Let V be the vector space of functions $f : N → R$. Prove that V is infinite dimensional.

My thoughts: A finite dimensional vector space over a countable field is necessarily countable: if $v_1$,…,$v_n$ is a basis, then every vector in $V$ can be written uniquely as $α_1$$v_1$+⋯+$α_n$$v_n$ for some scalars $α_1$,…,$α_n$$∈f$, so the cardinality of the set of all vectors is exactly $|$f$|^n$. But this is where I'm stuck. I am still unclear on how I can prove it is infinite dimensional.


Let $(f_n)_{n=1}^\infty$ be a sequence of functions defined as: $$f_n(i) = \begin{cases} 1, & \text{if $i = n$} \\ 0, & \text{if $i \ne n$} \end{cases}$$

Now let $\{i_1, \ldots, i_k\} \subseteq \mathbb{N}$ be a finite set of indices and $\alpha_{i_1}, \ldots, \alpha_{i_k}$ be scalars such that:

$$\alpha_{i_1}f_{i_1}, \ldots, \alpha_{i_k}f_{i_k} = 0$$

Evaluating both sides of the equality at $i_j$ for $j \in \{1, \ldots, k\}$ we obtain:

$$0 = 0(i_j) = (\alpha_{i_1}f_{i_1}, \ldots, \alpha_{i_k}f_{i_k})(i_j) = \alpha_{i_1}f_{i_1}(i_j) + \cdots + \alpha_{i_k}f_{i_k}(i_j) = \alpha_{i_j}$$

Thus, $\alpha_{i_1} = \cdots = \alpha_{i_k} = 0$ so the set $\{f_{i_1}, \ldots, f_{i_k}\}$ is linearly independent.

Therefore, the set $\{f_n : n \in \mathbb{N}\}$ is linearly independent since its every finite subset is linearly independent.

Hence, $V$ is infinite-dimensional.

Notice that this example works for arbitrary fields since $0$ and $1$ are present in every field.



$$ f_i(n) := \begin{cases} 1 & i = n\\ 0 & i \ne n\end{cases}. $$

Then the collection $\{f_i : i \in \mathbb{N}\}$ is an infinite linearly independent subset. (Make sure you can prove it.) Therefore, $V$ is infinite dimensional. (For suppose $V$ had dimension $n < \infty$, then any linearly independent set must have $\le n$ elements, a contradiction.)


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