Recall that the monoid H we call

A topological monoid is a set H such that

Let us choose any open set W containing e such that $W\subseteq U$. Note that W is also a neighborhood of e in H because it contains the open set $\left\{e\right\}$. Let’s define $V=W\cap W^{-1}$, where $W^{-1}=\{x^{-1}:x\in W\}$.

We will show that V is an open subset of H and contains e. To show that V is an open subset of H, it is enough to show that W and $W^{-1}$ are open subsets of H. We know that W is open by definition. To show that $W^{-1}$ is open, let’s use the fact that the action of monoid is continuous. Let $x\in W^{-1}$ and let $U_x\subseteq W^{-1}$ be any open set containing x. Then there exists an open set $V_x\subseteq H$ such that $x{V}_x\subseteq U_x$. Since $x\in W^{-1}$, we have $x=y^{-1}$ for some $y\in W$. So $yx{V}_x\subseteq y{U}_x\subseteq W$. But $yx=e$, so $V_x\subseteq W$. Therefore $x\in V_x\subseteq W\cap W^{-1}=V$. This means that V is the neighborhood of x in H. Since x was any element of $W^{-1}$, we get that $W^{-1}$ is open in H.

To show that V contains e, it is enough to note that $e\in W$ and $e\in W^{-1}$, since e is the neutral element of the monoid.

We will show that $VV\subseteq U$ and the closure of $V^{\prime }$ of the neighborhood of V is contained in U. Let $x,y\in V$ and show that $xy\in U$. We know that $x\in W$ and $y\in W^{-1}$, so $xy\in W{W}^{-1}$. But $W{W}^{-1}\subseteq U$, since for any $a\in W$ and $b\in W^{-1}$ we have $ab=c$ for some $c\in W$ (because $b=a^{-1}$ for some $a\in W$). So $xy\in U$. To show that the closure of $V^{\prime }$ of the neighborhood V is contained in U, it is enough to show that every boundary point of the set V belongs to U. Let $z\in H$ be the limit point of the set V and let $T\subseteq H$ be any open set containing z. We will show that $T\cap V\ne \varnothing$. Since z is the limit point of the set V, then there exists a sequence ${\left(x_n\right)}_{n=1}^{\infty }$ of the elements of the set V convergent to z. Since the action of the monoid is continuous, the sequence ${\left(x_n{x}_n\right)}_{n=1}^{\infty }$ of the elements of the set $VV$ also converges to $zz$. But we know that $VV\subseteq U$, so $zz\in U$. So $T\cap U\ne \varnothing$. Since T was any open set containing z, we obtain that z belongs to the interior of the set U. Therefore, the closure of $V^{\prime }$ of the neighborhood V is contained in U. □

The necessity of conditions (1), (2), (3) is obvious.

To prove that H is a topological monoid from conditions (1), (2), (3), it is enough to show that the action of the monoid is continuous with respect to the topology on H. In other words, we need to show that for any open sets U and V in H, the product $U\times V$ is an open subset of $H\times H$ and the image of this product by the monoid action is an open subset of H.

Let U and V be any open sets in H. We will show that $U\times V$ is an open subset of $H\times H$. Let $(x,y)\in U\times V$ and let $W_x\subseteq U$ and $W_y\subseteq V$ be any open sets containing x and y, respectively. Then $W_x\times W_y$ is an open subset of $U\times V$ containing $(x,y)$, because the Cartesian product of open sets is an open set. Since $(x,y)$ was any element of $U\times V$, we get that $U\times V$ is open in $H\times H$.

Let U and V be any open sets in H. We will show that the image of the product $U\times V$ by the monoid action is an open subset of H. Let $z\in UV$, where $UV=\{xy:x\in U,y\in V\}$. Then there exist $x\in U$ and $y\in V$ such that $z=xy$. Let’s take advantage of the fact that the shifts $x\mapsto hx$ and $x\mapsto xh$ are continuous mappings for each $h\in H$. Then there exist open sets $X\subseteq U$ and $Y\subseteq V$ such that $xX\subseteq U$ and $Yy\subseteq V$. Note that $Xy$ and $xY$ are also open sets because they are images of sets opened by continuous maps. Let’s define $Z=Xy\cap xY$. Note that Z is an open subset of $UV$ containing z because it is the intersection of open sets and $xy\in Xy\cap xY$. We will show that Z is also a neighborhood of z in H. Let $T\subseteq H$ be any open set containing z. We will show that $T\cap Z\ne \varnothing$. Let’s take advantage of the fact that the action of the monoid is continuous at the point $(e,e)$, where e is the neutral element of the monoid. Then there exist open sets $E_1$ and $E_2$ in H such that $e\in E_1$, $e\in E_2$ and $E_1\times E_2\subseteq T$. Since $x\in X\subseteq U$ and $y\in Y\subseteq V$, we have $x{E}_1\subseteq U$ and $E_{2}y\subseteq V$. But we know that $x{E}_{1}y=xy{E}_{1}y=xy{E}_{2}y=x{E}_{2}yy=x{E}_{2}y=z{E}_{2}y\subseteq T$. Therefore $xy\in T\cap Xy\subseteq T\cap Z$. This means that Z is the neighborhood of z in H. Since z was any element of $UV$, we get that $UV$ is open in H. □

The subset of the topological monoid H containing the neighborhood of neutral element is called the kernel of H.

The proof of the first part is obvious. In the second part, we define the topology on H by saying that U is an open set if for every $x\in U$ there exists a $V\in \Sigma$ such that $xV\subset U$. Checking all the required properties of such a defined family of sets is an easy exercise. □

The conditions $\left(a1\right)$ and $\left(a2\right)$ determine the topology. Indeed, let be given topologies $T_1$ and $T_2$ on $H/M$ satisfying $\left(a_1\right)$ and $\left(a2\right)$. Let us denote by $(H/M,T_i)$ the topological space $H/M$ with topology $T_i$ ($i=1,2$) and let $j:(H/M,T_1)\to (H/M,T_2)$ will be an identity map. Then, since $j\circ \pi :H\to (H/M,T_2)$ is a continuous mapping by condition $\left(a1\right)$ (applied to topology $T_2$), so by condition $\left(a2\right)$ (used for topology $T_1$) j is a continuous map. By exchanging the roles of $T_1$ and $T_2$ we get the continuity of $j^{-1}$. Therefore j is a homeomorphism and the topologies $T_1$ and $T_2$ are identical.

We will now show that there exists a topology satisfying $\left(a1\right)$ and $\left(a2\right)$. We define it as follows: Let U be a subset of $H/M$ that is open if and only if its counterimage ${\pi }^{-1}\left(U\right)$ is open in H. It is easy to see that all the conditions that the topology should meet are met here, and the continuity of the $\pi$ mapping also follows directly from the definition. To show that the condition $\left(a2\right)$ is satisfied, assume that $f\circ \pi :H\to P$ is a continuous mapping. Let V be an open set in P. Then $U={(f\circ \pi )}^{-1}\left(V\right)$ is an open set in H. So $\pi \left(U\right)$ is an open set in $H/M$, but $\pi \left(U\right)=f^{-1}\left(V\right)$, so f is a continuous mapping.

Let’s move on to proof $\left(b\right)$. Let U be an open set in H. Then $U_h$ for $h\in H$ is also an open set, so $UM={\bigcup }_{m\in M}{U}_m$ is an open set. But $UM={\pi }^{-1}\left(\pi \left(U\right)\right)$. Therefore, by the definition of quotient topology, the set $\pi \left(U\right)$ is open in $H/M$. □

It is enough to prove the continuity of the mapping $H/M\times H/M\to H/M$, $(h_{1}M,h_{2}M)\to h_1{h}_{2}M$. Let U be the neighborhood of element $h_1{h}_{2}M$ in $H/M$. Then ${\pi }^{-1}\left(U\right)$ is the neighborhood of $h_1{h}_2$ in H. There are $V_i$-neighborhoods $h_i$, $i=1,2$ in H such that $V_1{V}_2\subset {\pi }^{-1}\left(U\right)$. Therefore $\pi \left(V_1\right)$ and $\pi \left(V_2\right)$ are the neighborhoods of $h_{1}M$ and $h_{2}M$ and $\pi \left(V_1\right)\pi \left(V_2\right)=\pi \left(V_1{V}_2\right)\subset \pi \left({\pi }^{-1}\left(U\right)\right)=U$, which proves the continuity of our mapping. □

Since $\pi$ is open, $\pi \left(hM\right)=\left\{hM\right\}$ is an open set in $H/M$. Therefore, the points $H/M$ are open sets. □

Only the sufficiency of the continuity condition in e requires proof. Let V be the neighborhood of $\overline{e}$ in M. Then for $h\in H$ the set $\alpha \left(h\right)V$ is the neighborhood of the element $\alpha \left(h\right)$ in M. By assumption, there exists a neighborhood U of e in H such that $\alpha \left(U\right)\subset V$. Then $hU$ is the neighborhood of h in H and we have $\alpha \left(hU\right)=\alpha \left(h\right)\alpha \left(U\right)\subset \alpha \left(h\right)V$. This means that the homomorphism $\alpha$ is continuous at the point $h\in H$. □

The proof of this Proposition is an obvious modification (using Theorem 2.7 and its conclusions) of an analogous purely algebraic theorem. □

A topological monoid H is called compact if H is a compact topological space. H is called a locally compact monoid if there is a neighborhood of e whose closure is compact.

A topological monoid H is called connected if H is a connected topological space.

(1) Of course, every topological group is a topological monoid.

Let I be an idempotent ideal in R. Then $I^2=I$. Let $x\in I$. Then $x=x^2+x-x^2\in I^2+(x-x^2)H$.

Note that $I^2$ is an idempotent ideal because ${\left(I^2\right)}^2=I^4=I^2$. Furthermore, $x-x^2$ is idempotent because ${(x-x^2)}^2=x^2-2x^3+x^4=x^2-x^2+x^4=x-x^2$. Therefore I is the sum of the idempotent ideals $I^2$ and $(x-x^2)H$ for each $x\in I$. □

Let $A_1$, ⋯, $A_n$ be sums of idempotent ideals in H. Let $B=A_1\cap \cdots \cap A_n$. Let $C=I_1\cdots I_n$, where $I_j$ is the product of all idempotent ideals appearing in $A_j$. We will show that $B=C$.

We have $B\subseteq C$, because if $x\in B$, then x belongs to every $A_j$, therefore to every $I_j$, and therefore to $I_1\cdots I_n=C$.

We have $C\subseteq B$. If $x\in C$, then x is the sum of a finite number of elements of the form $i_1{i}_2\cdots i_n$, where $i_j\in I_j$. Each such element belongs to every $A_j$, because $A_j$ is the sum of the ideals of idempotent ideals, so it is closed to multiplication by the elements of these ideals. Therefore, x belongs to each $A_j$, and therefore to the intersection $A_1\cap \cdots \cap A_n$. So $B=C$. □

Let $I_1$, ⋯, $I_n$ be idempotent ideals in H. Let $J=I_1\cdots I_n$. We will show that J is an idempotent ideal.

It is easy to check that J is an ideal, because it is a product of ideals. To show that J is idempotent, it is enough to show that $J^2\subseteq J$.

Let $x\in J^2$. Then x is the sum of a finite number of elements of the form $j_1{j}_2$, where $j_1$, $j_2\in J$. Each such element $j_1{j}_2$ is the product of a finite number of elements of the form $i_1{i}_2\cdots i_n$, where $i_j\in I_j$. Since each $I_j$ is idempotent, then $i_j=i_j^2$. Therefore x is the product of a finite number of elements of the form $i_1^2{i}_2^2\cdots i_n^2$, where $i_j\in I_j$. But this means that $x\in J$, because J is closed to multiplication by elements of $I_j$. So $J^2\subseteq J$. □

The empty set is the sum of the empty family of idempotent ideals, so it is an open set. The entire monoid H is an idempotent ideal, because $H^2=H$, so it is also an open set.

Let $A_i$ for $i\in I$ (I as a set of indices) be a family of open sets, i.e. sums of idempotent ideals. Then the sum of $A_i$ for $i\in I$ is equal to the sum of all idempotent ideals appearing in $A_i$. Every idempotent ideal is the sum of idempotent ideals (from Lemma 3.1), so the sum $A_i$, $ı\in I$, is an open set.

Let $A_1$, ⋯, $A_n$ be open sets, i.e. sums of idempotent ideals. Then the intersection $A_1\cap \cdots \cap A_n$ is equal to the product of all idempotent ideals appearing in $A_1$, ⋯, $A_n$ (from Lemma 3.2). Every product of idempotent ideals is an idempotent ideal (from Lemma 3.3), so $A_1\cap \cdots \cap A_n$ is an open set. □

Let S denote the set of all ideals in H that are contained in I and do not contain any prime ideal. Note that S is nonempty because $\left(0\right)$ belongs to S. Note also that S is partially ordered with respect to inclusion. Let us now apply the Zorn’s Lemma to S. Therefore there is a maximal element M in S, i.e. there is no other element N in S such that M is proper contained in N. We will show that M is a prime ideal. Suppose that M is not prime. Then there are elements a and b in H such that $ab$ belongs to M, but neither a nor b belongs to M. Consider the ideals $M+\left(a\right)$ and $M+\left(b\right)$. It is easy to check that these are ideals in H that are contained in I and contain M. Moreover, neither of them contains the prime ideal, otherwise M would also contain this ideal, which contradicts the assumption that M belongs to S. Therefore, $M+\left(a\right)$ and $M+\left(b\right)$ belong to S. But this means that M is not maximal in S, which is contrary to the definition of M. Therefore M is a prime ideal, which completes the proof of the Lemma. □

Let H be a monoid and I be a square-free ideal in H. Suppose that I is not the sum of square-free ideals. Then there exists a prime ideal P in H such that P is contained in I and P is minimal with this property. From Lemma 3.5 we know that such an ideal P exists. Moreover, P is square-free because it is prime. Therefore $I=P+J$ for some ideal J in H. But then J is also square-free, because if $x^2$ belongs to J, then $x^2$ belongs to I, so x belongs to I, an x belongs to $P+J$, that is, $x=p+j$ for some p belonging to P and j belonging to J. Squaring both sides, we get $x^2=p^2+2pj+j^2$. Since $x^2$ belongs to J, then $p^2+2pj+j^2$ belongs to J. But $p^2$ belongs to P, so $p^2$ belongs to $P+J$, so $p^2$ belongs to J. Similarly, $2pj$ belongs to $P+J$, so $2pj$ belongs to J. Therefore $j^2$ belongs to J. But J is square-free, so j belongs to J. Therefore, x belongs to J. But this means that I is the sum of square-free ideals, which contradicts the assumption. Therefore I is the sum of square-free ideals, which completes the proof of the Lemma. □

Let $A_1$, ⋯, $A_n$ be sums of square-free ideals in H. Let $B=A_1\cap \cdots \cap A_n$. Let $C=I_1\cdots I_n$, where $I_j$ is the product of all square-free ideals appearing in $A_j$. We will show that $B=C$.

We have $B\subseteq C$, because if $x\in B$, then x belongs to every $A_j$, so to every $I_j$, and therefore to $I_1\cdots I_n=C$.

We have $C\subseteq B$. If $x\in C$, then x is the sum of a finite number of elements of the form $i_1{i}_2\cdots i_n$, where $i_j\in I_j$. Each such element belongs to every $A_j$, because $A_j$ is the sum of the ideals of square-free ideals, so it is closed to multiplication by elements of these ideals. Therefore, x belongs to each $A_j$, and therefore to the common part $A_1\cap \cdots \cap A_n$. So $B=C$. □

Let $I_1$, ⋯, $I_n$ be square-free ideals in H. Let $J=I_1\cdots I_n$. We will show that J is a square-free ideal.

It is easy to check that J is an ideal, because it is a product of ideals. To show that J is idempotent, it is enough to show that $J^2\subseteq J$.

Let $x\in J^2$. Then x is the sum of a finite number of elements of the form $j_1{j}_2$, where $j_1$, $j_2\in J$. Each such element $j_1{j}_2$ is the product of a finite number of elements of the form $i_1{i}_2\cdots i_n$, where $i_j\in I_j$. Since each $I_j$ is idempotent, then $i_j=i_j^2$. Therefore x is the product of a finite number of elements of the form $i_1^2{i}_2^2\cdots i_n^2$, where $i_j\in I_j$. But this means that $x\in J$, because J is closed to multiplication by elements of $I_j$. So $J^2\subseteq J$. □

The empty set is the sum of the empty family of square-free ideals, so it is an open set. The entire monoid H is a square-free ideal, because if $x^2\in H$, then $x\in H$, so H is also an open set.

Let $A_i$ for $i\in I$ (I as a set of indices) be a family of open sets, i.e. sums of square-free ideals. Then the sum of $A_i$ for $i\in I$ is equal to the sum of all square-free ideals appearing in $A_i$. Every square-free ideal is the sum of square-free ideals (Lemma 3.6), so the sum $A_i$, $ı\in I$, is an open set.

Let $A_1$, ⋯, $A_n$ be open sets, i.e. sums of square-free ideals. Then the intersection $A_1\cap \cdots \cap A_n$ is equal to the product of all square-free ideals appearing in $A_1$, ⋯, $A_n$ (Lemma 3.7). Every product of square-free ideals is a square-free ideal (Lemma 3.8), so $A_1\cap \cdots \cap A_n$ is an open set. □

It is enough to show that each element of the complement of B is the product of a finite number of elements from the idempotent ideals forming $A_1$, ⋯, $A_n$. Let x belongs to the complement of B. Then $x\notin B$, i.e. is not the sum of a finite number of elements from $A_1$, ⋯, $A_n$. Therefore, for every $i=1,\cdots ,n$, there exists an idempotent ideal $J_i$ such that $x\notin J_i$, and every element of $J_i$ belongs to $A_i$. Then $J_1\cdots J_n$ is the product of idempotent ideals, and $x\in J_1\cdots J_n$. Moreover, $J_1\cdots J_n$ is included in the complement of B, because if $y\in J_1\cdots J_n$, then $y\notin B$. Therefore, the complement of B is the product of idempotent ideals. □

Let I be the product of idempotent ideals in the monoid H. We will show that I is a closed set in ${Ł}_1$. It is enough to show that the complement of I is an open set in ${Ł}_1$.

Let $x\in H\setminus I$. Then $x\notin I$, i.e. is not the product of a finite number of elements from the idempotent ideals that creating I. Therefore, there exists an idempotent ideal J such that $x\notin J$, and every element of J belongs to I. Then J is an open set in ${Ł}_1$, because it is the sum of idempotent ideals, and $x\in J$. Furthermore, J is included in the complement of I. Therefore J is the neighborhood of x in the complement of I. Since x was any element of the complement of I, it means that the complement of I is an open set in ${Ł}_1$. □

Let I be the product of square-free ideals in the monoid H. We will show that I is a closed set in ${Ł}_2$. It is enough to show that the complement of I is an open set in ${Ł}_2$.

Let $x\in H\setminus I$. Then $x\notin I$, i.e. is not the product of a finite number of elements from the square-free ideals that creating I. Therefore, for every $i=1,\cdots ,n$, there exists a square-free ideal $J_i$ such that $x\notin J_i$, and every element of $J_i$ belongs to I. Then $J_1+\cdots +J_n$ is an open set in ${Ł}_2$ because it is the sum of square-free ideals, and $x\in J_1+\cdots +J_n$. Moreover, $J_1+\cdots +J_n$ is included in the complement of I, because if $y\in J_1+\cdots +J_n$, then $y\notin I$. Therefore $J_1+\cdots +J_n$ is the neighborhood of x in the complement of I. Since x was any element of the complement of I, it means that the complement of I is an open set in ${Ł}_2$. □

(a) ⇒ (b)

Assume $x\in \overline{A}$ in ${Ł}_1$. Let J be any idempotent ideal containing A. Then J is an open set in ${Ł}_1$ because it is the sum of idempotent ideals. Since $x\in \overline{A}$, then x belongs to every neighborhood of A, and therefore $x\in J$. So $x\in J+A$.

(b) ⇒ (a)

Assume that for any idempotent ideal J containing A, $x\in J+A$. We want to show that $x\in \overline{A}$. Let U be any neighborhood of x in ${Ł}_1$. Then U is the sum of idempotent ideals, i.e. $U=I_1+\cdots +I_n$ for some idempotent ideals $I_1$, ⋯, $I_n$. Let $J=I_1+\cdots +I_n+A$, i.e. $x\in U+A$. But $U+A=H$, because U is open and A is non-empty. So $x\in H$, or $x\in \overline{A}$.

(a) ⇔ (c)

The closure of the set A in the ${Ł}_1$–topology is the smallest closed set containing A, i.e. the intersection of all closed sets containing A. A closed set in the ${Ł}_1$–topology is the sum of idempotent ideals that are maximal in the sense of inclusion. Such a closed set can also be written as the product of all idempotent ideals (as we gave earlier) that contain it, but this is not necessary for the proof. Therefore $x\in \overline{A}$ if and only if x belongs to every idempotent ideal that contains A. □

(a) ⇒ (b)

Let J be a square-free ideal in H and let $x^2\in J$. Then $x\in J$, because J is square-free. So $x\in A+J$, which is an open set containing A. Since $x\in \overline{A}$, then $A+J$ must intersect A, i.e. $A\cap J\ne \varnothing$.

(b)⇒ (a)

Let U be any open set containing x. Then U is the sum of some square-free ideals, i.e. $U=I_1+\cdots +I_n$. Since $x\in U$, then $x\in I_k$ for some k. Then $x^2\in I_k$, so $A\cap I_k\ne \varnothing$. Therefore, from the condition $U\cap A\ne \varnothing$ we have $x\in \overline{A}$. □

(a)⇒(b)

Assume $x\in IntA$. Then there exists an open set U such that $x\in U$ and $U\subseteq A$. Since U is an open set in ${Ł}_1$, then U is an idempotent ideal, i.e. $U^2=U$. Therefore $x\in U^2$ and $U^2\subseteq A$. Let $I=U^2$. Then I is an idempotent ideal such that $x\in I$ and $I\subseteq A$.

(b)⇒ (a)

Assume there is an idempotent ideal I in H such that $x\in I$ and $I\subseteq A$. Then I is an open set in ${Ł}_1$ because $I^2=I$. Therefore $x\in I$ and $I\subseteq A$. Let us denote $U:=I$. Then U is an open set such that $x\in U$ and $U\subseteq A$. □

(a)⇒(b)

Assume $x\in IntA$. Then there exists an open set U such that $x\in U$ and $U\subseteq A$. Since U is an open set in ${Ł}_2$, then U is the sum of some square-free ideals, i.e. $U=I_1+\cdots +I_n$, where $I_1$, ⋯, $I_n$ are square-free ideals in H. Since $x\in U$, then $x\in I_k$, for some k. Then $x\in I_k$ and $I_k\subseteq A$. Let us denote $I:=I_k$. Then I is a square-free ideal such that $x\in I$ and $x\in I$, $I\subseteq A$.

(b)⇒(a)

Assume there is a square-free ideal I in H such that $x\in I$, $I\subseteq A$. Then I is an open set in ${Ł}_2$, because I is the sum of some square-free ideals, i.e. $I=J_1+\cdots +J_n$, where $J_1$, ⋯, $J_n$ are ideals square-free in H. Therefore $x\in I$, $I\subseteq A$. Let us denote $U:=I$. Then U is an open set such that $x\in U$ and $U\subseteq A$. □

We will perform the proof for idempotent ideals in the ${Ł}_1$–topology. The proof for square-free ideals in the ${Ł}_2$–topology will be analogous.

(a) ⇒ (b)

Assume $x\in FrA$. Then $x\in \overline{A}$ but $x\notin IntA$. Then by Theorem 3.16 x belongs to every idempotent ideal that contains A. From Theorem 3.17 it follows that there is no idempotent ideal such that x belongs to the idempotent ideal contained in A.

(b)⇒ (a)

Assume that x belongs to every idempotent ideal that contains A, but there is no idempotent ideal such that x belongs to that idempotent ideal that contains A. From Theorem 3.16 it follows that x belongs to the closure of the set A. From Theorem 3.17 it follows that x does not belong to the interior of the set A. Therefore x belongs to the boundary of the set A. □

Let $S=\mathbb{R}\left[x\right]/\left(x^2\right)$ be the set of all polynomials of degree at most 1. Then S is a Borel set in ${Ł}_1$, because it is the complement of the open set $\left(x^2\right)$ in ${Ł}_1$.

The set S is not Borelian in ${Ł}_2$ because it is not the sum of square-free ideals. Indeed, suppose that S is the sum of square-free ideals $I_1$, ⋯, $I_n$, where $I_k\subset \mathbb{R}\left[x\right]$. Then for each k we have $I_k=\{p\left(x\right)\in \mathbb{R}\left[x\right]:p{\left(x\right)}^2\in I_k\}$. Therefore, if $p\left(x\right)\in I_k$, then $p{\left(x\right)}^2\in I_k$. But then $p{\left(x\right)}^4$, $p{\left(x\right)}^8$, $p{\left(x\right)}^{16}$, $\cdots \in I_k$. Therefore, if $p\left(x\right)\ne 0$, then $I_k$ contains infinitely many powers of $p\left(x\right)$, which contradicts $I_k$ being an ideal. Therefore, any square-free ideal containing S must be trivial, i.e. equal to $\left\{0\right\}$ or $\mathbb{R}\left[x\right]$. Recall that if S is dense in $\mathbb{R}\left[x\right]$, this means that the closure of S in the topology ${Ł}_2$ is equal to $\mathbb{R}\left[x\right]$. This means that S has a nonempty intersection with all open sets in the topology ${Ł}_2$. On the other hand, if $I_1$, ⋯, $I_n$ are square-free ideals in $\mathbb{R}\left[x\right]$, then the sum $I_1+\cdots +I_n$ is closed in the topology ${Ł}_2$ because it is the product of square-free ideals. But if $I_1$, ⋯, $I_n$ are different from $\mathbb{R}\left[x\right]$, it means they are equal to $\left\{0\right\}$. Then the sum of $I_1+\cdots +I_n$ is also equal to $\left\{0\right\}$. Therefore, the sum of square-free ideals containing S cannot yield S because S is dense in $\mathbb{R}\left[x\right]$ and $\left\{0\right\}$ is not.

${Ł}_1$– and ${Ł}_2$–topologies are not equivalent. For example, let $H={\mathbb{Z}}_8$ and let $I=\{0,2,4,6\}$ be its ideal.

The ideal I is square-free, which is easy to show. But I is not idempotent because $I^2=\{0,4\}\ne I$.

The topology ${Ł}_2$ is stronger than ${Ł}_1$, because every set open in ${Ł}_1$ is open in ${Ł}_2$, but vice versa generally not.

This follows from the example 3.23 and from the property that every idempotent ideal is square-free. □

It is enough to show that if x, $y\in H$, where $x\ne y$, then there exists an idempotent/square-free ideal I such that $x\in I$, $y\notin I$ or vice versa.

Let J be the ideal generated by $x-y$. Then $J^2=J$, so J is an idempotent ideal. The ideal J is also square-free (because it is an idempotent ideal). Moreover, $x\in J$, $y\notin J$, because if it belonged, then $x-y\in J$, i.e. $x=y$, which is contrary to the assumption. Therefore J is the ideal sought. □

Let H be a monoid. Then $\left(H{,}_1\right)$ and $\left(H{,}_2\right)$ are not $T_1$–spaces.

It is enough to show that there are two distinct elements x, $y\in H$ for which there are no disjoint idempotent ideals I, J such that $x\in I$, $y\in J$. Let x, $y\in H$, x, $y\ne 0$. Then the ideals generated by x and y, i.e. $\left(x\right)$ and $\left(y\right)$, are idempotent ideals. Also, $x\in \left(x\right)$, $y\in \left(y\right)$. However, $\left(x\right)$ and $\left(y\right)$ are not disjoint, because $\left(x\right)+\left(y\right)$ is an idempotent ideal containing both x and y. So we cannot separate x and y by idempotent ideals.

The argument for square-free ideals is analogous. □

(a) ⇒ (b)

Let H be a factorial monoid. We will show that the topological space S is a compact, metrizable space and satisfies the second countability axiom.

To show that S is a compact space, we need to show that every open cover has a finite subcover. In a factorial monoid, each non-zero element can be represented as a product of prime elements. Since square-free ideals are generated by the products of different primes, there are only a finite number of different square-free ideals. Therefore, if we have an open cover of S, we can choose a finite set of square-free ideals that cover S. This shows that S is a compact space.

A space S is metrizable if there is a metric d such that the topology induced by d is equivalent to the ${Ł}_2$–topology. We can define the metric d on S as follows: for two square-free ideals I and J, let $d(I,J)$ be equal to the number of primes that must be added or removed to transform I in J. This metric is well defined because in a factorial monoid every element is uniquely represented as a product of prime elements, and therefore every square-free ideal is uniquely generated by the products of different prime elements.

To show that S satisfies the second countability axiom, we need to find a countable basis for the ${Ł}_2$–topology. In a factorial monoid, each square-free ideal is generated by a finite set of square-free elements. We can therefore take all possible finite combinations of square-free elements to generate a countable family of square-free ideals. This family will be the basis of the ${Ł}_2$–topology because any square-free ideal can be expressed as the sum of a finite number of ideals in this family.

(b)⇒(a)

We will show that each element $h\in H$ can be represented as a product of square-free elements. Since S is a compact topological space, every sequence of square-free ideals has a subsequence convergent to some square-free ideal. This means that each element h can be represented as a limit of a sequence of square-free elements, i.e. as a product of square-free elements.

Using the metrizability of S, we will show that prime elements in H correspond to isolated points in S. If p is a prime element in H, then $pH$ is a square-free ideal and is an isolated point in S. Thanks to metrizability, each such point is separate from the others, which means that the prime elements are uniquely defined.

Using the second countability axiom, we will show that there is a countable basis of primes in H. Each element $h\in H$ can be represented as a product of elements from this base, which means that this representation is unambiguous.

Finally, since each element $h\in H$ can be represented as a product of prime elements uniquely, then H is a factorial monoid. □

(a) ⇒ (b)

Assume that the monoid H satisfies the ACCP condition, i.e. each infinite chain of increasing ideals $I_1\subset I_2\subset \cdots$ stabilizes, i.e. there exists n such that $I_n=I_{n+1}=dots$. We will show that S is metrizable and compact.

A space S will be metrizable if there is a metric consistent with the ${Ł}_2$–topology. We can define a metric on S taking advantage of the fact that in an ACCP-monoid there are only a finite number of principal ideals that can generate square-free ideals. This metric can be defined, for example, as the minimum number of steps needed to go from one square-free ideal to another by adding or removing generators.

To show the compactness of S, we must show that every open cover S has a finite subcover. In an ACCP-monoid, where infinitely long chains of increasing principal ideals cannot be created, each square-free ideal is generated by a finite set of elements. This means that there are only a finite number of different square-free ideals, so every open cover S must have a finite subcover.

(b)⇒ (a)

We start with the assumption that S is a compact and metrizable space. From the properties of a compact space it follows that every sequence of elements from S has a convergent subsequence. In the context of the monoid H, this means that for every sequence of square-free ideals there is a subsequence that converges to some ideal in S.

The metricizability of S implies that there is a metric that defines the ${Ł}_2$–topology. From the definition of a metrizable space it follows that every Cauchy sequence converges, which in the context of the monoid H means that every sequence of square-free ideals that is "close" to being constant (in the sense of the metric) must stabilize.

These properties suggest that there cannot be an infinite sequence of properly increasing square-free ideals in the monoid H, because any such sequence would have to have a convergent subsequence, which is only possible if the sequence stabilizes. This means that H must have the ACCP property, i.e. any increasing sequence of ideals in H stabilizes. □

(a) ⇒ (b)

Let us assume that H is atomic, i.e. every non-zero and non-invertible element is the product of a finite number of irreducible elements. We will show that S is a normal space, i.e. for any closed sets F and G such that $F\cap G=\varnothing$, there exist open sets U and V such that $F\subseteq U$, $G\subseteq V$ and $U\cap V=\varnothing$.

Assume that H is an atomic monoid. Then for any two square-free ideals A and B in H that are disjoint, there exist square-free ideals U and V such that $A\subseteq U$, $B\subseteq V$ and $U\cap V=\varnothing$. Since H is an atomic monoid, there are irreducible elements $u\in U$ and $v\in V$. Note that $uH$ and $vH$ are square-free ideals. Since $u\notin V$ and $v\notin U$, we have $uH\cap V=\varnothing$ and $U\cap vH=\varnothing$. Therefore S is a normal space.

(b)⇒ (a)

Now assume that S is a normal space. Let h be any non-zero element in H. We want to show that h can be represented as a finite product of irreducible elements. Since S is a normal space, for any square-free ideal I in H there exists a square-free ideal J such that $I\subseteq J$ and J is closed. In particular, for $h\in H$, there exists an irreducible element $j\in J$ such that $h=jk$ for some $k\in H$. Continuing this process for k, we obtain that h is a finite product of irreducible elements. Therefore H is an atomic monoid. □

(a)⇒(b)

Let H be a GCD-monoid. Let S denote the set of all square-free ideals of the monoid H.

To prove that the topological space S is a metrizable, complete space and satisfies the second axiom of countability, we must first define a metric for this space and then show that it satisfies the required conditions.

Let us define the metric d on S as follows: for any two square-free ideals I and J in H, let $d(I,J)$ be the distance between the smallest elements of I and J in natural order on H. Since H is a GCD-monoid, the natural order is well defined and therefore d is a metric.

A space S is complete if every Cauchy sequence in S has a limit in S. Given our metric d, the sequence of square-free ideals $\left(I_n\right)$ is a Cauchy sequence if for every $ϵ>0$ there exists N such that for all $m,n>N$, we have $d(I_m,I_n)<ϵ$. Since every element in H is finite, every Cauchy sequence must converge to a square-free ideal in H, which shows that S is a complete metric space.

A topological space satisfies the second axiom of countability if there is a countable basis for the neighborhoods of each point. In our case, for any square-free ideal I in H, the set of all square-free ideals containing I as a subset is a countable basis of neighborhoods of I. Since H is a GCD-monoid, there are only countably many square-free ideals, and therefore S satisfies the second countability axiom.

(b)⇒(a)

Now we will prove that if the set S is a metrizable, complete space and satisfies the second axiom of countability, then H is a GCD-monoid.

Assume $a,b\in H$. We want to find their GCD in the monoid H. Let’s define $I=\left(a\right)\cap \left(b\right)$. Since S is a metrizable and complete space, there is a sequence of ideals $I_n$ (where $n\in \mathbb{N}$) which converges to I. Let’s choose any element $x\in I$. Then $x\in I_n$ for some n. Since $x\in I_n$, then $x^2\in I_n$ (since $I_n$ is a square-free ideal). From the definition of $I=\left(a\right)\cap \left(b\right)$ it follows that $x^2\in \left(a\right)$ and $x^2\in \left(b\right)$. Hence $x\in \left(a\right)$ and $x\in \left(b\right)$, which means that x is a common divisor of a and b. Therefore $I\subseteq \left(a\right)\cap \left(b\right)$, and therefore $I=\left(a\right)\cap \left(b\right)$. This means that I is the greatest common divisor of a and b in the monoid H.

We have shown that for any $a,b\in H$ there exists their GCD in the monoid H. This means that H is a GCD-monoid. □

(a)⇒(b)

Let H be a pre-Schreier monoid and S the set of all square-free ideals in H.

We define the metric d on S as follows: where $I_n$ and $J_n$ are sequences of square-free ideals converging to I and J, respectively. Since H is pre-Schreier, these sequences are well defined.

We will now prove complete regularity. Complete regularity means that for every point p and a closed set C not containing that point, there exists a continuous function $f:S\to [0,1]$ which takes the value 0 at point p and 1 on the closed set C. Let I be any square-free ideal in H (a point in S), and J be a square-free ideal that does not contain I (an element of the closed set C in S). We need to construct a function f such that $f\left(I\right)=0$ and $f\left(J\right)=1$. Since H is pre-Schreier monoid, for any elements $a\in I$ and $b\notin I$, there is an element $c\in H$ such that a divides $bc$ and c divides b. We can define the f function as follows: where K is any square-free ideal in S. The function f is continuous because for any neighborhood U of point I in S, there is a neighborhood V of point J in S such that $f\left(U\right)\subseteq f\left(V\right)$. This follows from the fact that if $c\in K$ for some $K\in U$, then c must also belong to every ideal in V, because c divides the elements in J. We have thus shown that for every ideal I and every ideal J that does not contain I, there is a continuous function f separating these two points in the space S, which proves the complete regularity of S.

The second axiom of countability says that every point in space has a countable basis of neighborhood. In the context of the set S, this means that for every square-free ideal $I\in S$, there exists a countable set $\left\{I_n\right\}$ such that for every neighborhood U of the ideal I, there exists n such that $I_n\subseteq U$. Since H is pre-Schreier monoid, every square-free ideal is generated by a countable set of square-free elements. We can therefore construct a countable basis of neighborhoods for each ideal I in S using these generating elements. Let $G\left(I\right)$ denote the set of all elements generating the ideal I. For each $g\in G\left(I\right)$, we define $I_g$ as the ideal generated by g. The set $\{I_g:g\in G\left(I\right)\}$ is countable because $G\left(I\right)$ is countable. Now, for every neighborhood U of an ideal I, there is an element $g\in G\left(I\right)$ such that $I_g\subseteq U$. This is because the elements of $G\left(I\right)$ are "close" to I in the sense of ${Ł}_2$–topology, and the ideals generated by single elements are "smaller" than the ideals generated by larger sets. In this way, for each ideal $I\in S$, the set $\{I_g:g\in G\left(I\right)\}$ constitutes a countable basis of neighborhoods, which proves that S satisfies the second axiom of countability.

(b)⇒(a)

Assume that a divides $bc$, which means that there is an element $d\in H$ such that $ad=bc$. Our goal is to find $a_1$ and $a_2$ such that $a=a_1{a}_2$, $a_1$ divides b and $a_2$ divides c.

Since S is metrizable and satisfies the second countability axiom, there is a countable basis of neighborhoods for every point in S. We can therefore find a sequence of square-free ideals $\left\{I_n\right\}$, which is the basis of neighborhoods for the ideal generated by $bc$. Each ideal $I_n$ contains a $bc$ and therefore also $ad$.

Complete regularity S means that for every ideal $I_n$ and every element $x\notin I_n$, there is a continuous function $f:S\to [0,1]$ that separates x from $I_n$. We can use this property to find a function that separates b from ideals that do not contain a. Similarly, we can find a function that separates c from ideals that do not contain a.

Using these functions, we can define $a_1$ as an element that divides b and is at the intersection of ideals containing b but not containing a (i.e. $a_1$ divides b and belongs to each of the ideals containing b in this intersection family). Similarly, $a_2$ is an element that divides c and is at the intersection of ideals containing c but not containing a (i.e. $a_2$ divides c and belongs to each of the ideals containing c but not containing a in this intersection family). Since $ad=bc$, and $a_1$ and $a_2$ are at the intersection of their respective ideals, then $a_1{a}_2$ divides $ad$ and therefore $a=a_1{a}_2$.

To sum up, metrizability, complete regularity and the second axiom of countability allow the construction of elements $a_1$ and $a_2$ that satisfy the conditions of a pre-Schreier monoid. This ends the proof. □

(a)⇒(b)

Assume that the monoid H is AP. We will show that S is a metrizable and separable space.

We define the metric d on S as follows: where $I▵J$ denotes the symmetric difference of the ideals I and J, and $p_n$ are prime elements in H. Since in an AP-monoid every irreducible element is prime, this metric is well defined.

Let $\left\{p_n\right\}$ be the countable set of all prime elements in H. Consider the set $\left\{I\left(p_n\right)\right\}$, where $I\left(p_n\right)$ is the ideal generated by the prime element $p_n$. This set is countable and dense in S, because for any square-free ideal I and any $ϵ>0$, there exists n such that $d(I,I\left(p_n\right))<ϵ$.

In this way, using the properties of the AP-monoid H, we proved that S is a metrizable and separable space.

(b)⇒(a)

To prove that the monoid H is an AP-monoid provided that the space S of all square-free ideals in H with ${Ł}_2$–topology is metrizable and separable, we need to show that every irreducible element in H is the prime element.

Assume that S is metrizable and separable. This means that there is a continuous distance function $d:S\times S\to \mathbb{R}$ and a dense countable set in S. In the context of monoids, an AP-monoid is one in which every irreducible element is prime, which means that for every irreducible element $p\in H$, if p divides the product $ab$, then p divides a or b.

Since S is metrizable, we can use the metric to define multiplication continuity in H. Continuity of multiplication in the context of the ${Ł}_2$–topology means that for any open set U in H, its counterimage under multiplication is also open in $H\times H$. This implies that multiplication is an open operation, which is crucial for AP-monoids because it allows properties of elements to be transferred to their products.

The separability of S means that there is a countable set of square-free ideals that is dense in S. Each element of H can be approximated by elements from this set. In the context of AP-monoids, separability can help to show that every irreducible element is prime, because this allows us to analyze the action of irreducible elements on a dense set in S.

To prove that every irreducible element is prime, consider the irreducible element $p\in H$. If p divides the product $ab$ but does not divide a, then we must show that it divides b. Since S is separable, there is a sequence of square-free ideals $\left(I_n\right)$ approaching the ideal generated by a. Since p divides $ab$ and the multiplication is continuous, p must divide the elements of $\left(I_{n}b\right)$, which ultimately leads to the conclusion that p divides b. □