# Find a basis for the subspace $W= \left \{ f \in V | f(x)=0 \text{ } \forall x \in X, \text{ except for a finite number of elements of } X\right \}$

I need your help to check if my proof is correct please. This is what I have to prove:

Let be $$X$$ a not empty set and let be $$F$$ a field. Consider the vector space given by the set $$V= \left \{ f : X \rightarrow F \right \}$$ with the usual operations. Find a basis for the subspace $$W= \left \{ f \in V | f(x)=0 \text{ } \forall x \in X, \text{ except for a finite number of elements of } X\right \}$$

My attempt:

With $$i,j \in \left \{ 1,2,...,n \right \}$$, we can define this functions:

$$\begin{equation*} f_{x_{j}}(x_i)= \left\{ \begin{array}{rcl} 1 & ~~~~~~~~~if & x_i=x_j, \text{ with } x_i,x_j\in X\\ 0 & ~~~~~~~~~&\text{another case} \end{array} \right. \end{equation*}$$

Let's see if the set $$S=\left \{ f_{x_1},f_{x_2},\cdots,f_{x_{n}} \right \}$$ is a basis of $$W$$. First, we note that if:

\begin{align} \lambda_1 f_{x_1}(x_i)+\lambda_2 f_{x_2}(x_i)+\cdots+\lambda_n f_{x_n}(x_i)=0 & &\text{ for any x_i \in X}\\ & &\lambda_i \in F \\& &i \in \left \{ 1,...,n \right \} \end{align} then, \begin{align} \lambda_1 +\lambda_2 +\cdots+\lambda_n=0 & &\text{ for any x_i \in X} \end{align} So, that implies that $$S=\left \{ f_{x_1},f_{x_2},\cdots,f_{x_{n}} \right \}$$ is linearly independent (1)

By the other hand, let be $$f \in W$$, then $$f(x_i)=0$$, except for a finite number of elements of $$X$$. Then, we can propose to write $$f$$ as:

\begin{align} f(x_i)=\lambda_1 f_{x_1}(x_i)+\lambda_2 f_{x_2}(x_i)+\cdots+\lambda_n f_{x_n}(x_i) \end{align}

since the quantity of $$\lambda_i$$ that satisfies that $$\lambda_i \neq 0$$ is the number of finite elements of $$X$$ where $$f(x_i)\neq 0$$, that implies that $$f \in span(S)$$. So, $$S$$ is a generating set of $$W$$. (2)

By (1) and (2), $$S$$ is a basis of $$W$$.

• Note that $X$ does not need to be finite but that is basically what you assume by writing $S$ that way. But even if it's infinite, just define $S$ in basically the same way ($S=\{f_x : x \in X\}$). Also, I don't understand completely what you are doing in (2). You are not really defining the $\lambda_i$. Take $f\in W$ and let $x_1 , \ldots x_n \in X$ the points with $f(x_j) \neq 0$ for all $j$. Then define $\lambda_j := f(x_j)$ and you are done. – Targon Feb 26 at 21:33
• You're being too loose with your notation of $x_i$. For instance, in your argument that $S$ spans $W$, the $x_i$ depend on $f$, so you're implicitly saying that $S$ depends on $f$. – Brian Moehring Feb 26 at 21:33
• $W$ has infinite dimension, so your $S$ can't be a basis.... – Surb Feb 26 at 21:33

You seems to have the main ideas. However your « proof » seems to mention a finite number of maps to be a basis of $$W$$ which isn’t the case if $$X$$ is infinite.

$$\mathcal B=\{f_x \mid x \in X\}$$

where

$$f_x(v)=\begin{cases} 1 & v =x\\ 0 & v \ne x \end{cases}$$ is indeed a basis of $$W$$. To prove that $$\mathcal B$$ is linearly independent, suppose that a finite linear combination of elements of $$\mathcal B$$ $$\lambda_1 f_{a_1} + \dots +\lambda_n f_{a_n}=0$$ always vanishes. Then by plugging in $$v=a_i$$ you get $$\lambda_i=0$$ as desired.

You were almost there to prove that $$\mathcal B$$ spans $$W$$.

Note: the cardinality of $$\mathcal B$$ is the one of $$X$$.

You have the right idea but you're not quite there. I'll define $$S = \{f_x | x \in X\}$$ where $$f_x: X \longrightarrow F$$ is defined by $$f_x(y) = 1$$ if $$x = y$$ and $$0$$ otherwise. This is like your definition, but you only allowed $$S$$ to be finite. But that won't work in case $$X$$ is infinite, so you need to consider all of these elements.

Now we show linear independence. Your proof is wrong, in part because showing $$\sum \lambda_i = 0$$ is not what you need for linear independence. You want all of the $$\lambda_i = 0$$. Let's suppose we have $$\sum_{i = 1}^n \lambda_i f_{x_i} = 0$$ for distinct $$x_i \in X$$. Now, let $$1 \leq j \leq n$$ and evaluate this at $$x_j$$. This yields $$\sum_i \lambda_i f_{x_i}(x_j) = 0$$. For every $$i \neq j$$ we have $$f_{x_i}(x_j) = 0$$. Furthermore, $$f_{x_j}(x_j) = 1$$. Hence, we have $$0 = \sum_i \lambda_i f_{x_i}(x_j) = \lambda_j$$. Thus, each coefficient $$\lambda_j = 0$$ so $$S$$ is linearly independent.

Now let's show $$S$$ spans $$W$$. Take some $$f: X \longrightarrow F$$ in $$W$$. We claim that $$f = \sum_{x \in X} f(x) f_x$$. First of all, why is this sum even well defined? Well the set of $$x$$ such that $$f(x) \neq 0$$ is finite by definition of $$W$$, so this sum only has finitely many nonzero elements. Now, let's prove that these are equal. Take a $$y \in X$$ and compute $$\left( \sum_x f(x) f_x \right)(y) = \sum_x f(x) f_x(y)$$. Again, $$f_x(y) = 0$$ if $$x \neq y$$ and $$1$$ if $$x = y$$. Thus, this sum equals $$f(y)$$. Hence, $$f = \sum_x f(x) f_x$$. The right hand side is a linear combination of elements of $$S$$ (recall that we showed that this is a finite sum), so $$f \in span(S)$$.

Hence, $$S \subseteq W$$ is linearly independent and spans, so it is a basis.