_1}+\cdots+\beta_k\overline{\alpha_k}=\overline{0}$ and it follows that each $\beta_i=0$ and by linear independence of $\{{\zeta_1,\cdots,\zeta_r}\}$ it follows that $\{{\alpha_1,\cdots,\alpha_k,\zeta_1,\cdots,\zeta_r}\}$ is linearly independent. Thus $Y$ can be extended to a basis of $V$.

PDT: I'm sorry, english is not my mother tongue.


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