Is $\mathbf{F^\infty}$ infinite dimensional

Is the following Proof Correct?

Note: In the proof below $\mathbf{F}$ is either $\mathbf{C}$ or $\mathbf{R}$ and (14) is the result a vector space $V$ is infinite dimensional if and only if there is a sequence of vectors $v_1,v_2,v_3,...$ such that $v_1,v_2,...,v_m$ is linearly independent for all $m\in\mathbf{Z^+}$.

Theorem. $\mathbf{F}^{\infty}$ is infinite dimensional.

Proof. Consider the sequence of vectors $\mathcal{S}:\mathbf{Z^+}\to V$ defined as follows. $$\mathcal{S}(j)=(a_1,a_2,a_3,...,a_j,...),\ \ a_j=1,\forall n\in\mathbf{Z^+}(n\neq j\implies a_n=0)$$ and the function $\mathcal{J}:\mathbf{Z^+}\to\ v_1,v_2,...,v_m$ defined as follows $$\mathcal{J}(j)=\mathcal{S}(1),\mathcal{S}(2),...,\mathcal{S}(j)$$ We now show by recourse to Mathematical Induction the truth of the following proposition.

$$\forall m\in\mathbf{Z^+}(\mathcal{J}(m)\text{ is linearly independent in } V)\tag{1}$$

Basis-Step: Evidently the vector $\mathcal{J}(1)=\mathcal{S}(1)=(1,0,0,...)$ constitutes a linearly independent list.

Inductive-Step: Assume for an arbitrary integer $k$ that $\mathcal{J}(k)$ is linearly independent in $V$. It is not difficult to see that $S(k+1)\not\in \operatorname{span}(\mathcal{J}(k))$ consequently $\mathcal{J}(k+1)$ is a linearly independent list in $V$. We have now proved $(1)$ this together with (14) implies that $\mathbf{F}^{\infty}$ is infinite dimensional.

$\blacksquare$

• Looks right! I would rather use the notation $\mathcal{J}:\mathbf{Z^+} \to P(V)$ where $P(V)$ is the power set of $V.$ Also you seem to have missed typing out "set" ; ($\mathcal{J}(m)$ is a linearly independent set in $V$.) – Bijesh K.S Jul 27 '17 at 12:32
• Thankyou for pointing – Atif Farooq Jul 27 '17 at 12:33