PlanetMath: dense ring of linear transformations

Note that the linear independence of the $v_i$ 's is essential in insuring the existence of a linear transformation $f$ . Otherwise, suppose $0=\sum d_iv_i$ where $d_1\neq 0$ . Pick $w_i$ 's so that they are linearly independent. Then $0=f(\sum d_iv_i)=\sum d_if(v_i)=\sum d_iw_i$ , contradicting the linear independence of the $w_i$ 's.

The notion of ``dense'' comes from topology: if $V$ is given the discrete topology and $\operatorname{End}_D(V)$ the compact-open topology, then $R$ is dense in $\operatorname{End}_D(V)$ iff $R$ is a dense ring of linear transformations of $V$ .

Proof. First, assume that $R$ is a dense ring of linear transformations of $V$ . Recall that the compact-open topology on $\operatorname{End}_D(V)$ has subbasis of the form $B(K,U):=\lbrace f\mid f(K)\subseteq U\rbrace$ , where $U$ is open and $K$ is compact in $V$ . Since $V$ is discrete, $K$ is finite. Now, pick a point $g\in \operatorname{End}_D(V)$ and let $$B=\bigcup_{\alpha\in I}\bigcap_{i=1}^{n(\alpha)}B(K_{i\alpha},U_{i\alpha})$$ be a neighborhood of $g$ , $I$ some index set. Then for some $\alpha\in I$ , $g\in \bigcap B(K_{i\alpha},U_{i\alpha})$ . This means that $g(K_{i\alpha})\subseteq U_{i\alpha}$ for all $i=1,\ldots,n(\alpha)$ . Since each $K_{i\alpha}$ is finite, so is $K:=\bigcup K_{i\alpha}$ . After some re-indexing, let $\lbrace v_1,\ldots, v_n\rbrace$ be a maximal linearly independent subset of $K$ . Set $w_j=g(v_j)$ , $j=1,\ldots,n$ . By assumption, there is an $f\in R$ such that $f(v_j)=w_j$ , for all $j$ . For any $v\in K$ , $v$ is a linear combination of the $v_j$ 's: $v=\sum d_jv_j$ , $d_j\in D$ . Then $f(v)=\sum d_jf(v_j)=\sum d_jg(_j)=g(v)\in U_{i\alpha}$ for some $i$ . This shows that $f(K_{i\alpha})\subseteq U_{i\alpha}$ and we have $f\in \bigcap B(K_{i\alpha},U_{i\alpha})\subseteq B$ .

Conversely, assume that the ring $R$ is a dense subset of the space $\operatorname{End}_D(V)$ . Let $v_1,\ldots,v_n$ be linearly independent, and $w_1,\ldots,w_n$ be arbitrary vectors in $V$ . Let $W$ be the subspace spanned by the $v_i$ 's. Because the $v_i$ 's are linearly independent, there exists a linear transformation $g$ such that $g(v_i)=w_i$ and $g(v)=0$ for $v\notin W$ . Let $K_i=\lbrace v_i\rbrace$ and $U_i=\lbrace w_i\rbrace$ . Then the $K_i$ 's are compact and the $U_i$ 's are open in the discrete space $V$ . Clearly $g\in\lbrace h\mid h(v_i)=w_i\rbrace=B(K_i,U_i)$ for each $i=1,\ldots,n$ . So $g$ lies in the neighborhood $B=\cap B(K_i,U_i)\subseteq \operatorname{End}_D(V)$ . Since $R$ is dense in $\operatorname{End}_D(V)$ , there is an $f\in R\cap B$ . This implies that $f(v_i)=w_i$ for all $i$ . $\qedsymbol$