I have a question on my proof of the following lemma, and I'd like to know if my answer is correct.

Lemma. Suppose $(X,T)$ is any completely normal topological space. Let's double the points of $X$, more precise, consider space $(Y,F)$, where $Y$ is the cartesian product of $X$ and $\{0,1\}$ (with trivial topology) and $F$ is product topology. Then $(Y,F)$ is also completely normal.

Proof: Suppose $A$ and $B$ are separable sets in $(Y,F)$, so $$\operatorname{clo}(A)\cap B=A\cap \operatorname{clo}(B)=\emptyset,$$ where $\operatorname{clo}(\cdot)$ is the closure operator. $A$ and $B$ are separable sets in $Y$ iff $$ A_1\equiv\{x\in X:(x,0)\in A\}\cup\{x\in X:(x,1)\in A\} $$ and $$ B_1\equiv \{x\in X:(x,0)\in B\}\cup\{x\in X:(x,1)\in B\} $$ are separable sets in $X$. So, there exist $U$ and $V$ disjoint open sets in $X$ such, that $A_1\subset U$ and $B_1\subset V$. Then $U\times\{0,1\}$ and $V\times\{0,1\}$ are disjoint open sets in $Y$ such that $A\subset U\times\{0,1\}$ and $V\subset V\times\{0,1\}$. So $(Y,F)$ is completely normal.

Is my proof correct? Thank you!

I have a question on my proof of the following lemma, and I'd like to know if my answer is correct.

Lemma. Suppose $(X,T)$ is any completely normal topological space. Let's double the points of $X$, more precise, consider space $(Y,F)$, where $Y$ is the cartesian product of $X$ and $\{0,1\}$ (with trivial topology) and $F$ is product topology. Then $(Y,F)$ is also completely normal.

Proof: Suppose $A$ and $B$ are separated sets in $(Y,F)$, so $$\operatorname{clo}(A)\cap B=A\cap \operatorname{clo}(B)=\emptyset,$$ where $\operatorname{clo}(\cdot)$ is the closure operator. $A$ and $B$ are separable sets in $Y$ iff $$ A_1\equiv\{x\in X:(x,0)\in A\}\cup\{x\in X:(x,1)\in A\} $$ and $$ B_1\equiv \{x\in X:(x,0)\in B\}\cup\{x\in X:(x,1)\in B\} $$ are separated sets in $X$. So, there exist $U$ and $V$ disjoint open sets in $X$ such, that $A_1\subset U$ and $B_1\subset V$. Then $U\times\{0,1\}$ and $V\times\{0,1\}$ are disjoint open sets in $Y$ such that $A\subset U\times\{0,1\}$ and $V\subset V\times\{0,1\}$. So $(Y,F)$ is completely normal.

Is my proof correct? Thank you!

I have a question on my proof of the following lemma, and I'd like to know if my answer is correct.

Lemma. Suppose $(X,T)$ is any completely normal topological space. Let's double the points of $X$, more precise, consider space $(Y,F)$, where $Y$ is the cartesian product of $X$ and $\{0,1\}$ (with trivial topology) and $F$ is product topology. Then $(Y,F)$ is also completely normal.

Proof: Suppose $A$ and $B$ are seperable sets in $(Y,F)$, so $$\operatorname{clo}(A)\cap B=A\cap \operatorname{clo}(B)=\emptyset,$$ where $\operatorname{clo}(\cdot)$ is the closure operator. $A$ and $B$ are separable sets in $Y$ iff $$ A_1\equiv\{x\in X:(x,0)\in A\}\cup\{x\in X:(x,1)\in A\} $$ and $$ B_1\equiv \{x\in X:(x,0)\in B\}\cup\{x\in X:(x,1)\in B\} $$ are separable sets in $X$. So, there exist $U$ and $V$ disjoint open sets in $X$ such, that $A_1\subset U$ and $B_1\subset V$. Then $U\times\{0,1\}$ and $V\times\{0,1\}$ are disjoint open sets in $Y$ such that $A\subset U\times\{0,1\}$ and $V\subset V\times\{0,1\}$. So $(Y,F)$ is completely normal.

Is my proof correct? Thank you!

I have a question on my proof of the following lemma, and I'd like to know if my answer is correct.

Lemma. Suppose $(X,T)$ is any completely normal topological space. Let's double the points of $X$, more precise, consider space $(Y,F)$, where $Y$ is the cartesian product of $X$ and $\{0,1\}$ (with trivial topology) and $F$ is product topology. Then $(Y,F)$ is also completely normal.

Proof: Suppose $A$ and $B$ are separable sets in $(Y,F)$, so $$\operatorname{clo}(A)\cap B=A\cap \operatorname{clo}(B)=\emptyset,$$ where $\operatorname{clo}(\cdot)$ is the closure operator. $A$ and $B$ are separable sets in $Y$ iff $$ A_1\equiv\{x\in X:(x,0)\in A\}\cup\{x\in X:(x,1)\in A\} $$ and $$ B_1\equiv \{x\in X:(x,0)\in B\}\cup\{x\in X:(x,1)\in B\} $$ are separable sets in $X$. So, there exist $U$ and $V$ disjoint open sets in $X$ such, that $A_1\subset U$ and $B_1\subset V$. Then $U\times\{0,1\}$ and $V\times\{0,1\}$ are disjoint open sets in $Y$ such that $A\subset U\times\{0,1\}$ and $V\subset V\times\{0,1\}$. So $(Y,F)$ is completely normal.

Is my proof correct? Thank you!

I have some question with my answer and I`m interested in if my answer is correct.

Suppose (X,T) is any completely normal topological space. Lets double the points of X, more precise, consider space (Y,F), where Y is the cartesian product of X and {0,1} (with trivial topology) and F is product topology. Than (Y,F) is also completely normal.

Proof: Suppose A and B is seperable sets in (Y,F), so cl(A)∩B=A∩cl(B)=∅, whete cl(*) is closure operator. A and B is seperable sets in Y iff A1≡{x∈X:(x,0)∈A}∪{x∈X:(x,1)∈A} and B1≡{x∈X:(x,0)∈B}∪{x∈X:(x,1)∈B} is seperable sets in X. So, There exist U and V disjoint open sets in X such, that A1⊂U and B1⊂V. Than U×{0,1} and V×{0,1} is disjoint open sets in Y such, that A⊂U×{0,1} and V⊂V×{0,1}. So, (Y,F) is completely normal.

Is my proof correct? Thank you!

I have a question on my proof of the following lemma, and I'd like to know if my answer is correct.

Lemma. Suppose $(X,T)$ is any completely normal topological space. Let's double the points of $X$, more precise, consider space $(Y,F)$, where $Y$ is the cartesian product of $X$ and $\{0,1\}$ (with trivial topology) and $F$ is product topology. Then $(Y,F)$ is also completely normal.

Proof: Suppose $A$ and $B$ are seperable sets in $(Y,F)$, so $$\operatorname{clo}(A)\cap B=A\cap \operatorname{clo}(B)=\emptyset,$$ where $\operatorname{clo}(\cdot)$ is the closure operator. $A$ and $B$ are separable sets in $Y$ iff $$ A_1\equiv\{x\in X:(x,0)\in A\}\cup\{x\in X:(x,1)\in A\} $$ and $$ B_1\equiv \{x\in X:(x,0)\in B\}\cup\{x\in X:(x,1)\in B\} $$ are separable sets in $X$. So, there exist $U$ and $V$ disjoint open sets in $X$ such, that $A_1\subset U$ and $B_1\subset V$. Then $U\times\{0,1\}$ and $V\times\{0,1\}$ are disjoint open sets in $Y$ such that $A\subset U\times\{0,1\}$ and $V\subset V\times\{0,1\}$. So $(Y,F)$ is completely normal.

Is my proof correct? Thank you!