Show product of two dense subsets is also dense

Question

Problem from Conway's "A course in point set topology": If for $k=1,2, A_k$ is a dense subset of the metric space $(X_k,d_k)$, then show that $A_1\times A_2$ is a dense subset of $X_1\times X_2$.

My attempted solution: Suppose $A_k\subseteq X_k$ is dense $\Rightarrow cl(A_k)=X_k\Rightarrow \forall r_k,\space B(x_k,r_k)\cap A_k\neq \varnothing$ so take $y_k\in B(x_k,r_k)\cap A_k$. Take $r=max\{r_1,r_2\}$ so that for $(y_1,y_2)\in B((x_1,x_2),r)\Rightarrow B((x_1,x_2),r)\cap A_1\times A_2\neq \varnothing$ whenever $(y_1,y_2)\in B((x_1,x_2),r)\cap A_1\times A_2\Rightarrow (y_1,y_2)\in cl(A_1)\times cl(A_2)\Rightarrow (y_1,y_2)\in X_1\times X_2$ since both $A_1$ & $A_2$ are both dense $\Rightarrow cl(A_1\times A_2)=X_1\times X_2$

Please drop a line if you think my proof is right, and of course, let me know how to improve

Answer 1

I’m afraid that your argument is so badly organized and confusingly written that I can’t tell whether it’s correct. From its conclusion it appears that you’re trying to prove that $\operatorname{cl}(A_1\times A_2)=X_1\times X_2$. You should begin by saying so, since there are other ways to prove that $A_1\times A_2$ is dense in $X_1\times X_2$.

To prove that $\operatorname{cl}(A_1\times A_2)=X_1\times X_2$, you could either prove the general result that $\operatorname{cl}(H\times K)=(\operatorname{cl}H)\times\operatorname{cl}K$ for arbitrary $H\subseteq X_1$ and $K\subseteq X_2$ or prove this special case. To do the latter you should start with an arbitrary point $\langle x_1,x_2\rangle\in X_1\times X_2$ and show that every open nbhd of this point intersects $A_1\times A_2$, something like this:

Let $p=\langle x_1,x_2\rangle\in X_1\times X_2$, and let $U$ be any open nbhd of $p$. Then there are real numbers $r_1,r_2>0$ such that $p\in B(x_1,r_1)\times B(x_2,r_2)\subseteq U$. $A_1$ is dense in $X_1$, so there is a $y_1\in B(x_1,r_1)\cap A_1$; similarly, there is a $y_2\in B(x_2,r_2)\cap A_2$. Clearly $$\langle y_1,y_2\rangle\in \big(B(x_1,r_1)\times B(x_2,r_2)\big)\cap(A_1\times A_2)\subseteq U\cap(A_1\times A_2)\;.$$ Thus, every open nbhd of $p$ contains a point of $A_1\times A_2$, so $p\in\operatorname{cl}(A_1\times A_2)$. And $p$ was an arbitrary point of $X_1\times X_2$ so $\operatorname{cl}(A_1\times A_2)=X_1\times X_2$.

Note that I’ve used a lot more words than you did: I use them as connective tissue to make the flow of logic clearer. I agree completely with Nate Eldredge’s comments: except in formal logic a proof should be written in ordinary prose, with symbols only where they are necessary or genuinely make things clearer. (For instance, the quadratic formula is much clearer when written in algebraic symbols than when it is expressed in words!)