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 $(1,1)$??
On
Hint. Note that the sequence $\left((\frac{1}{n},\frac{1}{n})\right)_{n\geq 1}$ is eventually contained in $\mathbb{R}^2\setminus ((1,\infty)\times(1,\infty))$.
On
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 $\mathbb{R}^2\setminus((1,\infty)\times(1,\infty))$, $\mathbb{R}^2$ is the only neighbourhood incidentally containing infinitely many points of $\{x_n\}$. points $\mathbb{R}^2\setminus((1,\infty)\times(1,\infty))$ is set of limit points of $\{x_n\}$.