Let $X\neq \emptyset$ be a set and $K$ a field.
For $a\in X$ let $e_a\in K^X$ the map from $X$ to $K$ with $e_a(a)=1_K$ and $e_a(x)=0_K$ for all $x\in X$ with $x\neq a$.
For $f\in K^X$ let $\text{Sup}(f):=\{x\in X\mid f(x)\neq 0_K\}\subseteq X$ the support of $f$.
(a) (i) Let $1\leq n\in \mathbb{N}$ and $X:=[n]$. Determine for $1\leq i\leq n$ the column vector to $e_i$.
(ii) Let $1\leq m,n\in \mathbb{N}$ and $X:=[m]\times [n]$. Determine for $1\leq i\leq m$, $1\leq j\leq n$ the matrix to $e_{(i,j)}$.
(iii) Let $X=\mathbb{N}$. Determine for $n\in \mathbb{N}$ the sequence to $e_n$.
(b) Show that:
(i) For all $f,g\in K^X$ it holds that $\text{Sup}(f+g)\subseteq \text{Sup}(f)\cup \text{Sup}(g)$ and the equality holds if $\text{Sup}(f)\cap \text{Sup}(g)=\emptyset$.
(ii) For all $f\in K$ and $\lambda\in K$ it holds that $\text{Sup}(\lambda f)\subseteq \text{Sup}(f)$ and the equality holds if $\lambda\neq 0_K$.
(c) Let $K^{(X)}:=\{f\in K^X\mid |\text{Sup}(f)|<\infty\}$ the set of all maps from $X$ to $K$ with finite support. Show that $K^{(X)}$ is a subspace of $K^X$ (i.e. that $K^{(X)}\leq_K K^X$) and that $K^{(X)}=K^X \iff |X|<\infty$.
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Let's start with (a).
What is it meant by column vector to $e_i$ ? Is it a vector with $1$ at the $i$-th component and $0$ everywhere else?
For (b)(i) I have done the following:
Let $f,g\in K^X$. We have that $\text{Sup}(f+g)=\{x\in X\mid (f+g)(x)\neq 0_K\}$.
It holds that $(f+g)(x)\neq 0_K\Rightarrow f(x)+g(x)\neq 0_K$. This implies that $f(x)\neq 0_K$ or $g(x)\neq 0_K$.
Therefore this implies that \begin{align*}&\{x\in X\mid (f+g)(x)\neq 0_K\}\subset \{x\in X\mid f(x)\neq 0_K\}\cup \{x\in X\mid g(x)\neq 0_K\} \\ & \Rightarrow \text{Sup}(f+g)\subseteq \text{Sup}(f)\cup \text{Sup}(g) \end{align*}
How can we show the part with the equality?
For (b)(ii) I have done the following:
Let $f\in K^X$ and $\lambda\in K$. We have that $\text{Sup}(\lambda f)=\{x\in X\mid (\lambda f)(x)\neq 0_K\}$.
It holds that $(\lambda f)(x)\neq 0_K\Rightarrow \lambda f(x)\neq 0_K$. This implies that $f(x)\neq 0_K$.
Therefore we get \begin{equation*}\{x\in X\mid (\lambda f)(x)\neq 0_K\}\subset \{x\in X\mid f(x)\neq 0_K\} \Rightarrow \text{Sup}(\lambda f)\subseteq \text{Sup}(f)\end{equation*}
How can we show the part with the equality?
Is everything correct so far?
Could you give me a hint for the questions (a) and **(c) ** ?
(b)(i): We want to show that if $\operatorname{Supp}(f) \cap \operatorname{Supp}(g) = \emptyset,$ then $\operatorname{Supp}(f + g) = \operatorname{Supp}(f) \cup \operatorname{Supp}(g).$ By your work, it suffices to show that $\operatorname{Supp}(f) \cup \operatorname{Supp}(g) \subseteq \operatorname{Supp}(f + g).$ On the contrary, consider an element $x \in X$ such that $f(x) \neq 0$ or $g(x) \neq 0$ and $(f + g)(x) = 0.$ Considering that $(f + g)(x) = 0,$ we have that $f(x) = -g(x).$ Either $f(x) \neq 0$ or $g(x) \neq 0,$ but in either case, we have that $f(x) \neq 0$ and $g(x) \neq 0$ so that $x \in \operatorname{Supp}(f) \cap \operatorname{Supp}(g)$ -- a contradiction.
(b)(ii): We want to show that if $\lambda \neq 0,$ then $\operatorname{Supp}(\lambda f) = \operatorname{Supp}(f).$ By your work, it suffices to show that $\operatorname{Supp}(f) \subseteq \operatorname{Supp}(\lambda f).$ On the contrary, consider an element $x \in X$ such that $f(x) \neq 0$ and $(\lambda f)(x) = 0.$ We have that $\lambda f(x) = 0.$ By hypothesis that $K$ is a field, it follows that $\lambda = 0$ or $f(x) = 0.$ Considering that $f(x) \neq 0,$ we conclude that $\lambda = 0$ -- a contradiction.
For part (a)(ii), use a similar idea to that of my comment above. Particularly, fix a pair $(i, j),$ and notice that $e_{i, j}(k, \ell) = 1$ whenever $k = i$ and $j = \ell$ and $e_{i, j}(k, \ell) = 0$ whenever $k \neq i$ or $\ell \neq j.$ Considering that $(i, j)$ is in the $i$th row and $j$th column, what will the resulting matrix look like?
For part (a)(iii), you should find your answer by a process analogous to part (a)(i).
For the first statement of part (c), it suffices to prove that $K^{(X)}$ is closed under addition (the sum of two elements with finite support has finite support) and scalar multiplication (the product of an element with finite support and a scalar $\lambda \in K$ has finite support). Use both parts of (b).