Find the limit of the sequence Find the limit of the sequence$$\left(\frac12,\frac12\right),\left(\frac23,\frac23\right),\ldots,\left(\frac n{n+1},\frac n{n+1}\right),\ldots$$
 A: $(1,0)$ is the unique limit point of this sequence. The "obvious limit" $(1,1)$ has a small vertical neighborhood that misses the entire sequence. On the other hand, every basis neighborhood of $(1,0)$ has to have some width to it, so will meet the sequence in question.
Edit: Here are some more details. Basis elements in the order topology are open intervals $(x_1\times y_1,x_2\times y_2)$, together with neighborhoods $(x\times y, 1\times 1]$ and $[0\times 0, x\times y)$. Here I am using $a\times b$ to denote an ordered pair so as not to confuse the notation with open intervals! So, for example the open set $(1\times \frac{1}{2}, 1\times 1]$ is a neighborhood of $1\times 1$ which misses the sequence. On the other hand, any neighborhood of $1\times 0$ would have to contain an open interval $(x_1\times y_1,x_2\times y_2)$ where $x_1\times y_1 < 1\times 0 <x_2\times y_2$. This implies $x_1< 1$. So this open interval will contain points $\frac{n}{n+1}\times\frac{n}{n+1}$ for all $n$ where $\frac{n}{n+1}>x_1$.
