Skip to main content
2 of 4
Math Jaxed for improved readability and possibly easy its migration to the Math.SE

If (X,T) Topological space is completely normal, that, if we double the point of X, the result space were also completely normal?

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!

VDGG
  • 73
  • 6