How can I prove that this space is not Urysohns frechet space? Define $f:\Bbb R\to Z,$ where $Z=(\Bbb R\setminus \Bbb N)\cup\{a\}$ for some $a\notin\Bbb R,$ by $$f(x)=\begin{cases}x & \text{if }x\in \Bbb R\setminus \Bbb N\\ a & \text{if }x \in\Bbb N.\end{cases}$$
We define the topology on $Z$ to be the family of all subsets of $Z$ whose preimage under $f$ is open in $\Bbb R$ (with usual topology).
I want to prove that $I\times Z$ is not a Urysohn frechet space, where $I =[0,1]$ is considered as a subspace of $\Bbb R$ (in the usual topology).
Can someone give me a hint?
 A: (Getting this off the Unanswered list.)
Let $X=\Bbb I\times Z$, $A=\{\langle x,y\rangle\in\Bbb I\times(\Bbb R\setminus\Bbb N):xy>1\}$, and $p=\langle 0,1\rangle$; I show first that $p\in\operatorname{cl}_XA$.
For each $\delta\in(0,1)$ and sequence $\epsilon=\langle\epsilon_k:k\in\Bbb N\rangle$ of real numbers in $(0,1)$ let
$$B(\delta,\epsilon)=\{p\}\cup\bigcup_{k\in\Bbb N}\Big(\big((k-\epsilon_k,k+\epsilon_k)\setminus\{k\}\big)\times[0,\delta)\Big)\;;$$
the set $\mathscr{B}$ of such $B(\delta,\epsilon)$ is a local base of open sets at the point $p$. Let $B(\delta,\epsilon)\in\mathscr{B}$ be arbitrary, and choose $n\in\Bbb N$ such that $n>\frac1\delta$. Then for any $x\in(n,n+\epsilon_n)$ we have $x>\frac1\delta$ and hence $x\delta>1$, Clearly there is a positive $y<\delta$ such that $xy>1$, so that $\langle x,y\rangle\in B(\delta,\epsilon)\cap A$, and therefore $p\in\operatorname{cl}_XA$.
Now suppose that $\langle p_n:n\in\Bbb N\rangle$ is a sequence in $A$ converging to $p$, where $p_n=\langle x_n,y_n\rangle$ for $n\in\Bbb N$; clearly the sequence $\langle y_n:n\in\Bbb N\rangle$ must converge to $0$ in $\Bbb I$, and hence $\langle x_n:n\in\Bbb N\rangle$ must diverge to $\infty$. By passing to a subsequence if necessary we may assume that $\langle x_n:n\in\Bbb N\rangle$ is strictly increasing. For each $k\in\Bbb N$ we can then choose $\epsilon_k\in(0,1)$ such that 
$$(k-\epsilon_k,k+\epsilon_k)\cap\{x_n:n\in\Bbb N\}=\varnothing\;.$$
Let $\epsilon=\langle\epsilon_k:k\in\Bbb N\rangle$; it’s not hard to see that $B(1,\epsilon)\cap\{p_n:n\in\Bbb N\}=\varnothing$, contradicting the assumption that $\langle p_n:n\in\Bbb N\rangle$ converges to $p$.
Thus, $p$ is in the closure of $A$ but not its sequential closure, and $X$ is not Fréchet-Uryson.
