Htykuut

Find the limit of sequence $x_{n} = (frac{1}{n},frac{1}{n})$ with respect to the topology $$tau = {(n,infty)times (n, infty): n in mathbb{N} } cup {emptyset,mathbb{R^2}}.$$

My work. Let $x$ be a limit.

If $x<0$, every number can be a limit, because in this case neighbourhood of $x$ is $mathbb{R}$.

If $x > 0$. For example, if $x = (0,0)$, neighbourhood looks like $(0,infty)times (0, infty)$. So, here is not limit for sequence $x_n$?? Or limit equals $(1,1)$??

Hint. Note that the sequence $left((frac{1}{n},frac{1}{n})right)_{ngeq 1}$ is eventually contained in $mathbb{R}^2setminus ((1,infty)times(1,infty))$.

Let's abstract a bit: the topology is just a decreasing sequence of open sets $U_1 supset U_2 supset U_3 supset ldots $ with empty intersection, and the compulsory $emptyset, X$ are open too. It's quite easy to check that this always gives a topology.

Now in our case, $U_n = (n,infty) times (n, infty)$ of course and the sequence
has the property:

$$forall n: x_n in U_0, x_n notin U_1$$

Now for any point $p$ not in $U_1$, the only open neighbourhoods it has are (possibly $U_0$) and in $X$, and in both cases, this contains the whole sequence. So $x_n to p$ for $p notin U_1$ and if $p in U_1$, then $U_1$ is a neighbourhood of $p$ such that no tail of the sequence lies in it (it entirely misses the sequence in fact), and so $x_n$ cannot converge to $p$.

So there lots of limits for this sequence : all points $(x,y) notin (1,infty) times (1,infty)$ i.e. all $(x,y)$ with $x le 1$ or $y le 1$.

The only neighbourhood of $x_n$ for $n>1$ is $mathbb{R}^2$. Since for any point in $mathbb{R}^2setminus((1,infty)times(1,infty))$, $mathbb{R}^2$ is the only neighbourhood incidentally containing infinitely many points of ${x_n}$. points $mathbb{R}^2setminus((1,infty)times(1,infty))$ is set of limit points of ${x_n}$.