This section is devoted to describing the triangularizable subalgebras of ${\mathrm{End}}_{k}(V)$. We begin with a technical lemma.

#### Lemma 6.1.

*Let **k* be a field, *V* a nonzero *k*-vector space, *R* a *k*-subalgebra of ${\mathrm{End}}_{k}\mathit{}\mathrm{(}V\mathrm{)}$, and *I* a topologically nilpotent ideal of *R*. Suppose that there exists $\mathrm{\{}{E}_{\alpha}\mathrm{\mid}\alpha \mathrm{\in}\mathrm{\Omega}\mathrm{\}}\mathrm{\subseteq}R$ such that the following conditions are satisfied.

(a)

${E}_{\alpha}{E}_{\alpha}-{E}_{\alpha}\in I$
* for all *
$\alpha \in \mathrm{\Omega}$
*, and *
${E}_{\alpha}{E}_{\beta}\in I$
* whenever *
$\alpha \ne \beta $.

(b)

*For every *
$T\in R$
* and *
$v\in V$
*, there exist *
${\alpha}_{1},\mathrm{\dots},{\alpha}_{n}\in \mathrm{\Omega}$,
${a}_{{\alpha}_{1}},\mathrm{\dots},{a}_{{\alpha}_{n}}\in k$
*, and *
$S\in I$
* such that*

$T(v)=\sum _{i=1}^{n}{a}_{{\alpha}_{i}}{E}_{{\alpha}_{i}}(v)+S(v).$

*Then **R* is triangularizable.

#### Proof.

Since *I* is topologically nilpotent, it is (strictly) triangularizable, by Theorem 4.1. Hence there exists a 1-dimensional *I*-invariant subspace $W\subseteq V$ (e.g., by Proposition 3.3), where necessarily $I(W)=\{0\}$. It follows that $U={\bigcap}_{T\in I}\mathrm{ker}(T)\ne \{0\}$. Moreover, for all $T\in R$ we have $T(U)\subseteq U$, since $ST\in I$ for all $S\in I$, and hence $ST(U)=\{0\}$.

We also note that ${E}_{\alpha}(U)\ne \{0\}$ for some $\alpha \in \mathrm{\Omega}$. For if it were the case that ${E}_{\alpha}(U)=\{0\}$ for all $\alpha \in \mathrm{\Omega}$, then the fact that $I(U)=\{0\}$ and condition (b) would imply that $R(U)=\{0\}$. Since *R* is assumed to be unital, this would mean that $U=\{0\}$, contrary to construction. Thus we may choose $u\in U\setminus \{0\}$ and $\beta \in \mathrm{\Omega}$ such that $v:={E}_{\beta}(u)\ne 0$. We claim that $\u3008v\u3009$ is *R*-invariant.

Since $I(U)=\{0\}$, the first relation in condition (a) gives $0={E}_{\beta}{E}_{\beta}(u)-{E}_{\beta}(u)$, and hence ${E}_{\beta}(v)=v$. Likewise, the second relation in condition (a) gives $0={E}_{\alpha}{E}_{\beta}(u)={E}_{\alpha}(v)$ for all $\alpha \in \mathrm{\Omega}\setminus \{\beta \}$. Moreover, $I(v)=0$, since $v={E}_{\beta}(u)\in U$, as noted in the first paragraph. Now let $T\in R$, and let ${\alpha}_{1},\mathrm{\dots},{\alpha}_{n}\in \mathrm{\Omega}$, ${a}_{{\alpha}_{1}},\mathrm{\dots},{a}_{{\alpha}_{n}}\in k$, and $S\in I$ be such that $T(v)={\sum}_{i=1}^{n}{a}_{{\alpha}_{i}}{E}_{{\alpha}_{i}}(v)+S(v)$, using condition (b). Then $T(v)={a}_{\beta}{E}_{\beta}(v)={a}_{\beta}v$ for some ${a}_{\beta}\in k$ (where ${a}_{\beta}=0$ if $\beta \notin \{{\alpha}_{1},\mathrm{\dots},{\alpha}_{n}\}$), showing that the 1-dimensional space $\u3008v\u3009$ is *R*-invariant.

Finally, it follows from Lemma 3.4 that the property of being a *k*-algebra with a topologically nilpotent ideal is inherited by quotients (see Definition 3.1). Moreover, it is easy to see that the property of having a subset satisfying conditions (a) and (b), with respect to the relevant topologically nilpotent ideal, is also inherited by quotients. Hence *R* is triangularizable, by Lemma 3.2.
∎

In the results that follow, we view subrings of ${\mathrm{End}}_{k}(V)$ as topological rings via the subspace topology induced by the function topology on ${\mathrm{End}}_{k}(V)$, and we view the product ${k}^{\mathrm{\Omega}}$ as a topological ring via the product topology, where each component copy of *k* is taken to be discrete. Also, see Section 2.1 for a review of the quotient topology and isomorphisms of topological algebras.

#### Proposition 6.2.

*Let **k* be a field, *V* a nonzero *k*-vector space, and *R* a *k*-subalgebra of ${\mathrm{End}}_{k}\mathit{}\mathrm{(}V\mathrm{)}$. Suppose there exist a topologically nilpotent ideal *I* of *R* and a set Ω such that $R\mathrm{/}I\mathrm{\cong}{k}^{\mathrm{\Omega}}$ as topological *k*-algebras. Then *R* is triangularizable, and $I\mathrm{=}\mathrm{TNil}\mathit{}\mathrm{(}R\mathrm{)}$.

#### Proof.

Let $\pi :R\to R/I$ be the canonical projection. Then, by the definition of the quotient topology, π is continuous. Moreover, it is a standard fact that π is necessarily an open map. (For if $\mathcal{\mathcal{U}}\subseteq R$ is an open set, then $\mathcal{\mathcal{U}}+I={\bigcup}_{i\in I}(\mathcal{\mathcal{U}}+i)$, and each $\mathcal{\mathcal{U}}+i$ is open in *R*, by the continuity of addition. Hence ${\pi}^{-1}(\pi (\mathcal{\mathcal{U}}))=\mathcal{\mathcal{U}}+I$ is open, and therefore so is $\pi (\mathcal{\mathcal{U}})$, again by the definition of the quotient topology.)

Since $R/I\cong {k}^{\mathrm{\Omega}}$ as topological *k*-algebras, there must exist $\{{E}_{\alpha}\mid \alpha \in \mathrm{\Omega}\}\subseteq R$, such that $\{{E}_{\alpha}+I\mid \alpha \in \mathrm{\Omega}\}\subseteq R/I$ is a set of orthogonal idempotents, and $R/I$ is the closure of the *k*-vector space spanned by this set. Now let $T\in R$, and let $\mathcal{\mathcal{U}}\subseteq R$ be an open neighborhood of *T*. Then, by the previous paragraph, $\pi (\mathcal{\mathcal{U}})$ is an open neighborhood of $\pi (T)$, and hence there exist ${\alpha}_{1},\mathrm{\dots},{\alpha}_{n}\in \mathrm{\Omega}$ and ${a}_{{\alpha}_{1}},\mathrm{\dots},{a}_{{\alpha}_{n}}\in k$ such that ${\sum}_{i=1}^{n}{a}_{{\alpha}_{i}}{E}_{{\alpha}_{i}}+I\in \pi (\mathcal{\mathcal{U}})$. It follows that ${\sum}_{i=1}^{n}{a}_{{\alpha}_{i}}{E}_{{\alpha}_{i}}+S\in \mathcal{\mathcal{U}}$ for some $S\in I$. Hence every open neighborhood of $T\in R$ contains an element of the form ${\sum}_{i=1}^{n}{a}_{{\alpha}_{i}}{E}_{{\alpha}_{i}}+S$. In particular, for all $v\in V$, there exist ${\alpha}_{1},\mathrm{\dots},{\alpha}_{n}\in \mathrm{\Omega}$, ${a}_{{\alpha}_{1}},\mathrm{\dots},{a}_{{\alpha}_{n}}\in k$, and $S\in I$ such that $T(v)={\sum}_{i=1}^{n}{a}_{{\alpha}_{i}}{E}_{{\alpha}_{i}}(v)+S(v)$. Thus $\{{E}_{\alpha}\mid \alpha \in \mathrm{\Omega}\}$ satisfies the conditions (a) and (b) of Lemma 6.1, and therefore *R* is triangularizable.

Next, we note that $I\subseteq \mathrm{TNil}(R)$, since every element of *I* is topologically nilpotent. To prove the opposite inclusion, let $T\in \mathrm{TNil}(R)$. Then for every open neighborhood $\mathcal{\mathcal{U}}\subseteq R$ of 0, there is some $n\in {\mathbb{Z}}^{+}$ such that ${T}^{n}\in \mathcal{\mathcal{U}}$. We claim that $T+I\in R/I$ enjoys the same property. Letting $\mathcal{\mathcal{U}}+I\subseteq R/I$ be an open neighborhood of *I*, where $\mathcal{\mathcal{U}}\subseteq R$, the continuity of the canonical projection $\pi :R\to R/I$ implies that $\mathcal{\mathcal{U}}+I$ is an open neighborhood of 0 in *R*. Hence there is some $n\in {\mathbb{Z}}^{+}$ such that ${T}^{n}\in \mathcal{\mathcal{U}}+I$, and therefore

${(T+I)}^{n}={T}^{n}+I\in \mathcal{\mathcal{U}}+I\subseteq R/I.$

But in $R/I\cong {k}^{\mathrm{\Omega}}$, the only element *r* having this property (that for every open neighborhood $\mathcal{\mathcal{U}}$ of 0, there is some $n\in {\mathbb{Z}}^{+}$ such that ${r}^{n}\in \mathcal{\mathcal{U}}$) is 0. Thus $T+I=I$, and hence $T\in I$. It follows that $I=\mathrm{TNil}(R)$.
∎

We observe that Example 5.5 shows that the conclusion of the previous proposition would cease to hold if we were to remove the hypothesis that *I* is topologically nilpotent. In that example, we constructed a subalgebra *R* of ${\mathrm{End}}_{k}(V)$ such that $R/\mathrm{rad}(R)\cong k$ as *k*-algebras, but *R* is not triangularizable and $\mathrm{rad}(R)$ is not topologically nilpotent. (Note that $R/\mathrm{rad}(R)$ and *k* in Example 5.5 are both discrete, and so the isomorphism between them is necessarily topological.)

We now turn to giving necessary conditions for subalgebras of ${\mathrm{End}}_{k}(V)$ to be triangularizable.

#### Proposition 6.3.

*Let **k* be a field, *V* a nonzero *k*-vector space, and *R* a *k*-subalgebra of ${\mathrm{End}}_{k}\mathit{}\mathrm{(}V\mathrm{)}$ triangular with respect to a well-ordered basis $\mathrm{(}\mathcal{B}\mathrm{,}\mathrm{\le}\mathrm{)}$. Define $\varphi \mathrm{:}R\mathrm{\to}{k}^{\mathcal{B}}$ by $\varphi \mathit{}\mathrm{(}T\mathrm{)}\mathrm{=}{\mathrm{(}{a}_{v}\mathrm{)}}_{v\mathrm{\in}\mathcal{B}}$, where ${a}_{v}\mathrm{\in}k$ is such that ${\pi}_{v}\mathit{}T\mathit{}\mathrm{(}v\mathrm{)}\mathrm{=}{a}_{v}\mathit{}v$, and ${\pi}_{v}\mathrm{\in}{\mathrm{End}}_{k}\mathit{}\mathrm{(}V\mathrm{)}$ denotes the projection onto $\mathrm{\u3008}v\mathrm{\u3009}$ with kernel $\mathrm{\u3008}\mathcal{B}\mathrm{\setminus}\mathrm{\{}v\mathrm{\}}\mathrm{\u3009}$. Then the following hold.

(1)

*The map *
$\varphi :R\to {k}^{\mathcal{\mathcal{B}}}$
* is a continuous *
*k*
*-algebra homomorphism.*

(2)

$\mathrm{TNil}(R)=\mathrm{ker}(\varphi )$
*, and this ideal is closed in the induced topology on *
*R*.

(3)

*If *
*R*
* is the *
*k*
*-subalgebra of *
${\mathrm{End}}_{k}(V)$
* consisting of all the transformations triangular with respect to *
$(\mathcal{\mathcal{B}},\le )$
*, then *
ϕ
* is open and surjective.*

*In particular, $R\mathrm{/}\mathrm{TNil}\mathit{}\mathrm{(}R\mathrm{)}$ is commutative.*

#### Proof.

(1) Let $S,T\in R$ and $v\in \mathcal{\mathcal{B}}$. Since *S* is triangular with respect to $\mathcal{\mathcal{B}}$, we can write $S(v)={b}_{v}v+{\sum}_{u<v}{b}_{u}u$ for some ${b}_{v},{b}_{u}\in k$ and $u\in \mathcal{\mathcal{B}}$. Since *T* is triangular with respect to $\mathcal{\mathcal{B}}$, we have

${\pi}_{v}TS(v)={\pi}_{v}T({b}_{v}v)+{\pi}_{v}T\left(\sum _{u<v}{b}_{u}u\right)={b}_{v}{\pi}_{v}T(v)={a}_{v}{b}_{v}v,$

where ${a}_{v}\in k$ is defined by ${\pi}_{v}T(v)={a}_{v}v$. It follows that $\varphi (TS)=\varphi (T)\varphi (S)$. It is also straightforward to show that ϕ is *k*-linear, and therefore ϕ is a *k*-algebra homomorphism.

Now let $\mathcal{\mathcal{U}}\subseteq {k}^{\mathcal{\mathcal{B}}}$ be a basic open set. Thus there exist ${v}_{1},\mathrm{\dots},{v}_{n}\in \mathcal{\mathcal{B}}$ and ${b}_{{v}_{1}},\mathrm{\dots},{b}_{{v}_{n}}\in k$ such that

$\mathcal{\mathcal{U}}=\{{({c}_{v})}_{v\in \mathcal{\mathcal{B}}}\mid {c}_{{v}_{i}}={b}_{{v}_{i}}\text{for}\mathrm{\hspace{0.25em}1}\le i\le n\}.$

Then

$\begin{array}{cc}\hfill {\varphi}^{-1}(\mathcal{\mathcal{U}})& =\{S\in R\mid {\pi}_{{v}_{i}}S({v}_{i})={b}_{{v}_{i}}{v}_{i}\text{for}\mathrm{\hspace{0.25em}1}\le i\le n\}\hfill \\ & =\bigcap _{i=1}^{n}\bigcup _{w\in \u3008\{u\in \mathcal{\mathcal{B}}\mid u<{v}_{i}\}\u3009}\{S\in R\mid S({v}_{i})={b}_{{v}_{i}}{v}_{i}+w\}\hfill \end{array}$

is open in the induced topology on *R*, from which it follows that ϕ is continuous.

(2) For any $T\in R$, we have

$\begin{array}{cc}\hfill \varphi (T)=0& \iff {\pi}_{v}T(v)=0\mathcal{\hspace{1em}}\text{for all}v\in \mathcal{\mathcal{B}}\hfill \\ & \iff T(v)\in \u3008\{u\in \mathcal{\mathcal{B}}\mid u<v\}\u3009\mathcal{\hspace{1em}}\text{for all}v\in \mathcal{\mathcal{B}}\hfill \\ & \iff T\in \mathrm{TNil}(R)\hspace{1em}(\text{by Proposition 2.4}),\hfill \end{array}$

and hence $\mathrm{TNil}(R)=\mathrm{ker}(\varphi )$. Since $\{0\}$ is closed in ${k}^{\mathcal{\mathcal{B}}}$ (which, as a product of discrete spaces, is Hausdorff), $\mathrm{ker}(\varphi )={\varphi}^{-1}(\{0\})$, and ϕ is continuous, we conclude that $\mathrm{ker}(\varphi )$ is closed in the induced topology on *R*.

(3) Suppose that *R* consists of all the transformations in ${\mathrm{End}}_{k}(V)$ triangular with respect to $\mathcal{\mathcal{B}}$, and let $\mathcal{\mathcal{U}}\subseteq R$ be an open subset. We shall show that $\varphi (\mathcal{\mathcal{U}})$ is open.

We may assume that

$\mathcal{\mathcal{U}}=\{S\in R\mid S({v}_{i})={w}_{i}\text{for}\mathrm{\hspace{0.25em}1}\le i\le n\}$

for some ${v}_{1},\mathrm{\dots},{v}_{n}\in \mathcal{\mathcal{B}}$ and ${w}_{1},\mathrm{\dots},{w}_{n}\in V$, since it is easy to see that every open set in *R* is a union of sets of this form. Write ${\pi}_{{v}_{i}}({w}_{i})={b}_{{v}_{i}}{v}_{i}$ ($1\le i\le n$) for some ${b}_{{v}_{i}}\in k$. Given a ${c}_{v}\in k$ for each $v\in \mathcal{\mathcal{B}}$, such that ${c}_{{v}_{i}}={b}_{{v}_{i}}$ for $1\le i\le n$, define $T\in {\mathrm{End}}_{k}(V)$ by $T({v}_{i})={w}_{i}$ for $1\le i\le n$, and $T(v)={c}_{v}v$ for all $v\in \mathcal{\mathcal{B}}\setminus \{{v}_{1},\mathrm{\dots},{v}_{n}\}$. Since *T* agrees with certain elements of *R* on $\{{v}_{1},\mathrm{\dots},{v}_{n}\}$, and each element of $\mathcal{\mathcal{B}}\setminus \{{v}_{1},\mathrm{\dots},{v}_{n}\}$ is an eigenvector of *T*, it is triangular with respect to $\mathcal{\mathcal{B}}$, and hence $T\in R$. Thus $T\in \mathcal{\mathcal{U}}$ and $\varphi (T)={({c}_{v})}_{v\in \mathcal{\mathcal{B}}}$. It follows that

$\varphi (\mathcal{\mathcal{U}})=\{{({c}_{v})}_{v\in \mathcal{\mathcal{B}}}\in {k}^{\mathcal{\mathcal{B}}}\mid {c}_{{v}_{i}}={b}_{{v}_{i}}\text{for}\mathrm{\hspace{0.25em}1}\le i\le n\},$

which is open in ${k}^{\mathcal{\mathcal{B}}}$. Hence ϕ is open, and it is surjective, by a simpler version of the same argument.

Finally, $R/\mathrm{TNil}(R)$ is isomorphic to a subalgebra of ${k}^{\mathcal{\mathcal{B}}}$, by (1) and (2), and is hence commutative.
∎

Combining the previous two propositions gives our main result.

#### Theorem 6.4.

*Let **k* be a field, *V* a nonzero *k*-vector space, and *R* a *k*-subalgebra of ${\mathrm{End}}_{k}\mathit{}\mathrm{(}V\mathrm{)}$. Then the following are equivalent.

*Moreover, if $\mathrm{(}\mathrm{2}\mathrm{)}$ holds and **R* is closed in the function topology, then $\mathrm{rad}\mathit{}\mathrm{(}R\mathrm{)}\mathrm{=}R\mathrm{\cap}\mathrm{rad}\mathit{}\mathrm{(}A\mathrm{)}$.

#### Proof.

Suppose that *R* is triangular with respect to some well-ordered basis $(\mathcal{\mathcal{B}},\le )$ for *V*. Let *A* be the *k*-subalgebra of ${\mathrm{End}}_{k}(V)$ consisting of all the transformations triangular with respect to $(\mathcal{\mathcal{B}},\le )$, and let $\varphi :A\to {k}^{\mathcal{\mathcal{B}}}$ be as in Proposition 6.3. Then, by that proposition, ϕ is a surjective open continuous *k*-algebra homomorphism, such that $\mathrm{ker}(\varphi )=\mathrm{TNil}(A)$. Hence $A/\mathrm{TNil}(A)\cong {k}^{\mathcal{\mathcal{B}}}$ as topological *k*-algebras (see, e.g., [11, Theorem 5.11]). Now, by Lemma 2.5, the closure $\mathrm{cl}(A)$ of *A* in the function topology on ${\mathrm{End}}_{k}(V)$ is triangular with respect to $\mathcal{\mathcal{B}}$. Hence $\mathrm{cl}(A)\subseteq A$, and so *A* is closed in ${\mathrm{End}}_{k}(V)$. Therefore, by Proposition 5.4, $\mathrm{TNil}(A)=\mathrm{rad}(A)$, and this ideal is topologically nilpotent, showing (2).

Now suppose that (2) holds. Then, by Proposition 6.2, *A* is triangularizable, and hence so is *R*, as a subset of *A*, proving (1). Moreover, by the same proposition, $\mathrm{rad}(A)=\mathrm{TNil}(A)$, and if *R* is closed, then also $\mathrm{rad}(R)=\mathrm{TNil}(R)$, by Proposition 5.4. Hence in this situation,

$\mathrm{rad}(R)=\mathrm{TNil}(R)=R\cap \mathrm{TNil}(A)=R\cap \mathrm{rad}(A),$

proving the final claim.
∎

The next example shows that for a closed triangularizable subalgebra *R* of ${\mathrm{End}}_{k}(V)$ it is not necessarily the case that $R/\mathrm{rad}(R)\cong {k}^{\mathrm{\Omega}}$ (as topological *k*-algebras) for some set Ω.

#### Example 6.5.

Let *k* be a countable field, $\mathrm{\Omega}={\mathbb{Z}}^{+}\cup \{\mathrm{\infty}\}$, and *V* a *k*-vector space with basis $\mathcal{\mathcal{B}}=\{{v}_{i}\mid i\in \mathrm{\Omega}\}$.
Then $\mathcal{\mathcal{B}}$ is well-ordered by the relation $\u2aaf$, where ${v}_{i}\u2aaf{v}_{j}$ if and only if either $i,j\in {\mathbb{Z}}^{+}$ and $i\le j$, or $j=\mathrm{\infty}$. For each $i\in {\mathbb{Z}}^{+}$, define ${E}_{i}\in {\mathrm{End}}_{k}(V)$ by

${E}_{i}({v}_{j})=\{\begin{array}{cc}{v}_{i}\hfill & \text{if}j\in \{i,\mathrm{\infty}\},\hfill \\ 0\hfill & \text{otherwise},\hfill \end{array}$

and extend linearly to all of *V*. Then $\{{E}_{i}\mid i\in {\mathbb{Z}}^{+}\}$ is a set of orthogonal idempotents, which is clearly triangular with respect to $(\mathcal{\mathcal{B}},\u2aaf)$. It follows that the *k*-subalgebra *R* of ${\mathrm{End}}_{k}(V)$ generated by this set is also triangular with respect to $(\mathcal{\mathcal{B}},\u2aaf)$.

Every element of *R* can be expressed in the form $a\cdot 1+{\sum}_{i=1}^{n}{a}_{i}{E}_{i}$ for some $n\in {\mathbb{Z}}^{+}$ and $a,{a}_{i}\in k$. Notice that

$\left(a\cdot 1+\sum _{i=1}^{n}{a}_{i}{E}_{i}\right)({v}_{\mathrm{\infty}})=a{v}_{\mathrm{\infty}}+\sum _{i=1}^{n}{a}_{i}{v}_{i},$

and so each element of *R* is completely determined by its action on ${v}_{\mathrm{\infty}}$. This implies that *R* is discrete in the topology induced on it by the function topology on ${\mathrm{End}}_{k}(V)$. The above computation also shows that if $a\ne 0$, then ${(a\cdot 1+{\sum}_{i=1}^{n}{a}_{i}{E}_{i})}^{m}({v}_{\mathrm{\infty}})\ne 0$ for all $m\in {\mathbb{Z}}^{+}$. Moreover, if $a=0$ but ${a}_{n}\ne 0$, then

${\left(a\cdot 1+\sum _{i=1}^{n}{a}_{i}{E}_{i}\right)}^{m}({v}_{n})={a}_{n}^{m}{v}_{n}\ne 0$

for all $m\in {\mathbb{Z}}^{+}$. Thus *R* has no nonzero topologically nilpotent elements, and hence $\mathrm{rad}(R)=0$, by Proposition 5.4.

Define $\varphi :R\to {k}^{\mathrm{\Omega}}$ as in Proposition 6.3. Specifically, $\varphi (a\cdot 1+{\sum}_{i=1}^{n}{a}_{i}{E}_{i})={({b}_{i})}_{i\in \mathrm{\Omega}}$, where ${b}_{i}={a}_{i}+a$ for $1\le i\le n$ and ${b}_{i}=a$ for all $j>n$. Thus ϕ is an injective *k*-algebra homomorphism (which is also continuous, by Proposition 6.3), with $\varphi (R)=\u3008{k}^{({\mathbb{Z}}^{+})}\cup \{1\}\u3009$, where ${k}^{({\mathbb{Z}}^{+})}$ is the direct sum of copies of *k* indexed by the elements of ${\mathbb{Z}}^{+}$ (with each element of this subring of ${k}^{\mathrm{\Omega}}$ understood to have 0 in the $\mathrm{\infty}$ coordinate). Therefore

$R/\mathrm{rad}(R)\cong R\cong \u3008{k}^{({\mathbb{Z}}^{+})}\cup \{1\}\u3009$

as *k*-algebras. In particular, $R/\mathrm{rad}(R)\ncong {k}^{\mathrm{\Delta}}$ for any set Δ, since such a Δ would need to be infinite, making ${k}^{\mathrm{\Delta}}$ uncountable, in contrast to *R* (given that *k* was assumed to be countable).

Next, let us show that *R* is closed. Let $\mathrm{cl}(R)$ denote the closure of *R* in the function topology on ${\mathrm{End}}_{k}(V)$, and let $T\in \mathrm{cl}(R)$. Then there exist $n\in {\mathbb{Z}}^{+}$ and $a,{a}_{1},\mathrm{\dots},{a}_{n}\in k$ such that $a\cdot 1+{\sum}_{i=1}^{n}{a}_{i}{E}_{i}\in R$ agrees with *T* on ${v}_{\mathrm{\infty}}$. Since each element of *R* is completely determined by its action on ${v}_{\mathrm{\infty}}$, and since every neighborhood of *T* must contain an element of *R*, it follows that $T=a\cdot 1+{\sum}_{i=1}^{n}{a}_{i}{E}_{i}$, which belongs to *R*, and hence $R=\mathrm{cl}(R)$.

We also observe that ϕ is not open. Since *R* is discrete, $\{{E}_{1}\}\subseteq R$ is open. On the other hand, the intersection of any open subset of ${k}^{\mathrm{\Omega}}$ with $\u3008{k}^{({\mathbb{Z}}^{+})}\cup \{1\}\u3009$ is infinite, and hence $\varphi (\{{E}_{1}\})$ cannot be open in the induced topology on $\u3008{k}^{({\mathbb{Z}}^{+})}\cup \{1\}\u3009$.

Finally, let *A* be the subalgebra of ${\mathrm{End}}_{k}(V)$ consisting of all the transformations triangular with respect to $(\mathcal{\mathcal{B}},\u2aaf)$, and extend ϕ to a map ${\varphi}^{\prime}:A\to {k}^{\mathrm{\Omega}}$ (again, defined as in Proposition 6.3). Then ${\varphi}^{\prime}$ is open and surjective, by Proposition 6.3(3), but ${\varphi}^{\prime}$ is not closed, since it sends the closed set *R* to $\u3008{k}^{({\mathbb{Z}}^{+})}\cup \{1\}\u3009$, whose closure in ${k}^{\mathrm{\Omega}}$ is ${k}^{\mathrm{\Omega}}$.

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