Spectral Closure Generates a Topology on a Ring I have been struggling with this problem for quite some time. Please can anybody help me with some of the details which I miss to complete a proof.
Theorem [Topology generated by Spectral Closure; Cvetkovic-Ilic - p 268] 
(1) The Spectral Closure $CL(K)=\{ a \in R: \forall\ Finite\ J \subseteq R, \exists\ a' \in K: 1-J(a-a') \subseteq R^{-1}\}$
generates a Topology $\tau=\{ K \subseteq R: CL(R \setminus K) = R \setminus K \}$ on a Ring $R$
(2) Addition and Multiplication are Jointly Continuous
(3) $(R,\tau)$ is Hausdorff if and only if $R$ is Semisimple
I have already proven part 1 and the forward implication of part 3. My real struggle is to prove the converse of part 3 and Joint continuity of addition and multiplication in part 2. My attempt for converse of part 3 is as below
Edit: Since the time of publishing this question, I have also managed to solve part 2 of the problem; i.e. Addition and Multiplication are Jointly Continuous Binary Operators
I want to show that $R$ Semisimple $\implies$ $(R,\tau)$ Hausdorff
Strategy: Contradiction
Suppose that $R$ is Semisimple, then $Rad(R)=\{0\}$
We show that $(R,\tau)$ is Hausdorff
Suppose by way of contradiction that $(R,\tau)$ is not Hausdorff, then;
$\exists\ a,b \in R, a\not=b:\forall\ A,B \in \tau;\ a\in A, b \in B, A \cap B \not= \emptyset \dots(*)$
Let $a,b \in R$ be two elements with the property $(*)$
Let $A,B \in \tau$ such that $a \in A$ and $b \in B$
We also have that $A \cap B \not = \emptyset$ so $\exists\ c \in A \cap B$ and $A \cap B \in \tau$
Notice;
$a\not=b$ iff $a-b\not=0$ iff $a-b \not \in Rad(R)$ since $Rad(R)=\{0\}$
$a-b \not \in Rad(R)$ iff $1-R(a-b) \not \subseteq R^{-1}$ iff
$\exists\ Finite J \subseteq R:\ 1-J(a-b) \not \subseteq R^{-1} \dots(\#)$
Notice;
$a \in A \in \tau$ iff $a \not \in CL(R\setminus A)=R\setminus A$
$b \in B \in \tau$ iff $ b \not \in CL(R\setminus B)=R\setminus B$
$c \in A \cap B \in \tau$ iff $ c \not \in CL(R\setminus A \cap B)=R\setminus A \cap B$
$a \not \in CL(R\setminus A)$ iff $\exists\ Finite J \subseteq R: \forall a' \in R\setminus A; 1-J(a-a') \not \subseteq R^{-1}$
$b \not \in CL(R\setminus B)$ iff $\exists\ Finite J \subseteq R: \forall b' \in R\setminus B; 1-J(b-b') \not \subseteq R^{-1}$
$c \not \in CL(R\setminus A\cap B)$ iff $\exists\ Finite J \subseteq R: \forall c' \in R\setminus A\cap B; 1-J(c-c') \not \subseteq R^{-1}$
We also have the identity that;
$CL(R\setminus A\cap B)=CL(R\setminus A \cup R\setminus B)=CL(R\setminus A) \cup CL(R \setminus B)$
I am not sure how to go about to derive a contradiction
I tried to compare the three expressions above with $(\#)$ without any success
I thought that contradiction would be the best way, since then I could pick elements a and b with the non-Huassdorf property, instead of constructing/finding sets in a direct prood.
Any ideas would be helpful. Thanx in advance
 A: So since I have not received any positive responses to my question, I was forced to solve the problem myself; I would however still appreciate any comments on how to improve my argument.
Preliminaries:
Lemma 1: $CL(\{a\})=a+Rad(R)$; where $Rad(R)$ is the Jacobson Radical; Moreover, if $R$ is Semisimple, then this identity reduces to $CL(\{a\})=\{a\} \quad \forall\ a \in R$
Lemma 2: $\langle R, \tau \rangle$ is Hausdorff if and only if $\{x\}=\bigcap\ \{CL(U):\ U \in  \mathcal{N}_x\} \quad \forall\ x \in R$ where $\mathcal{N}_x$ is the neighbourhood system at $x$
Lemma 3: $CL(H)=\bigcap\ \{K \subseteq R:\ H \subseteq K\ \text{and}\ K\ \tau \text{-closed} \}$
Proof:
Suppose that $R$ is Semisimple. We employ Lemma 2 to show that $\langle R, \tau \rangle$ is Huasdorff
Consider arbitrary $a \in R$. Then clearly $\{a\} \subseteq \bigcap\ \{K \subseteq R:\ \{a\} \subseteq K\ \text{and}\  \tau\text{-closed}\}$
It remains to show reverse inclusion $\bigcap\ \{K \subseteq R:\ \{a\} \subseteq K\ \text{and}\  \tau\text{-closed}\} \subseteq \{a\}$
Observe that $\{CL(U):\ U \in \mathcal{N}_a\} \subseteq \{K \subseteq R:\ \{a\} \subseteq K\ \text{and}\  \tau\text{-closed}\}$
so their intersection $\bigcap\ \{CL(U):\ U \in \mathcal{N}_a\} \subseteq \bigcap\ \{K \subseteq R:\ \{a\} \subseteq K\ \text{and}\  \tau\text{-closed}\} \dots (*)$
But from Lemma 3 we have that $CL(\{a\})=\bigcap\ \{K \subseteq R:\ \{a\} \subseteq K\ \text{and}\  \tau\text{-closed}\}$
and from hypothesis, $R$ is Semisimple, so Lemma 1 gives $CL(\{a\})=\{a\} \quad \forall\ a \in R$ thus, the right hand side of $(*)$ reduces;
$\bigcap\ \{CL(U):\ U \in \mathcal{N}_a\} \subseteq \{a\}$
Since $a \in R$ was arbitrary, this holds $\forall\ a \in R$ concluding that $\langle R, \tau \rangle$ is Huasdorff as desired
