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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 $W$.

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    $\begingroup$ 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. $\endgroup$ – Targon Feb 26 at 21:33
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    $\begingroup$ 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 $f$. $\endgroup$ – Brian Moehring Feb 26 at 21:33
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    $\begingroup$ $W$ has infinite dimension, so your $S$ can't be a basis.... $\endgroup$ – Surb Feb 26 at 21:33
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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 $X$.

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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.

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