Infinite Dimensional Vector Spaces

Given a basis, we can always represent $|a⟩\in \mathcal{V}$ as $|a⟩:=(a^1{a}^2{a}^3\cdots a^n{)}^T$ (with $n$ to infinity). So one obvious way to get an infinite dimensional vector space is to allow $\mathrm{\infty }$ sequences. $\Rightarrow$ $|a⟩:=(a^1\cdots {)}^T\in {\mathbb{C}}^{\mathrm{\infty }}$. The standard basis is then $|e_1⟩=(10\cdots {)}^T,|e_2⟩=(010\cdots {)}^T,\cdots$.

Could we even tell the difference between dimension of $\mathcal{V}$ being very large versus infinite (countable or uncountable).

Consider for two operators $[A,B]=\lambda \mathbb{I}$. Take the trace, $\text{Tr}([A,B])=\text{Tr}(AB)-\text{Tr}(BA)=\text{Tr}(AB)-\text{Tr}(AB)=0=\text{Tr}(\lambda \mathbb{I})=\lambda \text{Tr}(\mathbb{I})=\lambda \dim \mathcal{V}=\lambda n$. Thus, there are no operators on a finite dimensional vector space that has a commutator proportional to the Identity operator, Schor’s Lemma.

Consider $[x,p]=i\hslash$. Thus, they must be infinite dimensional. (If we were in a finite dimensional space then there would be no quantum mechanics)