TITLE: non sequential spaces
QUESTION [5 upvotes]: I am looking for a space which is not sequential. I tried to build this example:
Take $X$ to be a countable union of the folloing points:
 $X= (\bigcup_{n=1}^\infty( \bigcup_{k=1}^\infty (\frac1n,\frac1k)))\cup(\bigcup_{i=1}^\infty(0,\frac1i))\cup\{(0,0)\} $ ($n,k,i$ are natural numbrs). Define Basis neighborhoods of every point in $X$ to b:

if a point is of the form $(\frac1n,\frac1k)$, Then an open basis neighborhood of $x$ will be (the intersection of) vertical open segments containing $(\frac1n,\frac1k)$ (with $X$). 
if a point is of the form $(\frac1n,0)$, Then an open basis neighborhood of $x$ will be (the intersection of) open discs containing $(\frac1n,0)$ (with $X$).
an open basis neighborhood of $(0,0)$ is (the intersetion of) open disc containing $0$ in it's boundary union with $(0,0)$ itself (with $X$).

This is how it looks like:

I think that this is a topology and that $(X,T)$ is not sequential since take a subset $A$ of $X$ which is $A= (\bigcup_{n=1}^\infty( \bigcup_{k=1}^\infty (\frac1n,\frac1k)))$.
Then $(0,0)$ is contained in $\overline{A}$ but there is no sequence in $A$ which is converges to $(0,0)$, (because if non of the points are on $X-axis$, I can always find an open neighborhood of $(0,0)$ not containing it)
What do you think, am I right?
Thank you!!
Shir

REPLY [2 votes]: I have continued thinking about this example with respect to notions of sequential, Frechet Urysohn, and Countable generation. Wouldn't it be more correct that The space described above is sequential but not Frechet Urysohn? Because, if I get it correctly, Sequentiality does allow itterations of limits while Frechet Urysohn does not. And, as I understand it, every point of the form $(\frac1n,0)$ is a limit point of $A$, and $(0,0)$ is a limit point of the sequence $\{(\frac1n,0)\}$
What do you think?
Thanks!
Shir